JEE Work, Energy & Power Questions

The Carnot engine (CE) and the heat pump (HP) work in tandem, the work output of the CE being the work input of the HP.

Carnot engine

Efficiency of a Carnot engine is defined as $$\eta = 1 - \frac{T_C}{T_H}$$, where $$T_H$$ and $$T_C$$ are the absolute temperatures of the hot and cold reservoirs respectively.

Given $$\eta = 0.4$$ and $$T_H = 1000\,\text{K}$$, we have

$$0.4 = 1 - \frac{T_C}{1000}$$

$$\frac{T_C}{1000} = 1 - 0.4 = 0.6$$

$$T_C = 0.6 \times 1000 = 600\,\text{K}$$

Thus the cold-reservoir temperature of the Carnot engine is $$600\,\text{K}$$  →  Option B is correct.

The work obtained from the Carnot engine in one cycle is

$$W = \eta\,Q_H = 0.4 \times 150\,\text{J} = 60\,\text{J}$$

Hence the work delivered per cycle is $$60\,\text{J}$$  →  Option A is correct.

Heat pump

The entire $$60\,\text{J}$$ of work produced by the engine drives the heat pump. For a heat pump, the coefficient of performance (COP) is defined as

$$\text{COP} = \frac{Q_H^{(\text{HP})}}{W}$$

Given $$\text{COP} = 10$$ and $$W = 60\,\text{J}$$, the heat delivered to the hot reservoir of the pump is

$$Q_H^{(\text{HP})} = \text{COP}\times W = 10 \times 60\,\text{J} = 600\,\text{J}$$

The cold-reservoir heat extracted by the pump is

$$Q_C^{(\text{HP})} = Q_H^{(\text{HP})} - W = 600\,\text{J} - 60\,\text{J} = 540\,\text{J}$$

The problem states that the heat pump operates between $$T_H^{(\text{HP})}=300\,\text{K}$$ and an (as yet) unknown cold-reservoir temperature $$T_C^{(\text{HP})}$$. For an ideal (Carnot) heat pump,

$$\text{COP} = \frac{T_H}{T_H - T_C}$$

Substituting $$\text{COP}=10$$ and $$T_H = 300\,\text{K}$$,

$$10 = \frac{300}{300 - T_C}$$

$$300 - T_C = \frac{300}{10} = 30$$

$$T_C = 300 - 30 = 270\,\text{K}$$

Thus the cold-reservoir temperature of the heat pump is $$270\,\text{K}$$  →  Option C is correct.

Checking Option D

Option D claims that the heat supplied to the hot reservoir of the pump is $$540\,\text{J}$$ per cycle. The calculated value is $$Q_H^{(\text{HP})}=600\,\text{J}$$, so Option D is incorrect.

Correct statements: Option A, Option B, Option C.

Let the surface charge of the conducting sphere be uniformly distributed.
All the charge therefore resides on the outer surface.

Surface charge density:
$$\sigma = \frac{Q}{4\pi R^{2}}$$

Take the rotation axis as the z-axis. Consider a thin surface strip between polar angles $$\theta$$ and $$\theta + d\theta$$.

• Radius of the circular ring in this strip: $$r = R\sin\theta$$.
• Width of the strip: $$R\,d\theta$$.
• Area of the strip: $$dA = 2\pi R^{2}\sin\theta\,d\theta$$.
• Charge on the strip: $$dq = \sigma\,dA = \frac{Q}{4\pi R^{2}}\;(2\pi R^{2}\sin\theta\,d\theta) = \frac{Q}{2}\sin\theta\,d\theta.$$

The ring rotates with the sphere at angular speed $$\omega$$. Current in the ring:

$$dI = \frac{dq}{T} = \frac{dq\,\omega}{2\pi}.$$

Magnetic moment of a current loop: $$d\mu = dI \times (\text{area of ring})$$, and the loop area is $$\pi r^{2} = \pi R^{2}\sin^{2}\theta$$.

Therefore $$d\mu = \frac{dq\,\omega}{2\pi}\;(\pi R^{2}\sin^{2}\theta) = \frac{\omega R^{2}}{2}\;dq\,\sin^{2}\theta = \frac{\omega R^{2}}{2}\; \bigl(\frac{Q}{2}\sin\theta\,d\theta\bigr)\sin^{2}\theta = \frac{Q\omega R^{2}}{4}\sin^{3}\theta\,d\theta.$$

Total magnetic dipole moment:

$$\mu = \int_{0}^{\pi} d\mu = \frac{Q\omega R^{2}}{4}\int_{0}^{\pi} \sin^{3}\theta\,d\theta.$$

Using $$\int_{0}^{\pi}\sin^{3}\theta\,d\theta = \frac{4}{3}$$, we obtain

$$\mu = \frac{Q\omega R^{2}}{4}\left(\frac{4}{3}\right) = \frac{Q\omega R^{2}}{3}.$$

Moment of inertia of a solid sphere about a diameter:
$$I = \frac{2}{5}MR^{2}.$$

Angular momentum about the same axis:
$$L = I\omega = \frac{2}{5}MR^{2}\omega.$$

Ratio of magnitudes:

$$\frac{\mu}{L} = \frac{\dfrac{Q\omega R^{2}}{3}} {\dfrac{2}{5}MR^{2}\omega} = \frac{Q}{3}\;\frac{5}{2M} = \frac{5Q}{6M}.$$

The question writes this ratio as $$\alpha \dfrac{Q}{2M}$$, so

$$\alpha\,\frac{Q}{2M} = \frac{5Q}{6M} \;\;\Longrightarrow\;\; \alpha = \frac{5}{3} \approx 1.666.$$

Hence the required value lies in the range 1.65 - 1.67.

When 2 bodies of the same mass undergo elastic collision, their velocities get exchanged. 

In the given system, since ball A is faster than ball B, which is slower than ball C, first ball A and ball B will collide and exchange their velocities.

$$v_{A,\ final}=2$$ m/s and $$v_B'=5$$ m/s

Then, ball B and ball C will collide and exchange their velocities

$$v_{B,\ final\ }=4$$ m/s and $$v_{c,\ final\ }=5$$

Therefore, final velocities of the balls are $$v_{A,\ final}=2$$ m/s, $$v_{B,\ final\ }=4$$ m/s and $$v_{c,\ final\ }=5$$

A ball with kinetic energy $$KE$$ is projected at 60° from horizontal. At the highest point, only the horizontal component of velocity remains.

$$v_x = v\cos 60° = \frac{v}{2}$$

The kinetic energy at the highest point is

$$KE' = \frac{1}{2}mv_x^2 = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{4} \times \frac{1}{2}mv^2 = \frac{KE}{4}$$

Hence, the correct answer is Option 4: $$\frac{KE}{4}$$.

For an ideal monatomic gas, the ratio of specific heats is $$\gamma = \dfrac{5}{3}$$.

Case 1: Adiabatic branch $$W \rightarrow X$$
For a reversible adiabatic process, $$T V^{\,\gamma-1} = \text{constant}$$

Thus $$T_W V_W^{\,\gamma-1} = T_X V_X^{\,\gamma-1}$$ $$$ \Rightarrow \; T_X = T_W \left(\dfrac{V_W}{V_X}\right)^{\gamma-1} = T_W \left(\dfrac{64}{125}\right)^{\frac{2}{3}} $$$

Since $$64 = 4^3$$ and $$125 = 5^3$$, $$\left(\dfrac{64}{125}\right)^{\frac{2}{3}} = \left(\dfrac{4^3}{5^3}\right)^{\frac{2}{3}} = \left(\dfrac{4}{5}\right)^{2} = \dfrac{16}{25} = 0.64$$

Therefore, $$T_X = 0.64 \,T_W \qquad -(1)$$

Case 2: Isobaric branch $$X \rightarrow Y$$
For an isobaric process, $$P$$ is constant, so from the ideal-gas law $$PV = nRT$$ we have $$\dfrac{T}{V} = \text{constant} \;\Longrightarrow\; \dfrac{T_Y}{T_X} = \dfrac{V_Y}{V_X}$$

Given $$V_Y = 250\;\text{cm}^3$$ and $$V_X = 125\;\text{cm}^3$$, $$\dfrac{T_Y}{T_X} = \dfrac{250}{125} = 2 \;\Longrightarrow\; T_Y = 2\,T_X \qquad -(2)$$

Using $$(1)$$ in $$(2)$$, $$T_Y = 2 \times 0.64\,T_W = 1.28\,T_W$$

Hence the temperature rise along $$X \rightarrow Y$$ is $$\Delta T = T_Y - T_X = 1.28\,T_W - 0.64\,T_W = 0.64\,T_W$$

The heat absorbed in an isobaric process is $$Q_{XY} = n C_p \Delta T$$ where, for a monatomic gas, $$C_p = \dfrac{5}{2}R$$.

Therefore, $$Q_{XY} = n\left(\dfrac{5}{2}R\right)(0.64\,T_W) = \dfrac{5}{2} \times 0.64 \; (nRT_W)$$

The problem states that $$nRT_W = 1\;\text{J}$$, so $$Q_{XY} = \dfrac{5}{2} \times 0.64 \times 1 = 2.5 \times 0.64 = 1.6\;\text{J}$$

Hence, the heat absorbed by the gas along the path $$X \rightarrow Y$$ is 1.6 J.

Let the displacement from point P to point Q be $$\vec d$$. Then

$$\vec d = (6-3)\,\hat i + (10-4)\,\hat j = 3\,\hat i + 6\,\hat j\quad -(1)$$

The work done by the force $$\overrightarrow{F}=(2\hat i+3\hat j)\,$$N is given by

$$W = \overrightarrow{F}\cdot \vec d = 2\times3 + 3\times6 = 6 + 18 = 24\text{ J}\quad -(2)$$

Average power over the time interval $$\Delta t=4\,$$s is defined by

$$P_{\rm avg} = \frac{W}{\Delta t} = \frac{24}{4} = 6\text{ W}\quad -(3)$$

Since the force is constant, acceleration is

$$\vec a = \frac{\overrightarrow{F}}{m} = \frac{1}{4}(2\hat i +3\hat j) = 0.5\,\hat i + 0.75\,\hat j\;\mathrm{m/s^2}\quad -(4)$$

Assuming the body starts from rest, its velocity at $$t=4\,$$s is

$$\vec v = \vec a\,t = (0.5\times4)\,\hat i + (0.75\times4)\,\hat j = 2\,\hat i + 3\,\hat j\;\mathrm{m/s}\quad -(5)$$

Instantaneous power at $$t=4\,$$s is

$$P_{\rm inst} = \overrightarrow{F}\cdot \vec v = 2\times2 + 3\times3 = 4 + 9 = 13\text{ W}\quad -(6)$$

Hence the ratio of average power to instantaneous power is

$$P_{\rm avg} : P_{\rm inst} = 6 : 13$$

Final Answer: Option D; 6 : 13.

The string becomes slack at the highest point D,means the tension is zero. The centripetal force is provided entirely by gravity at this point.

$$ \frac{m v_D^2}{l} = mg $$

$$ v_D^2 = gl $$

$$ KE_D = \frac{1}{2} m v_D^2 = \frac{1}{2} mgl $$

Let the center O be the reference level for zero potential energy. The height of point D from O is $$l$$, so its potential energy is:

$$ U_D = mgl $$

According to the law of conservation of mechanical energy, the total energy $$E$$ at any point is constant.

$$ E = K_D + U_D $$

$$ E = \frac{1}{2} mgl + mgl = \frac{3}{2} mgl $$

Now, let us find the kinetic energy at point B ($$K_B$$).

The vertical depth of point B below O is $$l \cos 60^\circ$$.

$$ U_B = -mgl \cos 60^\circ = -mgl \left(\frac{1}{2}\right) = -\frac{1}{2} mgl $$

Using energy conservation at point B:

$$ E = K_B + U_B $$

$$ \frac{3}{2} mgl = K_B - \frac{1}{2} mgl $$

$$ K_B = \frac{3}{2} mgl + \frac{1}{2} mgl $$

$$ K_B = 2mgl $$

Next, let us find the kinetic energy at point C ($$K_C$$).

The vertical height of point C above O is $$l \cos 60^\circ$$.

$$ U_C = mgl \cos 60^\circ = mgl \left(\frac{1}{2}\right) = \frac{1}{2} mgl $$

Using energy conservation at point C:

$$ E = K_C + U_C $$

$$ \frac{3}{2} mgl = K_C + \frac{1}{2} mgl $$

$$ K_C = \frac{3}{2} mgl - \frac{1}{2} mgl $$

$$ K_C = mgl $$

Finally, we find the ratio of the kinetic energy at point B to that at point C.

$$ \text{Ratio} = \frac{K_B}{K_C} $$

$$ \text{Ratio} = \frac{2mgl}{mgl} $$

$$ \text{Ratio} = 2 $$

We need to evaluate the Assertion and Reason about central forces and potential energy.

Assertion (A): "In a central force field, the work done is independent of the path chosen."

Analysis: A central force is always directed along the line joining the particle to a fixed center, with magnitude depending only on the distance: $$\vec{F} = f(r)\hat{r}$$. Central forces are conservative forces -- the work done depends only on the initial and final positions, not on the path. This is because the curl of a central force is zero. Assertion (A) is TRUE.

Reason (R): "Every force encountered in mechanics does not have an associated potential energy."

Analysis: This statement says that not every mechanical force has an associated potential energy. This is TRUE -- non-conservative forces like friction and viscous drag do not have an associated potential energy function. Only conservative forces (like gravity, spring force, and central forces) have well-defined potential energy functions.

Relationship between A and R: While both statements are true, R does not explain A. The fact that "some forces don't have potential energy" does not explain why "central forces are path-independent." The reason A is true is that central forces are conservative (they have a potential energy function), which is essentially the opposite situation from what R describes.

The correct answer is Option 2: Both A and R are true but R is NOT the correct explanation of A.

Since $$\frac{dm}{dt} \propto \sqrt{v}$$, we write $$\frac{dm}{dt} = k\sqrt{v}$$ for some constant $$k$$.

The belt runs at constant speed $$v$$. The force needed to maintain constant speed when sand is being dropped is:

$$ F = v\frac{dm}{dt} $$

This is because the sand needs to be accelerated from rest to speed $$v$$.

The power delivered is:

$$ P = Fv = v^2 \frac{dm}{dt} = v^2 \cdot k\sqrt{v} = kv^{5/2} $$

Therefore: $$P \propto v^{5/2}$$

Squaring both sides: $$P^2 \propto v^5$$

The correct answer is Option 3: $$P^2 \propto v^5$$.

Given the force $$F = \alpha + \beta x^2$$, the work done is 5 J over a displacement of 1 m and α = 1 N, so we need to find β. Since the work is given by the integral of the force from 0 to 1, we have $$W = \int_0^1(\alpha+\beta x^2)dx = [\alpha x + \frac{\beta x^3}{3}]_0^1 = \alpha + \frac{\beta}{3}$$. Substituting the values gives $$5 = 1 + \frac{\beta}{3}$$, and therefore solving for β yields $$\beta = 12$$ N/m².

Thus the correct answer is Option 2: 12 N/m².

A ball of mass $$m = 100 \text{ g} = 0.1 \text{ kg}$$ is projected with velocity $$u = 20 \text{ m/s}$$ at $$\theta = 60°$$ with the horizontal.

Since the kinetic energy at the point of projection is $$KE_i = \frac{1}{2}mu^2 = \frac{1}{2} \times 0.1 \times 20^2 = \frac{1}{2} \times 0.1 \times 400 = 20 \text{ J}$$

At the highest point of projectile motion, the vertical component of velocity becomes zero and only the horizontal component remains. From this, the velocity at the highest point is $$v = u\cos\theta = 20 \times \cos 60° = 20 \times \frac{1}{2} = 10 \text{ m/s}$$

Next, the kinetic energy at the highest point is $$KE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.1 \times 10^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \text{ J}$$

Finally, the decrease in kinetic energy is calculated as $$\Delta KE = KE_i - KE_f = 20 - 5 = 15 \text{ J}$$

The correct answer is Option (2): $$\boxed{15 \text{ J}}$$.

Kinetic energy at any point on the circle can be calculated using conservation of energy.

$$KE_A\ =\ \frac{1}{2}mv_A^2\ =\ \frac{1}{2}\ \times\ 0.1\ \times\ 10^2\ =\ 5J$$

$$PE_A\ =\ 0\ $$

At point B:-

Height $$H_B\ =\ R\left(1-\cos30^{\circ\ }\right)\ =\ 2-\sqrt{\ 3}\ m$$

$$PE_B\ =\ mgH_B\ =\ 0.1\times10\times\left(2-\sqrt{\ 3}\ \right)$$

$$KE_A\ +\ PE_A\ =\ KE_{B\ }+\ PE_B$$ 

⇒ $$5+0=KE_B+2-\sqrt{\ 3}$$

$$\therefore\ KE_B\ =\ 3+\sqrt{\ 3}\ J$$

At point C:-

Height $$H_C\ =\ R\left(1+\cos\ 60^{\circ\ }\right)\ =\ 2\left(1+\frac{1}{2}\right)=3\ m$$

$$PE_C\ =\ mgH_C=0.1\times10\times3\ =3\ J\ $$

$$KE_A\ +\ PE_A\ =\ KE_C\ +\ PE_C$$

⇒ $$5+0=KE_C+3$$

$$\therefore\ KE_C\ =\ 5-3\ =2\ J$$

Ratio of Kinetic energy at point B and C is therefore, $$Ratio=\frac{\left(3+\sqrt{\ 3}\right)}{2}$$

When the length of the spring becomes R, the angle spring will form from the vertical will be $$\theta\ =60^{\circ\ }$$.

Since all forces acting on the body are conservative, total energy remains conserved.

Potential energy of the bead at the top due to gravity, $$PE_{g}=mg\left(2R\right)=2mgR$$

Potential energy of the bead at the top due to the spring, $$PE_{sp}=\frac{1}{2}kx^2\ =\frac{1}{2}k\left(2R-R\right)^2=\frac{1}{2}kR^2$$

⇒ Initial potential energy, $$PE_{initial}=PE_g+PE_{sp}=2mgR+\frac{1}{2}kR^2$$

When spring becomes the length R, the extension in the spring becomes 0. Therefore, Potential energy = gravitational potential energy

Final potential energy, $$PE_{final}=mgH_{final}\ =\ mg\left(R\cos60^{\circ\ }\right)=\frac{mgR}{2}$$

Loss in potential energy = Gain in kinetic energy

$$KE_{final}-KE_{initial}=PE_{initial}-PE_{final}$$

⇒ $$KE_{final}=PE_{initial}-PE_{final}\ =\ 2mgR+\frac{1}{2}kR^2-\frac{mgR}{2}=\frac{3}{2}mgR+\frac{1}{2}kR^2$$

$$\frac{1}{2}mv^2=\frac{3}{2}mgR+\frac{1}{2}kR^2$$

$$\therefore\ v=\sqrt{\ 3gR+\frac{kR^2}{m}}$$

The forces on the block are:
  • weight $$mg$$ downward
  • normal reaction $$N$$ upward
  • applied force $$\mathbf{F}$$ making $$45^{\circ}$$ with the horizontal
  • kinetic friction $$f_k$$ opposite to the motion along the surface.

Resolve the applied force:
  Horizontal component  $$F_x = F\cos 45^{\circ} = \dfrac{F}{\sqrt{2}}$$
  Vertical component    $$F_y = F\sin 45^{\circ} = \dfrac{F}{\sqrt{2}}$$

Vertical equilibrium (no vertical motion):
$$N + F_y = mg$$
so
$$N = mg - \frac{F}{\sqrt{2}} \quad -(1)$$

Friction coefficient is $$\mu = 0.25$$, hence
$$f_k = \mu N = \mu\left( mg - \frac{F}{\sqrt{2}} \right) \quad -(2)$$

The block is pulled steadily (horizontal acceleration zero). Therefore
horizontal forces balance:
$$F_x = f_k$$
$$\frac{F}{\sqrt{2}} = \mu\left( mg - \frac{F}{\sqrt{2}} \right) \quad -(3)$$

Solve equation $$-(3)$$ for $$F$$:
$$\frac{F}{\sqrt{2}} + \mu\frac{F}{\sqrt{2}} = \mu mg$$
$$F(1+\mu)\frac{1}{\sqrt{2}} = \mu mg$$
$$F = \frac{\mu mg\sqrt{2}}{1+\mu}$$

Insert the data $$m = 25\ \text{kg},\; g = 9.8\ \text{m s}^{-2},\; \mu = 0.25$$:
$$F = \frac{0.25 \times 25 \times 9.8 \times 1.414}{1 + 0.25}$$
$$F \approx \frac{86.6}{1.25} \approx 69.3\ \text{N}$$

Horizontal component of the applied force:
$$F_x = \frac{F}{\sqrt{2}} \approx \frac{69.3}{1.414} \approx 49.0\ \text{N}$$

Work done by the external force over the displacement $$s = 5\ \text{m}$$:
$$W = F_x\,s = 49.0 \times 5 \approx 245\ \text{J}$$

Therefore, the work done by the external force is $$245\ \text{J}$$.
Correct option: Option C

Potential energy $$U$$ is defined only for conservative forces. A conservative force satisfies the following statements:

• Work done by the force depends only on the initial and final positions, not on the actual path.
• For a conservative force $$\vec F_c$$ we can write $$\vec F_c = -\,\boldsymbol{\nabla} U$$, that is, the force equals the negative gradient of a scalar potential energy function $$U(x,y,z)$$.

Let us analyse each force in the options.

Case A: Coulomb’s force between two point charges is $$\vec F = k\,\frac{q_1 q_2}{r^2}\,\hat r$$. It is inverse-square and central, hence conservative. We can define the electrostatic potential energy $$U = k\,\dfrac{q_1 q_2}{r}$$ such that $$\vec F = -\,\dfrac{dU}{dr}\,\hat r$$. Therefore it can be expressed in terms of potential energy.

Case B: Gravitational force between two masses is $$\vec F = -G\,\dfrac{m_1 m_2}{r^2}\,\hat r$$, which is also inverse-square and conservative. The corresponding potential energy is $$U = -G\,\dfrac{m_1 m_2}{r}$$. Hence gravitational force is expressible through potential energy.

Case C: Frictional force (kinetic or static) depends on the nature of surfaces and usually acts opposite to the direction of motion or impending motion. Work done against friction depends on the path length, not merely on initial and final positions. Therefore friction is a non-conservative force. For a non-conservative force we cannot define a single-valued scalar function $$U$$ satisfying $$\vec F = -\,\boldsymbol{\nabla} U$$. So friction cannot be expressed in terms of potential energy.

Case D: A restoring force such as the spring force is $$\vec F = -k\,x\,\hat i$$. It is conservative; its potential energy is the elastic potential energy $$U = \tfrac12 k x^2$$ because $$\vec F = -\,\dfrac{dU}{dx}\,\hat i$$.

Only the frictional force fails to meet the criteria for conservative forces.

Hence the force that cannot be expressed in terms of potential energy is the frictional force → Option C (Option 3).

Let the particle be released from rest at height $$S$$ above the earth’s surface. Choose the earth’s surface as the reference level of zero potential energy.

Initial potential energy: $$U_i = mgS$$
Initial kinetic energy: $$K_i = 0$$ (because it is released from rest)

Total mechanical energy remains constant, so for any later instant
$$K + U = mgS$$  $$-(1)$$

At some height $$h$$ (measured from the surface) the kinetic energy is given to be three times the potential energy:
$$K = 3U$$  $$-(2)$$

Let the potential energy at this instant be $$U = mgh$$. Using $$-(2)$$, the kinetic energy at the same instant is
$$K = 3mgh$$.

Substitute these values of $$K$$ and $$U$$ into the energy conservation equation $$-(1)$$:
$$3mgh + mgh = mgS$$
$$4mgh = mgS$$
$$h = \frac{S}{4}$$.

Now compute the speed. Kinetic energy at height $$h$$ is also $$K = \frac12 mv^2$$. Equate this with $$3mgh$$:

$$\frac12 mv^2 = 3mgh$$
$$v^2 = 6gh$$
Insert $$h = \frac{S}{4}$$:
$$v^2 = 6g\left(\frac{S}{4}\right) = \frac{3gS}{2}$$
$$v = \sqrt{\frac{3gS}{2}}$$.

Therefore, the height of the particle above the earth’s surface and its speed at that instant are
$$h = \frac{S}{4}, \qquad v = \sqrt{\frac{3gS}{2}}$$.

Hence, the correct option is Option D.

The motion is along the +x-direction. Outside the rough strip the block is free; inside the strip (from $$x_i = 0.1\;{\rm m}$$ to $$x_f = 1.9\;{\rm m}$$) it experiences the position-dependent retarding force

$$F_r = -\,k\,x,$$ with $$k = 10\;{\rm N\,m^{-1}}$$. (The minus sign shows that the force is opposite to the direction of motion, so it removes kinetic energy from the block.)

Case 1 : Work done by the retarding force

The infinitesimal work done by the force over a small displacement $$dx$$ is $$dW = F_r\,dx = -k\,x\,dx.$$ To obtain the total work between the entry and exit points, integrate from $$x = 0.1$$ m to $$x = 1.9$$ m:

$$\begin{aligned} W &= \int_{0.1}^{1.9} (-k\,x)\,dx \\ &= -k \int_{0.1}^{1.9} x\,dx \\ &= -k\left[\frac{x^{2}}{2}\right]_{0.1}^{1.9} \\ &= -\frac{k}{2}\Bigl(x_f^{2}-x_i^{2}\Bigr). \end{aligned}$$

Insert the numerical values:

$$\begin{aligned} x_f^{2}-x_i^{2} &= (1.9)^{2}-(0.1)^{2}=3.61-0.01=3.60,\\[4pt] W &= -\frac{10}{2}\,(3.60) = -5 \times 3.60 = -18\;{\rm J}. \end{aligned}$$

The negative sign confirms that the force takes 18 J of mechanical energy out of the block.

Case 2 : Applying the work-energy theorem

The work-energy theorem states $$W = K_f - K_i,$$ where $$K_i$$ and $$K_f$$ are the initial and final kinetic energies, respectively.

Initial kinetic energy: $$K_i = \frac{1}{2} m v_i^{2} = \tfrac12 (1\;{\rm kg})(10\;{\rm m/s})^{2} = 50\;{\rm J}.$$

Therefore,

$$\begin{aligned} -18\;{\rm J} &= K_f - 50\;{\rm J}\\ K_f &= 50\;{\rm J} - 18\;{\rm J} = 32\;{\rm J}. \end{aligned}$$

Case 3 : Determining the final speed

Let the required speed be $$v_f$$. $$K_f = \frac{1}{2} m v_f^{2} \; \Longrightarrow\; \frac{1}{2}(1)\,v_f^{2} = 32.$$

Solve for $$v_f$$:

$$v_f^{2} = 64 \quad\Longrightarrow\quad v_f = 8\;{\rm m/s}.$$

Hence, when the block leaves the rough region its speed is 8 m s−1.

Final answer : 8 m/s (Option D)

The body moves in a vertical circle of radius $$R$$ under gravity $$g$$. The velocity at the top is given as $$v_{\text{top}} = n \sqrt{gR}$$, where $$n \geq 1$$ to ensure the body completes the circle.

The kinetic energy at the top is:

$$KE_{\text{top}} = \frac{1}{2} m v_{\text{top}}^2 = \frac{1}{2} m (n \sqrt{gR})^2 = \frac{1}{2} m n^2 gR$$

To find the kinetic energy at the bottom, use conservation of mechanical energy. Set the reference point for potential energy at the bottom of the circle.

At the bottom, height = 0, so potential energy = 0.

At the top, height = $$2R$$, so potential energy = $$mg \cdot 2R = 2mgR$$.

By conservation of energy:

Total energy at bottom = Total energy at top

$$\frac{1}{2} m v_{\text{bottom}}^2 + 0 = \frac{1}{2} m v_{\text{top}}^2 + 2mgR$$

Simplify:

$$\frac{1}{2} v_{\text{bottom}}^2 = \frac{1}{2} v_{\text{top}}^2 + 2gR$$

Multiply both sides by 2:

$$v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gR$$

Substitute $$v_{\text{top}} = n \sqrt{gR}$$:

$$v_{\text{bottom}}^2 = (n \sqrt{gR})^2 + 4gR = n^2 gR + 4gR = gR(n^2 + 4)$$

Kinetic energy at the bottom is:

$$KE_{\text{bottom}} = \frac{1}{2} m v_{\text{bottom}}^2 = \frac{1}{2} m gR (n^2 + 4)$$

The ratio of kinetic energy at the bottom to that at the top is:

$$\frac{KE_{\text{bottom}}}{KE_{\text{top}}} = \frac{\frac{1}{2} m gR (n^2 + 4)}{\frac{1}{2} m gR n^2} = \frac{n^2 + 4}{n^2}$$

Comparing with the options:

A. $$\frac{n^{2}}{n^{2}+4}$$

B. $$\frac{n^{2}+4}{n^{2}}$$

C. $$\frac{n+4}{n}$$

D. $$\frac{n}{n+4}$$

Option B matches the result.

Thus, the ratio is $$\frac{n^{2}+4}{n^{2}}$$.

We need to find the value of $$b$$ such that the work done is zero when force $$\vec{F} = 2\hat{i} + b\hat{j} + \hat{k}$$ causes displacement $$\vec{d} = \hat{i} - 2\hat{j} - \hat{k}$$.

Apply the work formula

Work done = $$\vec{F} \cdot \vec{d} = 0$$

Compute the dot product

$$W = (2)(1) + (b)(-2) + (1)(-1) = 2 - 2b - 1 = 1 - 2b$$

Set work equal to zero and solve

$$1 - 2b = 0$$

$$b = \frac{1}{2}$$

The answer is Option B: $$\frac{1}{2}$$.

The work done by $$\vec{F} = x^2 y\,\hat{i} + y^2\,\hat{j}$$ along a displacement from $$(0, 0)$$ to $$(4, 2)$$ is $$W = \int_C x^2 y\,dx + \int_C y^2\,dy$$.

Since the force acts in the plane $$x + y = 10$$, we substitute $$y = 10 - x$$ in the $$x$$-component of the force. The first integral becomes $$\int_0^4 x^2(10 - x)\,dx = \int_0^4 (10x^2 - x^3)\,dx = \left[\dfrac{10x^3}{3} - \dfrac{x^4}{4}\right]_0^4 = \dfrac{640}{3} - 64 = \dfrac{448}{3}$$.

The second integral evaluates directly as $$\int_0^2 y^2\,dy = \left[\dfrac{y^3}{3}\right]_0^2 = \dfrac{8}{3}$$.

Adding both contributions: $$W = \dfrac{448}{3} + \dfrac{8}{3} = \dfrac{456}{3} = \boxed{152}$$ Joule.

Work = $$\int_2^4 (3x^2+2x-5)dx = [x^3+x^2-5x]_2^4$$

At x=4: 64+16-20 = 60. At x=2: 8+4-10 = 2.

W = 60 - 2 = 58 J.

The answer is $$\boxed{58}$$ J.

A solid sphere and a hollow cylinder roll up an incline with the same initial velocity $$v$$. The ratio of the heights they reach is $$\frac{h_1}{h_2} = \frac{n}{10}$$. We need to find $$n$$.

Apply energy conservation for rolling bodies.

For a body rolling without slipping, the total kinetic energy converts to potential energy at the maximum height:

$$ \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh $$

Using $$\omega = v/r$$ (rolling condition): $$\frac{1}{2}mv^2\left(1 + \frac{I}{mr^2}\right) = mgh$$

$$ h = \frac{v^2}{2g}\left(1 + \frac{I}{mr^2}\right) $$

Calculate height for the solid sphere.

For a solid sphere: $$I = \frac{2}{5}mr^2$$, so $$\frac{I}{mr^2} = \frac{2}{5}$$

$$ h_1 = \frac{v^2}{2g}\left(1 + \frac{2}{5}\right) = \frac{v^2}{2g} \times \frac{7}{5} = \frac{7v^2}{10g} $$

Calculate height for the hollow cylinder.

For a hollow cylinder (thin-walled): $$I = mr^2$$, so $$\frac{I}{mr^2} = 1$$

$$ h_2 = \frac{v^2}{2g}(1 + 1) = \frac{v^2}{g} $$

Find the ratio.

$$ \frac{h_1}{h_2} = \frac{7v^2/(10g)}{v^2/g} = \frac{7v^2}{10g} \times \frac{g}{v^2} = \frac{7}{10} $$

Therefore $$n = 7$$.

The answer is $$\boxed{7}$$.

Potential Energy at point A, $$PE_A=mgH_A\ =mgR\left(1+\sin45^{\circ\ }\right)\ =m\times\ 10\times\ \left(1+\frac{1}{\sqrt{\ 2}}\right)\ =\ 240m\ J$$

Potential Energy at point B = 0

From conservation of energy,

$$PE_A+KE_A=PE_B+KE_B$$

⇒ $$KE_B=PE_A+KE_A-PE_B\ =240m+0-0=240m$$

⇒ $$\frac{1}{2}mv_B^2=240m\ ⇒\ v_B^2\ =480$$

$$\therefore\ v_B=\sqrt{\ 480}=21.91\ $$ m/s

Work done by all the forces on the particle = change in kinetic energy of the particle

$$KE\ =\ \frac{1}{2}mv^2\ =\ \frac{1}{2}m \left(\alpha \sqrt{\ x}\right)^2\ =\ \frac{1}{2}m\alpha\ ^2x$$

$$KE_{initial}\ =\ 0$$

$$KE_{final}\ =\ \frac{1}{2}m\alpha\ ^2d$$

$$\therefore\ W\ =\ KE_{final}-KE_{initial}\ =\ \frac{1}{2}m\alpha\ ^2d$$

A stationary particle breaks into two parts. We need the ratio $$K_B : K_A$$.

By conservation of momentum, the initial momentum is zero since the particle is stationary, so $$m_A v_A = m_B v_B$$.

The kinetic energies of the parts are given by $$K_A = \frac{1}{2}m_A v_A^2, \quad K_B = \frac{1}{2}m_B v_B^2$$.

Thus, the ratio of the kinetic energies is $$\frac{K_B}{K_A} = \frac{m_B v_B^2}{m_A v_A^2} = \frac{m_B v_B}{m_A v_A} \times \frac{v_B}{v_A}$$. Since $$m_B v_B = m_A v_A$$ from conservation of momentum, the first factor equals 1, giving $$\frac{K_B}{K_A} = 1 \times \frac{v_B}{v_A} = \frac{v_B}{v_A}$$.

Therefore $$K_B : K_A = v_B : v_A$$.

The correct answer is Option (1): $$v_B : v_A$$.

Given,
bob of pendulum is released from horizontal position,
Length of Pendulum ,L= $$10$$ m
$$g = 10 \text{ m s}^{-2}$$
At horizontal position, the bob is at height h=L above the lowest point.

Energy of bob at Initial Position is Ei = P.E = mgh

where: m = mass of the bob - $$g = 10 \text{ m s}^{-2}$$ (acceleration due to gravity) - h= $$10$$ m (length of the pendulum) So,

                                 P.E= m$$\times\ $$g$$\times\ 10$$

Now Lets Calculate Energy Dissipated:-

                                 K.E=$$\ \frac{\ 1}{2}\times\ m\times\ v^2$$
 Since 10% of the energy is dissipated due to air resistance, the energy available for conversion to kinetic energy (KE) is:

                                 K.E(final)=$$90\ \%\ $$ of $$m\times\ g\times\ h$$ = $$\ \frac{\ 1}{2}\times\ m\times\ v^2$$

                                 K.E(final)=$$\ \frac{\ 90}{100}\times\ m\times\ 10\times\ 10$$= $$\ \frac{\ 1}{2}\times\ m\times\ v^2$$
                                 On cancelling m on both sides,

                                 K.E(final)=$$\ \frac{\ 90}{100}\times\ 10\times\ 10$$= $$\ \frac{\ 1}{2}\times\ v^2$$         

                                 $$v^2=2\times\ \ \frac{\ 90}{100}\times\ 10\times\ 10$$

                                  $$v^2=2\times\ \ \frac{\ 90}{100}\times\ 100$$

                                  $$v^2=180$$

                                   $$v=\sqrt{\ 180}$$

                                   $$v=6\sqrt{\ 5}$$

Therefore, Final velocity v=$$6\sqrt{\ 5}$$
                  Correct answer is A

The relationship between kinetic energy and momentum is:

$$KE = \frac{p^2}{2m}$$, so $$p = \sqrt{2mKE}$$.

Since both bodies have equal kinetic energy:

$$\frac{p_1}{p_2} = \frac{\sqrt{2m_1 KE}}{\sqrt{2m_2 KE}} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{4}{25}} = \frac{2}{5}$$

The ratio of magnitudes of their linear momenta is $$2 : 5$$, which corresponds to Option (3).

The kinetic energy at bottom is the potential energy when it was at the top of inclined plane

$$W_{KE}$$= mgh

=$$5 \times 10 \times (10\sin30^\circ)$$

=$$250\text{ J} $$

Friction acts opposite to the motion of block,so this is negative work

$$W_{fr} $$=$$ \mu mgd $$

=$$ 0.5 \times 5 \times 10 \times 2 $$

=$$ 50\text{ J}$$

the leftover energy now, is stored completely in spring as final kinetic energy of mass is 0.

$$W_{spring} $$= $$\frac{1}{2}(100)x^2 $$=$$ 50x^2 $$

250 - 50 = $$50x^2 $$

200=$$50x^2 $$

4=$$x^2 $$

$$x$$=2

We need to find how displacement depends on time for a body under constant power. The constant power relation is $$P = Fv = mav = \text{constant}$$ so $$ma \cdot v = P \Rightarrow m\frac{dv}{dt} \cdot v = P$$ which gives $$mv \, dv = P \, dt$$. Integrating: $$\frac{mv^2}{2} = Pt$$ (starting from rest) and thus $$v = \sqrt{\frac{2Pt}{m}}$$.

Using $$v = \frac{ds}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$$ we find $$s = \sqrt{\frac{2P}{m}} \int_0^t t^{1/2} dt = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3}t^{3/2}$$, so $$s \propto t^{3/2}$$.

Displacement is proportional to $$t^{3/2}$$, which matches Option B. Therefore, the answer is Option B.

$$\vec{F}=(6t)\hat{i}+(6t^2)\hat{j}$$. $$m=2$$ kg. $$\vec{a}=\vec{F}/m=(3t)\hat{i}+(3t^2)\hat{j}$$.

$$\vec{v}=\int\vec{a}dt=\frac{3t^2}{2}\hat{i}+t^3\hat{j}$$.

Power: $$P=\vec{F}\cdot\vec{v}=(6t)(3t^2/2)+(6t^2)(t^3)=9t^3+6t^5$$.

The answer is Option (4): $$(9t^3+6t^5)$$ W.

$$KE_i = \frac{1}{2}(0.05)(100^2) = 250$$ J. $$KE_f = \frac{1}{2}(0.05)(40^2) = 40$$ J.

Loss = $$250 - 40 = 210$$ J. Percentage loss = $$\frac{210}{250} \times 100 = 84\%$$.

The correct answer is Option (1): 84%.

At equilibrium, the net moment on the bar would be zero.

Weight acts on the center of the rod (at length L/2 from either end).

Let the reaction force acting on the man's shoulder be R. This is the weight experienced by the man.

Balancing the Moment at the end of the rod on the ground:-

Moment due to the weight of the bar = W $$\times\ $$ $$\frac{L}{2}$$ $$\times\ $$ $$\cos\theta\ $$ (clockwise)

Moment due to reaction force on the shoulder = R $$\times\ $$ $$L$$ $$\times\ $$ $$\cos\theta\ $$ (anticlockwise)

At equilibrium, both of these moments would be equal

⇒ W $$\times\ $$ $$\frac{L}{2}$$ $$\times\ $$ $$\cos\theta\ $$ $$=$$ R $$\times\ $$ $$L$$ $$\times\ $$ $$\cos\theta\ $$

$$\therefore\ \ R\ =\ \frac{W}{2}$$ is the weight experienced by the man

A bullet of mass 0.01 kg moving at 200 m/s embeds into a 1 kg pendulum bob. In this perfectly inelastic collision, linear momentum is conserved: $$m_{\text{bullet}}u=(m_{\text{bullet}}+m_{\text{bob}})v$$. Substituting gives $$0.01\times200=(0.01+1)\times v\,,\quad2=1.01\,v\,,\quad v=\frac{2}{1.01}\approx1.98\approx2\text{ m/s}\,.$$

After the collision, the combined system converts kinetic energy to potential energy during its rise: $$\frac{1}{2}(m_{\text{bullet}}+m_{\text{bob}})v^2=(m_{\text{bullet}}+m_{\text{bob}})gh\,.$$ Cancelling the mass yields $$h=\frac{v^2}{2g}\,.$$ Substituting $$v=2$$ m/s and $$g=10$$ m/s$$^2$$ gives $$h=\frac{4}{20}=0.20\text{ m}\,.$$

The correct answer is Option B: 0.20 m.

We need to find the magnitude of the x-component of force at point P(1, 2, 3) given the potential energy function $$U = 2x^2 + 3y^3 + 2z$$.

The force is the negative gradient of the potential energy:

$$\vec{F} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)$$

Therefore, the x-component of force is:

$$F_x = -\frac{\partial U}{\partial x}$$

$$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x$$

(The terms $$3y^3$$ and $$2z$$ are treated as constants when differentiating with respect to $$x$$.)

$$F_x = -4x$$

At $$x = 1$$:

$$F_x = -4(1) = -4 \text{ N}$$

The magnitude of the x-component of force is $$|F_x| = 4$$ N.

The correct answer is Option (3): 4 N.

Three bodies A, B, C with equal kinetic energies and masses 400 g, 1.2 kg, 1.6 kg. Find the ratio of their momenta.

Recall the relation between momentum and kinetic energy. $$KE = \frac{p^2}{2m}$$, so $$p = \sqrt{2mKE}$$.

Since KE is the same for all three: $$p_A : p_B : p_C = \sqrt{m_A} : \sqrt{m_B} : \sqrt{m_C}$$

Substitute the masses. $$m_A = 0.4$$ kg, $$m_B = 1.2$$ kg, $$m_C = 1.6$$ kg. So $$= \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6}$$

Multiply all by $$\sqrt{10}$$ (equivalently, consider $$4:12:16$$ inside the roots): $$= \sqrt{4} : \sqrt{12} : \sqrt{16} = 2 : 2\sqrt{3} : 4 = 1 : \sqrt{3} : 2$$

The correct answer is Option (2): $$1 : \sqrt{3} : 2$$.

We are given that the kinetic energy of a body becomes 36 times its original value and asked to determine the percentage increase in momentum.

The relationship between kinetic energy ($$KE$$) and momentum ($$p$$) is $$KE = \frac{p^2}{2m}$$, where $$m$$ is the mass of the body, and rearranging gives $$p = \sqrt{2m \cdot KE}$$.

Letting the initial kinetic energy be $$KE$$ with corresponding momentum $$p$$ leads to $$p = \sqrt{2m \cdot KE}$$.

When the kinetic energy increases to $$KE' = 36 \, KE$$, the new momentum becomes $$p' = \sqrt{2m \cdot KE'} = \sqrt{2m \cdot 36 \, KE} = \sqrt{36} \cdot \sqrt{2m \cdot KE} = 6p$$, since $$\sqrt{36} = 6$$ and $$\sqrt{2m \cdot KE} = p$$.

The increase in momentum is then $$\Delta p = p' - p = 6p - p = 5p$$.

Therefore, the percentage increase in momentum is $$\frac{\Delta p}{p} \times 100 = \frac{5p}{p} \times 100 = 500\%$$.

The correct answer is Option (4): 500%.

All have same momentum $$p$$. $$KE = \frac{p^2}{2m}$$.

Smallest mass has maximum KE. Particle A has mass $$m/2$$ (smallest).

The correct answer is Option (2): A.

A ball is released from a height h. By applying energy conservation, its speed just before striking the ground is given by $$ v = \sqrt{2gh} $$.

After impact, the ball rebounds to a height of h/2. The potential energy at the initial height is $$E_i = mgh$$, while at the rebound height it is $$E_f = mg(h/2) = mgh/2$$.

The energy lost during the collision is therefore $$\Delta E = mgh - mgh/2 = mgh/2$$, which corresponds to a percentage loss of $$\frac{\Delta E}{E_i} \times 100 = \frac{mgh/2}{mgh} \times 100 = 50\%$$.

Hence, the correct answer is Option 1: $$50\%,\ \sqrt{2gh}$$.

Statement I:

When the speed of liquid is zero everywhere, the fluid is in hydrostatic condition. In this case, pressure variation depends only on depth. The relation between pressure difference and height difference is given by:

$$P_1-P_2\ =\ \rho\ g\left(h_2-h_1\right)$$

This is the standard hydrostatic pressure relation, so Statement I is correct.

Statement II:

In a venturi tube, fluid is flowing, so Bernoulli’s equation must be applied:

$$P_1+\ \frac{\ 1}{2}\rho\ v_1^2=\ P_2\ +\ \ \frac{\ 1}{2}\ \rho\ v_2^2$$

From Bernoulli’s equation:

Also, from the manometer height difference:

$$P_1-P_2\ =\ \rho\ gh$$

Combining these gives:

$$P_1+\ \frac{\ 1}{2}\rho\ v_1^2=\ P_2\ +\ \ \frac{\ 1}{2}\ \rho\ v_2^2$$

$$\ \ \ P_1-P_2=\frac{\ 1}{2}\ \rho\ v_2^2-\frac{\ 1}{2}\rho\ v_1^2=\rho gh$$

$$\frac{\ 1}{2}\ \rho\ v_2^2-\frac{\ 1}{2}\rho\ v_1^2=\rho gh$$

$$\ \ \rho\ v_2^2-\ \rho\ v_1^2=2\rho gh$$

$$\ v_2^2-\ \ \ v_1^2=2 gh$$

But in the statement it is given as:

$$\ v_1^2-\ \ \ v_2^2=2 gh$$

which has the wrong sign.

Hence, Statement II is incorrect.

Two vectors $$\vec{P}$$ and $$\vec{Q}$$ are perpendicular if their dot product is zero:

$$\vec{P} \cdot \vec{Q} = 0$$

Given:

$$\vec{P} = \hat{i} + 2m\hat{j} + m\hat{k}$$

$$\vec{Q} = 4\hat{i} - 2\hat{j} + m\hat{k}$$

Computing the dot product:

$$\vec{P} \cdot \vec{Q} = (1)(4) + (2m)(-2) + (m)(m) = 0$$

$$4 - 4m + m^2 = 0$$

$$m^2 - 4m + 4 = 0$$

$$(m - 2)^2 = 0$$

$$m = 2$$

The correct answer is Option 2: $$m = 2$$.

Find the loss of potential energy in the last second of fall for a 0.4 kg mass dropped from height P, taking 8 seconds to reach ground.

$$h_8 = \frac{1}{2}g t^2 = \frac{1}{2} \times 10 \times 64 = 320 \text{ m}$$

$$h_7 = \frac{1}{2} \times 10 \times 49 = 245 \text{ m}$$

$$\Delta h = h_8 - h_7 = 320 - 245 = 75 \text{ m}$$

$$\Delta PE = mg\Delta h = 0.4 \times 10 \times 75 = 300 \text{ J}$$

The answer is $$\boxed{300}$$.

Work done by a variable force: $$W = \int_{x_1}^{x_2} F \, dx$$

$$W = \int_2^4 5x \, dx = 5 \times \frac{x^2}{2}\bigg|_2^4 = \frac{5}{2}(16 - 4) = \frac{5}{2} \times 12 = 30 \text{ J}$$

The work done is $$\mathbf{30}$$ J.

Given: Mass = 5 kg, angle = 30°, acceleration = 1 m/s², starts from rest, g = 10 m/s².

First, we find the net pulling force.

Along the incline, the pulling force must overcome gravity and provide acceleration:

$$F = m(a + g\sin\theta) = 5(1 + 10 \times \sin 30°) = 5(1 + 5) = 30 \text{ N}$$

Next, we find velocity at t = 10 s.

$$v = u + at = 0 + 1 \times 10 = 10 \text{ m/s}$$

From this, we power delivered.

$$P = Fv = 30 \times 10 = 300 \text{ W}$$

A body of mass $$m = 1$$ kg starts from rest under the force $$\vec{F} = (t\hat{i} + 3t^2\hat{j})$$ N, so the acceleration is $$\vec{a} = \frac{\vec{F}}{m} = t\hat{i} + 3t^2\hat{j}$$ m/s².

Since the initial velocity is $$\vec{v}(0)=0$$, integrating the acceleration from 0 to $$t$$ gives the velocity as $$ \vec{v} = \int_0^t \vec{a}\,dt = \frac{t^2}{2}\hat{i} + t^3\hat{j}. $$

Next, at $$t = 2$$ s, we obtain $$ \vec{v}(2) = \frac{4}{2}\hat{i} + 8\hat{j} = 2\hat{i} + 8\hat{j} $$ m/s and $$ \vec{F}(2) = 2\hat{i} + 12\hat{j} $$ N, so the power is $$ P = \vec{F}\cdot\vec{v} = (2)(2) + (12)(8) = 4 + 96 = 100 \text{ W}. $$ Therefore the correct answer is 100 W.

We have a body of mass $$m = 5$$ kg with initial momentum $$p_i = 10$$ kg m/s, acted upon by a force $$F = 2$$ N for $$t = 5$$ s.

The initial velocity is $$u = \frac{p_i}{m} = \frac{10}{5} = 2$$ m/s, so the initial kinetic energy is:

$$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(5)(4) = 10$$ J

Now, using the impulse-momentum theorem to find the final momentum:

$$p_f = p_i + Ft = 10 + 2 \times 5 = 20 \text{ kg m/s}$$

So the final velocity is $$v = \frac{p_f}{m} = \frac{20}{5} = 4$$ m/s, and the final kinetic energy is:

$$K_f = \frac{1}{2}mv^2 = \frac{1}{2}(5)(16) = 40$$ J

Hence, the increase in kinetic energy is:

$$\Delta K = K_f - K_i = 40 - 10 = 30 \text{ J}$$

So, the answer is $$30$$ J.

The tube is a vertical circle of radius $$R = 15 \text{ cm} = 0.15 \text{ m}$$. Take the bottom‐most point of the circle as the reference level of gravitational potential energy (GPE).

Initial (block released at the top)
Height of the top above the bottom is the diameter, $$h = 2R = 0.30 \text{ m}$$. Initial speed is $$v_0 = 22 \text{ m s}^{-1}$$.

Initial kinetic energy $$K_i = \tfrac12 m v_0^{2} = \tfrac12 (1) (22)^{2} = \tfrac12 (484) = 242 \text{ J}$$.

Initial gravitational potential energy $$U_i = mgh = (1)(10)(0.30) = 3 \text{ J}$$.

Hence the initial total mechanical energy $$E_i = K_i + U_i = 242 + 3 = 245 \text{ J}$$.

Final (block finally comes to rest at the bottom)
Final speed $$v_f = 0$$, so $$K_f = 0$$. At the bottom, $$U_f = 0$$ by our choice of reference. Therefore the final mechanical energy $$E_f = 0 + 0 = 0 \text{ J}$$.

Work done by the tube (friction)
For non-conservative forces, $$W_{\text{tube}} = E_f - E_i$$.

Substituting, $$W_{\text{tube}} = 0 - 245 = -245 \text{ J}$$.

The negative sign shows that the tube removes energy from the block. Thus the magnitude of the work done by the tube on the block is $$245 \text{ J}$$.

Answer: $$245 \text{ J}$$.

We have initial position $$\vec{r_i} = 2\hat{i} + 3\hat{j} - 4\hat{k}$$, final position $$\vec{r_f} = 5\hat{i} - 2\hat{j} + \hat{k}$$, and force $$\vec{F} = 5\hat{i} + 2\hat{j} + 7\hat{k}$$ N.

The displacement vector is:

$$\vec{d} = \vec{r_f} - \vec{r_i} = (5-2)\hat{i} + (-2-3)\hat{j} + (1-(-4))\hat{k} = 3\hat{i} - 5\hat{j} + 5\hat{k}$$

Now, work done is the dot product of force and displacement:

$$W = \vec{F} \cdot \vec{d} = (5)(3) + (2)(-5) + (7)(5)$$

$$W = 15 - 10 + 35 = 40 \text{ J}$$

So, the answer is $$40$$ J.

We need to find the minimum power delivered by the motor of an elevator carrying a maximum load of 1400 kg, moving upward with a uniform speed of 3 m/s, against a frictional force of 2000 N.

We begin by determining the total weight of the system. The total mass is 1400 kg (600 kg passengers + 800 kg elevator). Taking $$g = 10$$ m/s$$^2$$ gives:
$$W = mg = 1400 \times 10 = 14000 \text{ N}$$

Next, we determine the total force the motor must exert. Since the elevator moves upward at uniform speed, the acceleration is zero ($$a = 0$$), so by Newton's second law the net force is zero and the motor must exert an upward force equal to the sum of the weight (downward) and friction (opposing motion):
$$F_{motor} = W + F_{friction} = 14000 + 2000 = 16000 \text{ N}$$

Then, using the power formula for a constant force applied in the direction of motion at constant velocity, namely $$P = F \times v$$, where $$F$$ is the force and $$v$$ is the velocity, and substituting in the values gives:
$$P = 16000 \times 3 = 48000 \text{ W} = 48 \text{ kW}$$

Therefore, the minimum power delivered by the motor is 48 kW.

We need to find the percentage increase in kinetic energy when momentum is increased by 50%.

Relationship between kinetic energy and momentum: $$KE = \frac{p^2}{2m}$$

New momentum. If momentum increases by 50%: $$p' = 1.5p = \frac{3p}{2}$$

New kinetic energy: $$KE' = \frac{(p')^2}{2m} = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = 2.25 \times KE$$

Percentage increase: $$\% \text{ increase} = \frac{KE' - KE}{KE} \times 100 = (2.25 - 1) \times 100 = 125\%$$

The correct answer is 125%.

$$f_k = \mu mg$$

$$f_k = 0.04 \times 500 \times 10 = 200 \text{ N}$$

$$W = f_k \times d$$

$$W = 200 \text{ N} \times 4000 \text{ m} = 800,000 \text{ J}$$

$$W = 800 \text{ kJ}$$

A body of mass 1 kg collides head-on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses direction and moves at 2 m/s.

Since the collision is elastic and head-on, the velocity of the first body after collision is given by $$v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1$$ where $$u_1$$ is the initial velocity of the first body.

Substituting $$m_1 = 1$$ kg and $$m_2 = 3$$ kg, and noting that after collision the smaller body reverses direction with speed 2 m/s so $$v_1 = -2$$ m/s (taking the initial direction as positive), we get:

$$-2 = \frac{1 - 3}{1 + 3} \times u_1$$

This simplifies to:

$$-2 = \frac{-2}{4} \times u_1$$

$$-2 = -\frac{u_1}{2}$$

Therefore,

$$u_1 = 4$$ m/s

Now we verify this result using conservation of momentum. The velocity of the second body after collision is:

$$v_2 = \frac{2 m_1}{m_1 + m_2} u_1 = \frac{2}{4} \times 4 = 2$$ m/s

Checking momentum: Before collision the total momentum is $$1 \times 4 = 4$$ kg·m/s, and after collision it is $$1 \times (-2) + 3 \times 2 = -2 + 6 = 4$$ kg·m/s ✓

Checking kinetic energy: Before collision the total kinetic energy is $$\frac{1}{2}(1)(16) = 8$$ J, and after collision it is $$\frac{1}{2}(1)(4) + \frac{1}{2}(3)(4) = 2 + 6 = 8$$ J ✓

Therefore, the initial speed of the smaller body is $$\mathbf{4}$$ m/s.

The work done by a variable force is calculated by integration:

$$W = \int_{x_1}^{x_2} F \, dx$$

We are given that $$F = (2 + 3x)$$ N, displacement from $$x = 0$$ to $$x = 4$$ m

$$W = \int_0^4 (2 + 3x) \, dx$$

$$W = \left[2x + \frac{3x^2}{2}\right]_0^4$$

$$W = \left(2(4) + \frac{3(16)}{2}\right) - 0$$

$$W = 8 + 24 = 32 \text{ J}$$

The work done is $$32$$ J.

Solution :

Given :

Mass of lift,

$$M = 500\text{ kg}$$

Initial speed,

$$u = 2\text{ m s}^{-1}$$

Acceleration downward,

$$a = 2\text{ m s}^{-2}$$

Distance fallen,

$$s = 6\text{ m}$$

Using equation of motion :

$$v^2 = u^2 + 2as$$

$$= (2)^2 + 2(2)(6)$$

$$= 4 + 24$$

$$= 28$$

Kinetic energy at the end :

$$K = \frac{1}{2}Mv^2$$

$$= \frac{1}{2}(500)(28)$$

$$= 250 \times 28$$

$$= 7000\text{ J}$$

$$= 7\text{ kJ}$$

Final Answer :

$$7\text{ kJ}$$

An object of mass m is placed on a smooth horizontal surface and acted upon by a force F making an angle $$\theta\ $$ with the horizontal, where $$θ=kx$$.

Since the surface is smooth, there is no friction. The vertical component of the force is balanced by the normal reaction, so only the horizontal component contributes to acceleration.

$$F\cos\ \theta\ =ma$$

Now relate acceleration to velocity using:

$$F\cos\left(\ kx\right)\ =m\times\ \frac{\ dv}{dt}$$

$$F\cos\left(\ kx\right)\ =m\times\ \frac{\ dv}{dt}\times\ \ \frac{\ dx}{dx}$$

$$F\cos\left(\ kx\right)\ =m\times\ \frac{\ dv}{dx}\times\ \ \frac{\ dx}{dt}$$

$$F\cos\left(\ kx\right)\ =m\times\ v\times\ \frac{\ dv}{dx}$$

Now separate variables and integrate both sides, taking limits from initial position (where velocity is zero) to any position xxx.

$$F\cos\left(\ kx\ \right)\times\ dx\ \times\ \frac{\ 1}{m} =v\times\ dv$$

$$\int\ F\cos\left(\ kx\ \right)\times\ dx\ \times\ \frac{\ 1}{m}\ =\int\ v\times\ dv$$

$$\ \ \frac{\ F}{k}\sin\ \left(kx\right)\times\ \frac{\ 1}{m}\ =\ \frac{\ v^2}{2}$$

Now express kinetic energy using velocity.

$$\ \ \frac{\ F}{k}\sin\ \left(kx\right)=\ \frac{\ mv^2}{2}=\ K.E$$

Finally, use the relation $$\theta = kx$$ to rewrite the expression in terms of $$\theta\ $$ and compare with the given form.

$$\ \frac{\ F}{k}\sin\theta\ =\frac{n}{k}\sin\theta$$

$$\ \frac{\ 2}{k}\sin\theta\ =\frac{n}{k}\sin\theta$$ (as F = 2)

Thus, the value of n is 2.

$$W = \int_{y_i}^{y_f} F(y) \, dy$$

$$W = \int_{2}^{5} (5 + 3y^2) \, dy$$

$$W = \left[ 5y + \frac{3y^3}{3} \right]_2^5$$

$$W = \left[ 5y + y^3 \right]_2^5$$

$$W = 132 \text{ J}$$

For a string fixed at both ends, the frequency of the $$n^{\text{th}}$$ harmonic is given by
$$f_n = \frac{n\,v}{2L}$$
where $$v$$ is the wave speed on the string and $$L$$ is its length.

Successive harmonics correspond to $$n$$ and $$n+1$$, so the difference between their frequencies is
$$\Delta f = f_{n+1} - f_n = \frac{(n+1)v}{2L} - \frac{nv}{2L} = \frac{v}{2L}$$ $$-(1)$$

The two given successive frequencies are 750 Hz and 1000 Hz, hence
$$\Delta f = 1000 - 750 = 250\ \text{Hz}$$

Substituting $$L = 1\ \text{m}$$ into $$(1)$$:
$$v = 2L\,\Delta f = 2 \times 1 \times 250 = 500\ \text{m s}^{-1}$$

The wave speed on a stretched string is also related to the tension $$T$$ and the linear mass density $$\mu$$ by
$$v = \sqrt{\frac{T}{\mu}}$$ $$\Longrightarrow$$ $$T = \mu v^{2}$$ $$-(2)$$

The string’s mass is $$2 \times 10^{-5}\ \text{kg}$$ and its length is $$1\ \text{m}$$, so
$$\mu = \frac{\text{mass}}{\text{length}} = \frac{2 \times 10^{-5}}{1} = 2 \times 10^{-5}\ \text{kg m}^{-1}$$

Substituting $$\mu$$ and $$v = 500\ \text{m s}^{-1}$$ into equation $$(2)$$:
$$T = (2 \times 10^{-5}) \times (500)^{2}$$
$$T = 2 \times 10^{-5} \times 250000$$
$$T = 5\ \text{N}$$

Therefore, the tension in the string is 5 N.

Given:

Mass of both balls = equal

Radius of quarter circle = 20 cm = 0.2 m

Initial velocity of ball P=0 (starts from rest)

Initial velocity of ball Q=0(at rest)

Acceleration due to gravity $$g = 10$$ m s$$^{-2}$$

Collision is perfectly elastic

Friction is neglected

A small ball P starts from rest and slides down a smooth quarter circular path. Since friction is neglected, mechanical energy is conserved during the motion.

As the ball moves from the top point to the bottom, it loses gravitational potential energy which gets converted into kinetic energy.

Initial P.E of the ball P = $$m\times\ g\times\ h$$ 

Final K.E of the ball is $$\ \frac{\ 1}{2}m\times v^2$$

As both should be equal,

$$\ \frac{\ 1}{2}m\times v^2$$ = $$m\times\ g\times\ h$$

$$v_p=\sqrt{\ 2gh}$$

$$v_p=\sqrt{\ 2\times\ 0.2\times\ 10}$$

$$v_p=\sqrt{\ 4}$$

$$v_p=2 m s^{-1}$$

At the lowest point, the velocity of the ball is horizontal. The ball then collides elastically with another ball Q of equal mass which is initially at rest.

$$v_q=0$$ and $$v_p=2 m s^{-1}$$

Since the collision is perfectly elastic and the masses are equal, the velocities are exchanged after collision.

Thus, after collision, ball P comes to rest and ball Q moves with the velocity that ball P had just before collision.

$$v_q=2 m s^{-1}$$ and $$v_p=0$$ 

Hence, the required velocity of ball Q is $$2$$ m s$$^{-1}$$

Statement I: By work-energy theorem, $$Fd = KE$$. Same KE and same retarding force F → same distance d. True.

Statement II: Direction changes from east to north → velocity direction changed → acceleration is NOT zero (centripetal acceleration exists). False.

Statement I is correct but Statement II is incorrect.

We need to identify the correct statements about work done in various situations.

(A) Work done by a man lifting a bucket out of a well:

The man applies an upward force, and the displacement is upward. Since force and displacement are in the same direction, the work done by the man is positive. The statement says negative — INCORRECT.

(B) Work done by gravitational force in lifting a bucket out of a well:

Gravitational force acts downward, while the displacement is upward. Since force and displacement are in opposite directions, the work done by gravity is negative. — CORRECT. ✓

(C) Work done by friction on a body sliding down an incline:

Friction acts up the incline (opposing motion), while the body moves down the incline. Since force and displacement are in opposite directions, the work done by friction is negative. The statement says positive — INCORRECT.

(D) Work done by applied force on a body moving on rough horizontal plane with uniform velocity:

For uniform velocity on a rough surface, the applied force equals the friction force and is in the direction of motion. Therefore, the work done by the applied force is positive, not zero. — INCORRECT.

(E) Work done by air resistance on an oscillating pendulum:

Air resistance always opposes the direction of motion. Since force and displacement are always in opposite directions, the work done by air resistance is negative. — CORRECT. ✓

The correct statements are B and E.

The correct answer is Option 1: B and E only.

A bullet of mass 75 g is fired at a stationary wooden bob of mass 50 g on a pendulum of length 2 m. The bullet emerges with speed v/3 and the bob just completes the vertical circle.

First, for the bob to just complete the vertical circle, the minimum speed at the bottom must satisfy the energy condition. At the top of the circle, the minimum speed requires:

$$mg = \frac{mv_{top}^2}{L} \implies v_{top}^2 = gL$$

Using energy conservation from bottom to top (height = 2L):

$$\frac{1}{2}mv_{bob}^2 = \frac{1}{2}mv_{top}^2 + mg(2L)$$

$$v_{bob}^2 = gL + 4gL = 5gL$$

Substituting values:

$$v_{bob}^2 = 5 \times 10 \times 2 = 100$$

$$v_{bob} = 10 \text{ m/s}$$

Next, applying conservation of momentum for the bullet-bob collision gives:

$$m_{bullet} \cdot v = m_{bullet} \cdot \frac{v}{3} + m_{bob} \cdot v_{bob}$$

Substituting the masses and speeds,

$$75 \times 10^{-3} \cdot v - 75 \times 10^{-3} \cdot \frac{v}{3} = 50 \times 10^{-3} \times 10$$

$$75 \times 10^{-3} \cdot \frac{2v}{3} = 0.5$$

$$50 \times 10^{-3} \cdot v = 0.5$$

$$v = \frac{0.5}{0.05} = 10 \text{ m/s}$$

Therefore, the value of v is 10 m/s.

Given,

• Mass, m=100 g ,=0.1 kg ,m = 100
• Height, h=10 cm, h =0.1 
• $$g = 10$$ m s$$^{-2}$$
• Compression of spring =$$\ \frac{\ h}{2}$$=0.05
  
  The ball is dropped from a height  above the platform. After reaching the platform, it continues to move downward and compresses the spring by an additional distance$$\ \frac{\ h}{2}$$

Thus, the total downward displacement of the ball from its initial position to the point of maximum compression is $$h\ +\ \ \frac{\ h}{2}\ =\ \ \frac{\ 3h}{2}$$

During this motion, the ball loses gravitational potential energy corresponding to this total displacement.

P.E = $$m\times\ g\times\ h_{total}$$

P.E = $$m\times\ g\times\ \ \frac{\ 3h}{2}$$

At the point of maximum compression, the velocity of the ball becomes zero, and all the lost gravitational potential energy is stored as elastic potential energy in the spring.

$$U=\ \frac{\ 1}{2}\times\ k\times\ x^2$$

$$U=\ \frac{\ 1}{2}\times\ k\times\ \left(\frac{\ h}{2}\right)^2$$

Applying conservation of energy, the loss in gravitational potential energy is equal to the gain in spring potential energy.

P.E = U
$$m\times\ g\times\ \ \frac{\ 3h}{2} = \ \frac{\ 1}{2}\times\ k\times\ \left(\frac{\ h}{2}\right)^2$$ 

Now substitute the given values of mass, gravity, and height into the equation.

$$0.1\times\ 10\times\ 0.15\ =\ \ \frac{\ 1}{2}\times\ k\times\ \left(\ \frac{\ 0.1}{2}\right)^2$$

On simplifying the expression, we obtain the value of the spring constant.

$$\ 0.15\ =\ k\frac{\ 0.01}{8}$$

$$k\ =\ \ \frac{\ 1.2}{0.01}$$

$$k\ =\ \ \ 120 N m ^{-1}$$

Hence, the spring constant is  120 N m$$^{-1}$$

We need to find the value of the integer $$n$$ given that a block of mass $$m$$ with initial kinetic energy $$E$$ compresses a spring by a distance of 25 cm when its speed is halved.

The initial kinetic energy of the block before hitting the spring is given by: $$E = \frac{1}{2} m v^2$$, where $$v$$ is the initial speed of the block.

When the block compresses the spring and its speed is halved, its new speed becomes $$\frac{v}{2}$$. The final kinetic energy of the block at this instant is: $$E_{final} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{E}{4}$$.

According to the law of conservation of mechanical energy, the loss in kinetic energy of the block is equal to the gain in elastic potential energy stored in the spring: $$\Delta KE = U_{spring}$$.

The energy stored in the spring when compressed by a distance $$x$$ is given by $$\frac{1}{2} k x^2$$, where $$k$$ is the spring constant. Substituting the energy values, we get: $$E - \frac{E}{4} = \frac{1}{2} k x^2$$, which simplifies to: $$\frac{3}{4} E = \frac{1}{2} k x^2$$.

Solving the expression for the spring constant $$k$$ yields: $$k = \frac{3 E}{2 x^2}$$.

We are given that the compression distance is $$x = 25\text{ cm} = 0.25\text{ m} = \frac{1}{4}\text{ m}$$. Substituting this value into the formula gives: $$k = \frac{3 E}{2 \left(\frac{1}{4}\right)^2} = \frac{3 E}{2 \times \frac{1}{16}} = \frac{3 E}{\frac{1}{8}} = 24 E$$.

Comparing this with the given expression for the spring constant $$k = n E$$, we find that $$n = 24$$.

Therefore, the value of $$n$$ is 24.

From conservation of energy,

Loss in Potential Energy (PE) = Gain in Kinetic Energy (KE)

$$PE_{initial}\ =\ mgh_i$$

From figure, $$h_i\ =\ l\left(1-\sin\left(30^{\circ\ }\right)\right)$$ = $$250\left(1-\frac{1}{2}\right)\ =\ \frac{250}{2}\ =\ 125\ cm\ =\ 1.25\ m$$

Therefore, $$PE_i\ =\ 0.2\ \times\ 10\ \times\ 1.25\ =\ 2.5\ J$$

$$PE_{final}\ =\ 0\ $$

$$KE_{initial}\ =\ 0\ $$ (as the bob is at rest)

Therefore, $$PE_{initial}\ -\ PE_{final}\ =\ KE_{final}\ -\ KE_{initial}$$

$$\therefore\ \ KE_{final}\ =\ PE_{initial}\ =\ 2.5\ J$$

$$\therefore\ \ \frac{1}{2}mv_{\max}^2\ =\ 2.5\ J\ $$

$$\therefore\ \ v_{\max}\ =\ \sqrt{\frac{\left(\ 2\times\ 2.5\right)}{0.2}\ }=\ 5$$m/s

Let the convex lens be taken as the origin and its principal axis as the positive $$x$$-axis. Distances measured to the left of the lens are negative.

The focal length is $$f = 10\text{ cm}$$ and the foot of the rod on the axis is kept at an object distance
$$u_A = -\frac{40}{3}\text{ cm}$$.

Object end A (on the principal axis)
Applying the thin-lens formula $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$,

$$\frac{1}{v_A} - \frac{1}{-\frac{40}{3}} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_A} + \frac{3}{40} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_A} = \frac{1}{40} \\ \Rightarrow v_A = 40\text{ cm}$$

The transverse magnification for this point is
$$m_A = \frac{v_A}{u_A} = \frac{40}{-\frac{40}{3}} = -3$$.
Since $$y_A = 0$$, the image coordinate is $$y'_A = m_A y_A = 0$$.

Second end B of the rod
The rod has length $$L = 2\text{ cm}$$ and is inclined at
$$\theta = \frac{2\pi}{3} = 120^{\circ}$$ to the axis.
Its components are $$\Delta x = L\cos\theta = 2(-\tfrac12) = -1\text{ cm},\; \Delta y = L\sin\theta = 2\!\left(\tfrac{\sqrt3}{2}\right) = \sqrt3\text{ cm}$$.

Hence the coordinates of B are
$$x_B = u_A+\Delta x = -\frac{40}{3}-1 = -\frac{43}{3}\text{ cm},\; y_B = 0+\Delta y = \sqrt3\text{ cm}$$.

Using the lens formula again for B,

$$\frac{1}{v_B} - \frac{1}{-\frac{43}{3}} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_B} + \frac{3}{43} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_B} = \frac{13}{430} \\ \Rightarrow v_B = \frac{430}{13}\text{ cm}$$

The magnification for B is
$$m_B = \frac{v_B}{u_B} = \frac{\tfrac{430}{13}}{-\tfrac{43}{3}} = -\frac{30}{13}$$.

Therefore the image coordinate of B is
$$y'_B = m_B y_B = -\frac{30}{13}\,\sqrt3\text{ cm}$$,
while $$x'_B = v_B = \frac{430}{13}\text{ cm}$$.

Length and orientation of the image rod
The differences between image coordinates are
$$\Delta y' = y'_B - y'_A = -\frac{30\sqrt3}{13}\text{ cm} \\ \Delta x' = x'_B - x'_A = \frac{430}{13} - 40 = \frac{430 - 520}{13} = -\frac{90}{13}\text{ cm}$$.

The magnitude of the height is indeed
$$|\Delta y'| = \frac{30\sqrt3}{13}\text{ cm},$$
matching the data given.

The angle $$\alpha$$ that the image makes with the axis is obtained from

$$\tan\alpha = \left|\frac{\Delta y'}{\Delta x'}\right| = \frac{30\sqrt3/13}{90/13} = \frac{\sqrt3}{3} = \tan\!\left(\frac{\pi}{6}\right).$$

Hence $$\alpha = \frac{\pi}{6}\text{ rad}$$. Writing $$\alpha = \frac{\pi}{n}$$ gives $$n = 6$$.

Answer (numerical): 6

A block of mass $$m = 0.5$$ kg moving with initial speed $$v_i = 12$$ m/s compresses a spring by $$x = 30$$ cm $$= 0.3$$ m and slows to a final speed $$v_f = \frac{12}{2} = 6$$ m/s.

By the work-energy theorem, the decrease in kinetic energy of the block is stored as elastic potential energy in the spring:

$$\frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2 = \frac{1}{2}kx^2$$

Substituting the values gives the initial kinetic energy $$KE_i = \frac{1}{2}(0.5)(12)^2 = \frac{1}{2}(0.5)(144) = 36 \text{ J}$$ and the final kinetic energy $$KE_f = \frac{1}{2}(0.5)(6)^2 = \frac{1}{2}(0.5)(36) = 9 \text{ J}$$, so the loss in kinetic energy is $$36 - 9 = 27 \text{ J}$$.

Equating this to the spring’s potential energy, $$\frac{1}{2}kx^2 = 27$$, and using $$x = 0.3$$ m leads to $$\frac{1}{2}k(0.3)^2 = 27$$ or $$\frac{1}{2}k(0.09) = 27$$, from which $$k = \frac{27 \times 2}{0.09} = \frac{54}{0.09} = 600 \text{ Nm}^{-1}$$.

The spring constant is 600 Nm$$^{-1}$$.

A block of mass $$m = 2$$ kg has initial speed $$v_0 = 4$$ m s$$^{-1}$$. It encounters a rough surface from $$x = 0.5$$ m to $$x = 1.5$$ m with retarding force $$F = -kx$$, where $$k = 12$$ N m$$^{-1}$$. The initial kinetic energy is $$KE_i = \frac{1}{2}mv_0^2 = \frac{1}{2}(2)(4^2) = 16 \text{ J}$$.

The work done by the retarding force is $$W = \int_{0.5}^{1.5} F \, dx = -\int_{0.5}^{1.5} 12x \, dx = -12 \left[\frac{x^2}{2}\right]_{0.5}^{1.5}$$ which gives $$W = -6\left[(1.5)^2 - (0.5)^2\right] = -6[2.25 - 0.25] = -6 \times 2 = -12 \text{ J}$$. Applying the work-energy theorem, $$KE_f = KE_i + W = 16 - 12 = 4 \text{ J}$$ so that $$\frac{1}{2}mv^2 = 4$$ and hence $$v = \sqrt{\frac{2 \times 4}{2}} = \sqrt{4} = 2 \text{ m s}^{-1}$$. The correct answer is Option A.

A particle of mass $$m$$ moves in a circular path of constant radius $$r$$ with centripetal acceleration $$a_c = k^2rt^2$$. Since centripetal acceleration is also given by $$a_c = \frac{v^2}{r}$$, substituting the given expression yields $$\frac{v^2}{r} = k^2rt^2$$, which leads to $$v^2 = k^2r^2t^2$$ and hence $$v = krt$$.

Differentiating this velocity with respect to time gives the tangential acceleration $$a_t = \frac{dv}{dt} = kr$$, so the corresponding tangential force is $$F_t = ma_t = mkr$$.

The power delivered by this force is $$P = F_t \times v = mkr \times krt = mk^2r^2t$$. Since the centripetal force does no work (it is perpendicular to the velocity), only the tangential force contributes to the power. Therefore, the power delivered to the particle is $$mk^2r^2t$$.

The correct answer is Option C.

Here 4 graphs are given and were asked to be arranged in descending order of work done

We know that work done in an F-x graph is equal to the area under the curve.

• Area above x-axis → positive work
• Area below x-axis → negative work
• Net work = (positive area − negative area)

Graph (1);-

Here work done $$W_1$$ is area of this graph which is combination of 2 triangles 

One from $$0\ to\ x_0$$ with a Force of -F
Another from $$x_0\ to\ x_1$$ with a Force of F

Here both are triangles, so area is

 $$Net\ Area\ =\ W_1\ =\ \ \frac{\ 1}{2}\times\ -F\times\ \left(x_0-0\right)\ \ +\ \ \frac{\ 1}{2}\times\ F\times\ \left(x_1-x_0\right)$$

$$Net\ Area\ =\ W_1\ =\ \ -\frac{\ 1}{2}Fx_0\ \ +\ \ \frac{\ 1}{2}\ F\ \left(x_1-x_0\right)$$

Graph (2);-

Here work done $$W_2$$ is area of this graph which is combination of 2 triangles and one rectangle

One triangle from $$0\ to\ x_0$$ with a Force of -F
Another triangle from $$x_0\ to\ x_1$$ with a Force of F
Rectangle from $$x_1to\ x_2$$ with a Force of F

So area is

$$Net\ Area\ =\ W_2\ =\ \ \frac{\ 1}{2}\times\ -F\times\ \left(x_0-0\right)\ \ +\ \ \frac{\ 1}{2}\times\ F\times\ \left(x_1-x_0\right)\ +\ F\times\ \left(x_2-x_1\right)$$

$$Net\ Area\ =\ W_2\ =\ \ -\frac{\ 1}{2}Fx_0\ \ +\ \ \frac{\ 1}{2}\ F\ \left(x_1-x_0\right)\ +\ F\left(x_2-x_1\right)$$

Graph (3);-

Here work done $$W_3$$ is area of this graph which is combination of 3  triangles and one rectangle

First triangle from $$0\ to\ x_0$$ with a Force of -F
Second from $$x_0\ to\ x_1$$ with a Force of F
Third from $$x_2\ to\ x_3$$ with a Force of F
Rectangle from $$x_1to\ x_2$$ with a Force of F


So area is,

$$Net\ Area\ =\ W_3\ =\ \ \frac{\ 1}{2}\times\ -F\times\ \left(x_0-0\right)\ \ +\ \ \frac{\ 1}{2}\times\ F\times\ \left(x_1-x_0\right)\ +\ F\times\ \left(x_2-x_1\right)\ +\ \ \frac{\ 1}{2}\times\ F\times\ \left(x_3-x_2\right)$$

$$Net\ Area\ =\ W_3\ =\ \ -\frac{\ 1}{2}Fx_0\ \ +\ \ \frac{\ 1}{2}\ F\ \left(x_1-x_0\right)\ +\ F\left(x_2-x_1\right)\ +\ \ \frac{\ 1}{2}\ F\ \left(x_3-x_2\right)$$

Graph (4);-

Here work done $$W_4$$ is area of this graph which is combination of 3 triangles and one rectangle

One from $$0\ to\ x_0$$ with a Force of F
Another from $$x_0\ to\ x_1$$ with a Force of -F
Rectangle from $$x_1to\ x_2$$ with a Force of -F
Another from $$x_2\ to\ x_3$$ with a Force of F

So area is,

$$Net\ Area\ =\ W_4\ =\ \ \frac{\ 1}{2}\times\ F\times\ \left(x_0-0\right)\ \ +\ \ \frac{\ 1}{2}\times\ -F\times\ \left(x_1-x_0\right)\ +\ -F\times\ \left(x_2-x_1\right)\ +\ \ \frac{\ 1}{2}\times\ F\times\ \left(x_3-x_2\right)$$

$$Net\ Area\ =\ W_4\ =\ \ \frac{\ 1}{2}\ F\ \left(x_0-0\right)\ \ \ -\ \ \ \frac{\ 1}{2}\ F\ \left(x_1-x_0\right)\ +\ -F\ \left(x_2-x_1\right)\ +\ \ \frac{\ 1}{2}\ F\ \left(x_3-x_2\right)$$

A bullet of mass $$m = 0.2 \text{ kg}$$ moving at $$v = 10 \text{ m s}^{-1}$$ embeds in a sandbag of mass $$M = 9.8 \text{ kg}$$. This is a perfectly inelastic collision.

Conservation of momentum yields:

$$mv = (m + M)V$$

$$0.2 \times 10 = (0.2 + 9.8) \times V$$

$$2 = 10V$$

$$V = 0.2 \text{ m s}^{-1}$$

We now calculate the kinetic energies before and after the collision.

Initial kinetic energy (only the bullet is moving):

$$KE_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.2 \times 10^2 = 10 \text{ J}$$

Final kinetic energy (combined system):

$$KE_f = \frac{1}{2}(m + M)V^2 = \frac{1}{2} \times 10 \times (0.2)^2 = \frac{1}{2} \times 10 \times 0.04 = 0.2 \text{ J}$$

The loss in kinetic energy is therefore:

$$\Delta KE = KE_i - KE_f = 10 - 0.2 = 9.8 \text{ J}$$

Hence, the correct answer is Option B: $$9.8 \text{ J}$$.

We are given a body of mass $$m = 0.5 \text{ kg}$$ with velocity $$v = (3x^2 + 4) \text{ m/s}$$, and we need to find the net work done as it moves from $$x = 0$$ to $$x = 2 \text{ m}$$.

By the Work-Energy Theorem, the net work done equals the change in kinetic energy, so we have $$W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)$$.

At the initial position $$x = 0$$ the velocity is $$v_i = 3(0)^2 + 4 = 4 \text{ m/s}$$, and at the final position $$x = 2$$ it becomes $$v_f = 3(2)^2 + 4 = 12 + 4 = 16 \text{ m/s}$$.

Substituting these values into the work expression yields $$W = \frac{1}{2} \times 0.5 \times (16^2 - 4^2)$$, which simplifies to $$W = 0.25 \times (256 - 16)$$ and then to $$W = 0.25 \times 240$$, giving $$W = 60 \text{ J}$$.

Hence, the correct answer is Option B.

Two bodies of masses $$8 \text{ kg}$$ and $$2 \text{kg}$$ have equal kinetic energy and we need to find the ratio of their momenta.

Kinetic energy can be expressed in terms of momentum as $$KE = \dfrac{p^2}{2m}$$. Rearranging gives $$p = \sqrt{2m \cdot KE}$$.

Since $$KE_1 = KE_2 = KE$$, substituting yields $$p_1 = \sqrt{2 \times 8 \times KE} = \sqrt{16 \cdot KE}$$ and $$p_2 = \sqrt{2 \times 2 \times KE} = \sqrt{4 \cdot KE}$$.

From this, $$\dfrac{p_1}{p_2} = \dfrac{\sqrt{16 \cdot KE}}{\sqrt{4 \cdot KE}} = \sqrt{\dfrac{16}{4}} = \sqrt{4} = 2$$ so that $$p_1 : p_2 = 2 : 1$$.

The correct answer is Option B: $$2:1$$.

Given the potential energy function $$U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$$ where $$A$$ and $$B$$ are positive constants, the force is the negative derivative of potential energy with respect to $$r$$, so $$F = -\frac{dU}{dr}$$. Computing the derivative gives $$\frac{dU}{dr} = \frac{d}{dr}\left(\frac{A}{r^{10}}\right) - \frac{d}{dr}\left(\frac{B}{r^{5}}\right) = -\frac{10A}{r^{11}} + \frac{5B}{r^{6}}$$. Therefore, $$F = -\frac{dU}{dr} = \frac{10A}{r^{11}} - \frac{5B}{r^{6}}$$.

At equilibrium, the net force is zero ($$F = 0$$), leading to $$\frac{10A}{r^{11}} = \frac{5B}{r^{6}}$$. Multiplying both sides by $$r^{11}$$ yields $$10A = 5B \cdot r^{5}$$, so $$r^{5} = \frac{10A}{5B} = \frac{2A}{B}$$ and hence $$r = \left(\frac{2A}{B}\right)^{\frac{1}{5}}$$. The equilibrium distance between the two atoms is $$\left(\frac{2A}{B}\right)^{1/5}$$. The correct answer is Option C.

Sand is dropped from a stationary dropper at a rate of $$\frac{dm}{dt} = 0.5 \text{ kg s}^{-1}$$ onto a conveyor belt moving with velocity $$v = 5 \text{ m/s}$$. When the sand falls on the belt it has zero horizontal velocity, so the belt must accelerate each sand particle from 0 to $$v = 5$$ m/s, requiring a continuous force to maintain the belt speed against the momentum gained by the falling sand.

The rate of change of momentum of the sand gives the force required: $$F = v \cdot \frac{dm}{dt}$$ because each unit of sand gains momentum $$v \cdot dm$$ in time $$dt$$. Substituting the values, $$F = 5 \times 0.5 = 2.5 \text{ N}.$$

The power supplied by the motor is the force multiplied by the belt’s velocity: $$P = F \times v = v^2 \cdot \frac{dm}{dt}$$ so $$P = (5)^2 \times 0.5 = 25 \times 0.5 = 12.5 \text{ W}.$$

Note: The kinetic energy gained by the sand per second is $$\tfrac{1}{2}\,\frac{dm}{dt}\,v^2 = \tfrac{1}{2} \times 0.5 \times 25 = 6.25\text{ W},$$ and the remaining $$12.5 - 6.25 = 6.25\text{ W}$$ is dissipated as heat due to friction between the sand and the belt. The total power supplied by the motor must therefore be $$12.5\text{ W}$$ to account for both the kinetic energy gain and the frictional losses.

The correct answer is Option D: $$12.5 \text{ W}$$.

Water falls from a height $$h = 40$$ m at a rate of $$\frac{dm}{dt} = 9 \times 10^4$$ kg per hour. 50% of gravitational potential energy is converted to electrical energy.

Calculate the gravitational potential energy released per hour.

$$PE = mgh = 9 \times 10^4 \times 10 \times 40 = 3.6 \times 10^7 \text{ J per hour}$$

Convert to power (energy per second).

$$P_{total} = \frac{3.6 \times 10^7}{3600} = 10000 \text{ W} = 10^4 \text{ W}$$

Calculate the electrical power available (50% efficiency).

$$P_{electrical} = 0.5 \times 10^4 = 5000 \text{ W}$$

Find the number of 100 W lamps that can be lit.

$$n = \frac{5000}{100} = 50$$

The correct answer is Option B.

We are given: $$\vec{F} = 4x\hat{i} + 3y^2\hat{j}$$ N. The particle moves from point $$(1, 2)$$ to point $$(2, 3)$$.

Apply the Work-Energy theorem: the change in kinetic energy equals the work done by the force:

$$ \Delta KE = W = \int \vec{F} \cdot d\vec{r} $$

Check if the force is conservative: $$\frac{\partial F_x}{\partial y} = \frac{\partial (4x)}{\partial y} = 0$$ and $$\frac{\partial F_y}{\partial x} = \frac{\partial (3y^2)}{\partial x} = 0$$.

Since $$\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$$, the force is conservative, and the work done is path-independent.

Calculate the work done: $$ W = \int_1^2 4x\,dx + \int_2^3 3y^2\,dy $$

$$ W = \left[2x^2\right]_1^2 + \left[y^3\right]_2^3 $$

$$ W = (2 \times 4 - 2 \times 1) + (27 - 8) $$

$$ W = (8 - 2) + (19) = 6 + 19 = 25 \text{ J} $$

Therefore, the kinetic energy changes by $$25$$ J.

The correct answer is Option A.

We have the kinetic energy expressed in terms of momentum as $$K = \frac{p^2}{2m}$$.

If the momentum is increased by 20%, the new momentum is $$p' = 1.2\,p$$. The new kinetic energy is $$K' = \frac{(1.2\,p)^2}{2m} = \frac{1.44\,p^2}{2m} = 1.44\,K$$.

The percentage increase in kinetic energy is $$(1.44 - 1) \times 100\% = 44\%$$.

Hence, the correct answer is Option 3.

A stone tied to a string of length $$L$$ is whirled in a vertical circle. At the lowest position, it has speed $$u$$. We need to find the magnitude of change in velocity when the string becomes horizontal.

Find the speed at the horizontal position.

Using energy conservation from the lowest point to the point where the string is horizontal (height = $$L$$):

$$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$$

$$v^2 = u^2 - 2gL$$

Find the change in velocity vector.

At the lowest point, the velocity is horizontal (say, to the right): $$\vec{v_1} = u\hat{i}$$

At the horizontal position (string horizontal means the stone is at the same height as the center), the velocity is directed vertically (upward or downward). Since the stone moves in a circle, at the 90° position the velocity is vertical: $$\vec{v_2} = v\hat{j}$$

Calculate $$|\Delta \vec{v}|$$.

$$|\Delta \vec{v}| = |\vec{v_2} - \vec{v_1}| = \sqrt{u^2 + v^2}$$

Substituting $$v^2 = u^2 - 2gL$$:

$$|\Delta \vec{v}| = \sqrt{u^2 + u^2 - 2gL} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$$

Comparing with $$\sqrt{x(u^2 - gL)}$$, we get $$x = 2$$.

The correct answer is Option A.

First, we note that a particle of mass $$m$$ moving with velocity $$u$$ strikes a stationary particle of mass $$5m$$ in a head-on elastic collision, and we wish to find the percentage of kinetic energy transferred.

Next, we apply conservation of momentum, which gives $$mu = mv_1 + 5mv_2.$$ Dividing by $$m$$ yields $$u = v_1 + 5v_2 \quad \cdots(1).$$

Since the collision is elastic, conservation of kinetic energy implies $$\tfrac{1}{2}mu^2 = \tfrac{1}{2}mv_1^2 + \tfrac{1}{2}(5m)v_2^2.$$ Dividing by $$\tfrac{1}{2}m$$ leads to $$u^2 = v_1^2 + 5v_2^2 \quad \cdots(2).$$

Now, using the coefficient of restitution condition for a perfectly elastic collision, the relative velocity of separation equals the relative velocity of approach: $$v_2 - v_1 = u \quad \cdots(3).$$

Substituting from equation (3) into equation (1) allows us to solve for the velocities. Since $$v_1 = v_2 - u$$, equation (1) becomes $$u = (v_2 - u) + 5v_2 = 6v_2 - u,$$ so $$2u = 6v_2$$ and therefore $$v_2 = \frac{u}{3}.$$ Then equation (3) gives $$v_1 = \frac{u}{3} - u = \frac{-2u}{3}.$$

Next, we calculate the kinetic energy transferred to the initially stationary particle of mass $$5m$$. The initial kinetic energy of the system is $$KE_i = \frac{1}{2}mu^2,$$ and the final kinetic energy of the $$5m$$ particle is $$KE_2 = \frac{1}{2}(5m)\left(\frac{u}{3}\right)^2 = \frac{1}{2}(5m)\left(\frac{u^2}{9}\right) = \frac{5mu^2}{18}.$$

Therefore, the percentage of kinetic energy transferred is $$\text{Percentage} = \frac{KE_2}{KE_i} \times 100 = \frac{\frac{5mu^2}{18}}{\frac{mu^2}{2}} \times 100 = \frac{5}{18} \times 2 \times 100 = \frac{10}{18} \times 100 = \frac{5}{9} \times 100 \approx 55.6\%.$$ Thus, the correct answer is Option C.

The condition $$\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$$ gives $$AB\cos\theta = AB\sin\theta$$, where $$\theta$$ is the angle between the two vectors. Dividing both sides by $$AB\cos\theta$$ yields $$\tan\theta = 1$$, so $$\theta = 45°$$.

The magnitude of the difference of two vectors is given by $$|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2\vec{A}\cdot\vec{B} = A^2 + B^2 - 2AB\cos\theta$$.

Substituting $$\theta = 45°$$, so $$\cos 45° = \frac{1}{\sqrt{2}}$$, we get $$|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2AB \cdot \frac{1}{\sqrt{2}} = A^2 + B^2 - \sqrt{2}\,AB$$.

Therefore, $$|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - \sqrt{2}\,AB}$$.

We need to determine the maximum compression in the spring after block $$C$$ collides with block $$A$$.

1. Analyze Phase 1: The Collision

From the layout shown on the, block $$C$$ of mass $$m$$ moving with velocity $$v$$ makes a head-on collision with an identical stationary block $$A$$ (mass $$m$$).

Assuming the collision is perfectly elastic (standard for such idealized physics problems):

• When two identical masses collide elastically, they exchange their velocities.
• Therefore, immediately after the collision, block $$C$$ comes to a complete rest ($$v_C = 0$$).
• Block $$A$$ instantly acquires the entire initial velocity of block $$C$$, starting its motion with velocity $$v_A = v$$. Block $$B$$ remains stationary ($$v_B = 0$$) at this exact moment.

2. Analyze Phase 2: Maximum Compression of the Spring

As block $$A$$ moves to the right, it begins to compress the spring connecting it to block $$B$$. This compression exerts a forward force on block $$B$$ (speeding it up) and a backward force on block $$A$$ (slowing it down).

The spring reaches its maximum compression ($$x_{\text{max}}$$) at the exact instant when both blocks $$A$$ and $$B$$ are moving with the exact same common velocity ($$V_c$$). At this point, relative motion between them temporarily stops.

• Conservation of Linear Momentum:
  Since the floor is smooth and no external horizontal forces act on the $$A-B$$ spring system, momentum is conserved:

  $$m \cdot v = (m + m) \cdot V_c$$

  $$m v = 2m V_c \implies V_c = \frac{v}{2}$$

• Conservation of Mechanical Energy:
  The total kinetic energy of block $$A$$ immediately after the collision is converted into the combined kinetic energy of both blocks plus the elastic potential energy stored in the compressed spring:

  $$\frac{1}{2} m v^2 = \frac{1}{2} (2m) V_c^2 + \frac{1}{2} K x_{\text{max}}^2$$

3. Solve for Maximum Compression ($$x_{\text{max}}$$)

Substitute $$V_c = \frac{v}{2}$$ into the energy conservation equation:

$$\frac{1}{2} m v^2 = \frac{1}{2} (2m) \left(\frac{v}{2}\right)^2 + \frac{1}{2} K x_{\text{max}}^2$$

Multiply the entire equation by 2 to clear the fractions:

$$m v^2 = 2m \left(\frac{v^2}{4}\right) + K x_{\text{max}}^2$$

$$m v^2 = \frac{1}{2} m v^2 + K x_{\text{max}}^2$$

Isolate the spring term:

$$K x_{\text{max}}^2 = m v^2 - \frac{1}{2} m v^2$$

$$K x_{\text{max}}^2 = \frac{1}{2} m v^2$$

$$x_{\text{max}}^2 = \frac{m v^2}{2K}$$

Taking the square root gives us the maximum compression:

$$x_{\text{max}} = v \sqrt{\frac{m}{2K}}$$

Conclusion

The maximum compression in the spring is $$v \sqrt{\frac{m}{2K}}$$, which corresponds to Option A.

We start with the definition of mechanical work done by a variable force. In vector form, the small (differential) work done $$dW$$ when the particle is displaced by an element $$d\vec r$$ is given by the dot-product formula

$$dW \;=\; \vec F \,\cdot\, d\vec r.$$

To obtain the total work $$W$$ along a path, we integrate this expression:

$$W \;=\; \displaystyle\int \vec F \,\cdot\, d\vec r.$$

In the present situation the force is purely in the $$\hat{j}$$ (i.e., $$y$$) direction and has the magnitude

$$\vec F \;=\; (5y + 20)\,\hat{j}\;\text{N}.$$

The particle is moved only along the $$y$$-axis, from the initial point $$y = 0\;\text{m}$$ to the final point $$y = 10\;\text{m}$$. Hence the displacement element is purely $$d\vec r = dy\,\hat{j}$$. The dot product of the force with this displacement is therefore simply the product of their magnitudes because both vectors are parallel:

$$\vec F \,\cdot\, d\vec r = (5y + 20)\,\hat{j}\;\cdot\; dy\,\hat{j} = (5y + 20)\,dy.$$

Now substitute this result into the integral for work:

$$W = \displaystyle\int_{y=0}^{y=10} (5y + 20)\,dy.$$

We split the integral into two separate, easier integrals:

$$W = \int_{0}^{10} 5y\,dy \;+\; \int_{0}^{10} 20\,dy.$$

Let us evaluate each part one by one.

For the first integral, we use the power rule $$\int y\,dy = \dfrac{y^{2}}{2}\,.$$ Hence

$$\int_{0}^{10} 5y\,dy \;=\; 5 \left[\dfrac{y^{2}}{2}\right]_{0}^{10} = 5 \left(\dfrac{10^{2}}{2} - \dfrac{0^{2}}{2}\right) = 5 \left(\dfrac{100}{2}\right) = 5 \times 50 = 250 \;\text{J}.$$

For the second integral, $$\int 20\,dy = 20y$$. So

$$\int_{0}^{10} 20\,dy \;=\; 20\,[y]_{0}^{10} = 20\,(10 - 0) = 20 \times 10 = 200 \;\text{J}.$$

Finally, we add the two contributions to obtain the total work:

$$W = 250 \;\text{J} + 200 \;\text{J} = 450 \;\text{J}.$$

So, the answer is $$450$$.

We are given that $$\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$$.

From the properties of the cross product, we know that $$\vec{Q} \times \vec{P} = -(\vec{P} \times \vec{Q})$$. Substituting this into the given equation: $$\vec{P} \times \vec{Q} = -(\vec{P} \times \vec{Q})$$.

This gives $$2(\vec{P} \times \vec{Q}) = \vec{0}$$, which means $$\vec{P} \times \vec{Q} = \vec{0}$$.

The cross product $$\vec{P} \times \vec{Q} = |\vec{P}||\vec{Q}|\sin\theta \; \hat{n}$$, where $$\theta$$ is the angle between the two vectors. For this to be zero (assuming neither vector is a zero vector), we need $$\sin\theta = 0$$.

In the range $$0° < \theta < 360°$$, $$\sin\theta = 0$$ at $$\theta = 180°$$. (Note: $$\theta = 0°$$ is excluded by the given constraint.)

Therefore, the angle between $$\vec{P}$$ and $$\vec{Q}$$ is $$\theta = 180°$$.

The kinetic energy of a body can be expressed in terms of its linear momentum $$p$$ as $$K.E. = \frac{p^2}{2m}$$, since $$p = mv$$ gives $$K.E. = \frac{1}{2}mv^2 = \frac{p^2}{2m}$$.

Given that solids $$A$$ and $$B$$ have equal linear momentum, i.e., $$p_A = p_B = p$$, their kinetic energies are $$(K.E.)_A = \frac{p^2}{2m_A} = \frac{p^2}{2 \times 1} = \frac{p^2}{2}$$ and $$(K.E.)_B = \frac{p^2}{2m_B} = \frac{p^2}{2 \times 2} = \frac{p^2}{4}$$.

The ratio of their kinetic energies is $$\frac{(K.E.)_A}{(K.E.)_B} = \frac{p^2/2}{p^2/4} = \frac{4}{2} = \frac{2}{1}$$.

Since the ratio is given as $$\frac{A}{1}$$, the value of $$A$$ is $$2$$.

The correct answer is 2.

Let us denote the original (total) mass of the block by $$m$$ and its initial speed by $$u = 40\ \text{m s}^{-1}$$. Since the surface is smooth, there is no external horizontal force, so the linear momentum of the system will be conserved at the instant the block splits.

After splitting, the block forms two equal parts, each of mass $$\dfrac{m}{2}$$. One part is observed to move in the same original direction with speed $$60\ \text{m s}^{-1}$$. We shall call the speed of the other part $$v$$ (also in the original direction, because nothing in the statement suggests reversal).

Conservation of linear momentum states:

$$\text{Initial momentum} = \text{Final total momentum}.$$

Mathematically,

$$m\,u = \frac{m}{2}\,(60) + \frac{m}{2}\,v.$$

Substituting $$u = 40\ \text{m s}^{-1}$$, we have

$$m\,(40) = \frac{m}{2}\,(60) + \frac{m}{2}\,v.$$

Cancelling the common factor $$m$$ on both sides:

$$40 = 30 + \frac{v}{2}.$$

Re-arranging,

$$40 - 30 = \frac{v}{2}\quad\Longrightarrow\quad 10 = \frac{v}{2}.$$

So,

$$v = 20\ \text{m s}^{-1}.$$

Now we evaluate the kinetic energies.

The initial kinetic energy of the single block is given by the familiar formula $$K = \dfrac12 m u^{2}$$:

$$K_{\text{initial}} = \frac12\,m\,(40)^{2} = \frac12\,m\,(1600) = 800\,m.$$

The final kinetic energy is the sum of the kinetic energies of the two fragments:

$$K_{\text{final}} = \frac12\left(\frac{m}{2}\right)(60)^{2} + \frac12\left(\frac{m}{2}\right)(20)^{2}.$$

Simplifying step by step,

$$K_{\text{final}} = \frac{m}{4}\bigl(60^{2} + 20^{2}\bigr) = \frac{m}{4}\bigl(3600 + 400\bigr) = \frac{m}{4}\,(4000) = 1000\,m.$$

The change in kinetic energy is therefore

$$\Delta K = K_{\text{final}} - K_{\text{initial}} = 1000\,m - 800\,m = 200\,m.$$

The fractional change (ratio of the change to the original value) is

$$\frac{\Delta K}{K_{\text{initial}}} = \frac{200\,m}{800\,m} = \frac14.$$

The problem states that this fractional change can be written in the form $$x : 4$$. Comparing, we identify

$$x = 1.$$

So, the answer is $$1$$.

1. Conservation of Energy

As the block slides from the top of the hemisphere (radius $$R = 3\text{ m}$$) down to a height $$h$$, it drops a vertical distance of $$R - h$$. Equating its loss in potential energy to its gain in kinetic energy gives:

$$mg(R - h) = \frac{1}{2}mv^2 \implies v^2 = 2g(R - h)$$

2. Condition for Losing Contact

The centripetal force equation along the radial direction is:

$$mg\cos\theta - N = \frac{mv^2}{R}$$

The block leaves the surface when the normal force drops to zero ($$N = 0$$). Substituting $$\cos\theta = \frac{h}{R}$$ yields:

$$mg\left(\frac{h}{R}\right) = \frac{mv^2}{R} \implies v^2 = gh$$

3. Solve for Height ($$h$$)

Equating the two expressions for $$v^2$$:

$$2g(R - h) = gh$$

$$2R - 2h = h \implies 3h = 2R \implies h = \frac{2}{3}R$$

Plugging in $$R = 3\text{ m}$$:

$$h = \frac{2}{3} \times 3 = 2\text{ m}$$

Conclusion

The block will lose contact with the hemisphere at a height of 2 m.

We have a uniform chain of total length $$L = 3\ \text{m}$$ and total mass $$M = 3\ \text{kg}$$. Because the chain is uniform, its linear mass density is given by the formula $$\lambda = \dfrac{M}{L}$$. Substituting the given values, $$\lambda = \dfrac{3\ \text{kg}}{3\ \text{m}} = 1\ \text{kg m}^{-1}$$.

At the beginning, $$2\ \text{m}$$ of the chain rests on the smooth table and $$1\ \text{m}$$ overhangs. The mass that is hanging is therefore $$m_{\text{hang}} = \lambda \times 1\ \text{m} = 1\ \text{kg}$$, while the mass on the table is $$m_{\text{table}} = \lambda \times 2\ \text{m} = 2\ \text{kg}$$.

We choose the level of the tabletop as our reference level for gravitational potential energy (GPE). Hence any element of chain that is below the tabletop has negative potential energy.

Initial gravitational potential energy
Only the hanging part contributes because the portion on the table is at the reference level. For a uniform hanging segment of length $$1\ \text{m}$$, its centre of mass lies halfway down, i.e. at a vertical distance $$h_i = 0.5\ \text{m}$$ below the table. The GPE of this part is obtained from the standard formula $$U = m g h$$. Since the height is downward (negative), $$h = -0.5\ \text{m}$$. Thus

$$U_i = m_{\text{hang}}\; g\; h_i = (1\ \text{kg})(10\ \text{m s}^{-2})(-0.5\ \text{m}) = -5\ \text{J}.$$

The portion on the table has $$h = 0$$, so its GPE is zero. Therefore the total initial potential energy of the chain is $$U_i = -5\ \text{J}$$.

Final gravitational potential energy
When the chain has completely slipped off, the whole length $$3\ \text{m}$$ hangs vertically. For any uniform body, the centre of mass lies at its midpoint. Hence the centre of mass is now at a distance $$h_f = 1.5\ \text{m}$$ below the tabletop, i.e. $$h_f = -1.5\ \text{m}$$. Applying the same formula,

$$U_f = M g h_f = (3\ \text{kg})(10\ \text{m s}^{-2})(-1.5\ \text{m}) = -45\ \text{J}.$$

Change in potential energy
The change is

$$\Delta U = U_f - U_i = (-45\ \text{J}) - (-5\ \text{J}) = -40\ \text{J}.$$

The negative sign means the chain has lost $$40\ \text{J}$$ of gravitational potential energy.

The table is smooth, so there is no friction and hence no non-conservative work. By conservation of mechanical energy, the loss in potential energy equals the gain in kinetic energy. Therefore,

$$K = -\Delta U = 40\ \text{J}.$$

Hence, the correct answer is Option 40.

First, we note that the only mechanical energy initially present in the engine-wagon system is its translational kinetic energy. The mass of the entire system is given as $$m = 40\,000\ \text{kg}$$ and its initial speed is stated as $$72\ \text{km h}^{-1}$$. We convert this speed into the SI unit metres per second:

$$72\ \text{km h}^{-1} = 72 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 20\ \text{m s}^{-1}.$$

Now we write the formula for kinetic energy:

$$\text{Kinetic energy} = \frac12 m v^{2}.$$

Substituting the numerical values, we obtain

$$\frac12 \times 40\,000\ \text{kg} \times (20\ \text{m s}^{-1})^{2} = \frac12 \times 40\,000 \times 400 = 20\,000 \times 400 = 8\,000\,000\ \text{J}.$$

Thus the initial kinetic energy is $$8.0 \times 10^{6}\ \text{J}.$$

The problem statement tells us that, while the brakes are applied, 90 % of this energy is dissipated as heat and other losses due to friction. Therefore only 10 % of the original kinetic energy remains to be stored as elastic potential energy in the spring (shock absorber) when it is fully compressed.

We calculate this remaining energy:

$$E_{\text{spring}} = 0.10 \times 8.0 \times 10^{6}\ \text{J} = 0.8 \times 10^{6}\ \text{J} = 8.0 \times 10^{5}\ \text{J}.$$

The spring is stated to compress by $$x = 1.0\ \text{m}$$. The formula for elastic potential energy stored in a spring is

$$E_{\text{spring}} = \frac12 k x^{2},$$

where $$k$$ is the spring (force) constant that we need to determine. We already know $$E_{\text{spring}} = 8.0 \times 10^{5}\ \text{J}$$ and $$x = 1.0\ \text{m}$$. Inserting these into the formula gives

$$8.0 \times 10^{5} = \frac12 k (1.0)^{2}.$$

Solving for $$k$$, we multiply both sides by 2:

$$k = 2 \times 8.0 \times 10^{5} = 1.6 \times 10^{6}\ \text{N m}^{-1}.$$

Finally, we express this value in the form asked for in the question, namely as a multiple of $$10^{5}\ \text{N m}^{-1}$$:

$$k = 16 \times 10^{5}\ \text{N m}^{-1}.$$

Hence, the correct answer is Option 16.

The potential energy of the diatomic molecule is given by $$U = \frac{\alpha}{r^{10}} - \frac{\beta}{r^5} - 3$$.

At equilibrium, the net force on the molecule is zero, which means $$\frac{dU}{dr} = 0$$.

Differentiating with respect to $$r$$:

$$\frac{dU}{dr} = -\frac{10\alpha}{r^{11}} + \frac{5\beta}{r^6} = 0$$

$$\frac{10\alpha}{r^{11}} = \frac{5\beta}{r^6}$$

$$\frac{2\alpha}{r^5} = \beta$$

$$r^5 = \frac{2\alpha}{\beta}$$

$$r = \left(\frac{2\alpha}{\beta}\right)^{\frac{1}{5}}$$

Comparing with the given form $$\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$$, where $$\frac{a}{b} = \frac{1}{5}$$, we get $$a = 1$$.

The answer is $$a = 1$$.

We are given two particles with masses $$m_1 = 4$$ g and $$m_2 = 16$$ g moving with equal kinetic energies. We need to find the ratio of their linear momenta.

The kinetic energy of a particle is related to its momentum by $$K = \frac{p^2}{2m}$$, which gives $$p = \sqrt{2mK}$$.

Since both particles have equal kinetic energies ($$K_1 = K_2 = K$$):

$$p_1 = \sqrt{2m_1 K} = \sqrt{2 \times 4 \times K} = \sqrt{8K}$$

$$p_2 = \sqrt{2m_2 K} = \sqrt{2 \times 16 \times K} = \sqrt{32K}$$

The ratio of their momenta is: $$\frac{p_1}{p_2} = \frac{\sqrt{8K}}{\sqrt{32K}} = \sqrt{\frac{8}{32}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$.

So $$p_1 : p_2 = 1 : 2$$. Comparing with the given ratio $$n : 2$$, we get $$n = 1$$.

We know that “work” is defined as the dot product of force and displacement. Mathematically, the magnitude of work done by a force $$\vec F$$ in producing a displacement $$\vec d$$ is given by the formula

$$W = F\,d \,\cos\theta,$$

where $$F$$ is the magnitude of the force, $$d$$ is the magnitude of the displacement and $$\theta$$ is the angle between the directions of $$\vec F$$ and $$\vec d$$.

Person $$A$$ applies a force $$F_A$$ at an angle $$45^\circ$$ with the displacement. Therefore the work done by $$A$$ is

$$W_A \;=\; F_A \, d \, \cos 45^\circ.$$

Person $$B$$ applies a force $$F_B$$ at an angle $$60^\circ$$ with the displacement. Hence the work done by $$B$$ is

$$W_B \;=\; F_B \, d \, \cos 60^\circ.$$

According to the statement of the problem, both persons perform the same amount of work in moving the body through the same distance $$d$$. So we equate the two expressions:

$$W_A \;=\; W_B.$$

Substituting the detailed expressions for $$W_A$$ and $$W_B$$, we get

$$F_A \, d \, \cos 45^\circ \;=\; F_B \, d \, \cos 60^\circ.$$

The distance $$d$$ is common on both sides, so we cancel it out:

$$F_A \,\cos 45^\circ \;=\; F_B \,\cos 60^\circ.$$

Now we isolate the required ratio $$\dfrac{F_A}{F_B}$$ by dividing both sides by $$F_B \cos 45^\circ$$:

$$\frac{F_A}{F_B} \;=\; \frac{\cos 60^\circ}{\cos 45^\circ}.$$

We recall the standard cosine values:

$$\cos 60^\circ = \frac{1}{2}, \qquad \cos 45^\circ = \frac{1}{\sqrt{2}}.$$

Substituting these numerical values, we obtain

$$\frac{F_A}{F_B} \;=\; \frac{\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}}.$$

To simplify the complex fraction, we multiply numerator and denominator appropriately:

$$\frac{F_A}{F_B} \;=\; \frac{1}{2} \times \frac{\sqrt{2}}{1} \;=\; \frac{\sqrt{2}}{2}.$$

Notice that $$\dfrac{\sqrt{2}}{2}$$ can also be written as $$\dfrac{1}{\sqrt{2}}$$, because

$$\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}.$$

The problem statement expresses the ratio in the form $$\dfrac{1}{\sqrt{x}}$$, so by direct comparison we identify

$$x = 2.$$

Hence, the correct answer is Option 2.

When the ball collides with the spring and compresses it maximally, all the kinetic energy of the ball is converted into the potential energy stored in the spring. At maximum compression, the ball momentarily comes to rest.

Using energy conservation: $$\frac{1}{2}mv^2 = \frac{1}{2}kx_c^2$$, where $$x_c$$ is the compression of the spring. Substituting the given values: $$\frac{1}{2}(4)(10)^2 = \frac{1}{2}(100)x_c^2$$, which gives $$200 = 50 x_c^2$$, so $$x_c^2 = 4$$ and $$x_c = 2$$ m.

The natural length of the spring is 8 m. After compression by 2 m, the length of the compressed spring is $$x = 8 - 2 = 6$$ m.

The value of $$x$$ is $$6$$.

Let the mass of the pendulum bob be $$m$$ and the length of the string be $$L = 0.50\ \text{m}$$. At its lowest position the bob has speed $$v_0 = 3\ \text{m s}^{-1}$$. We take the gravitational potential energy to be zero at this lowest point.

The principle we use is the conservation of mechanical energy, which states:

$$\text{Total mechanical energy at any position} = \text{constant.}$$

Hence,

$$\tfrac12 m v_0^{\,2} + 0 = \tfrac12 m v^{\,2} + m g h,$$

where

• $$v$$ is the speed of the bob when the string makes an angle $$\theta = 60^\circ$$ with the vertical,
• $$h$$ is the vertical height gained by the bob relative to the lowest point.

The height $$h$$ is obtained from simple geometry of the circle traced by the pendulum. When the string of length $$L$$ makes an angle $$\theta$$ with the vertical, the bob rises from the lowest point by

$$h = L - L\cos\theta = L(1 - \cos\theta).$$

Substituting $$L = 0.50\ \text{m}$$ and $$\theta = 60^\circ$$, and recalling that $$\cos 60^\circ = \tfrac12$$, we get

$$h = 0.50(1 - \tfrac12) = 0.50 \times 0.50 = 0.25\ \text{m}.$$

Now we substitute $$h$$ into the energy conservation equation:

$$\tfrac12 m (3)^{2} = \tfrac12 m v^{\,2} + m(10)(0.25).$$

Simplifying term by term, first compute the left side kinetic energy:

$$\tfrac12 m (3)^{2} = \tfrac12 m \times 9 = 4.5 m.$$

The potential energy term is

$$m g h = m \times 10 \times 0.25 = 2.5 m.$$

So the equation becomes

$$4.5 m = \tfrac12 m v^{\,2} + 2.5 m.$$

Subtract $$2.5 m$$ from both sides:

$$4.5 m - 2.5 m = \tfrac12 m v^{\,2}.$$

That gives

$$2.0 m = \tfrac12 m v^{\,2}.$$

Divide both sides by $$\tfrac12 m$$ (which is equivalent to multiplying by 2):

$$v^{\,2} = 4.$$

Finally, take the positive square root (speed is positive):

$$v = 2\ \text{m s}^{-1}.$$

So, the answer is $$2\ \text{m s}^{-1}$$.

When a constant power $$P$$ is delivered to a body of mass $$m$$ starting from rest, the work-energy theorem tells us that all power goes into kinetic energy: $$P = \frac{d}{dt}\!\left(\frac{1}{2}mv^2\right) = mv\frac{dv}{dt}$$.

Rearranging: $$mv\,dv = P\,dt$$. Integrating the left side from $$v=0$$ to $$v$$ and the right side from $$0$$ to $$t$$: $$\int_0^v mv'\,dv' = \int_0^t P\,dt'$$, which gives $$\frac{1}{2}mv^2 = Pt$$.

So the velocity at time $$t$$ is $$v = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$$, i.e., $$v \propto t^{1/2}$$.

The distance $$x$$ is found by integrating velocity: $$x = \int_0^t v\,dt' = \int_0^t \sqrt{\frac{2P}{m}}\,(t')^{1/2}\,dt' = \sqrt{\frac{2P}{m}} \cdot \frac{(t')^{3/2}}{3/2}\Bigg|_0^t = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3}\,t^{3/2}$$.

Therefore $$x \propto t^{3/2}$$.

The initial kinetic energy of the ball is $$K_i = \frac{1}{2}mv^2 = \frac{1}{2}(0.5)(20)^2 = 100$$ J. After deflection the ball retains 5% of this energy, so $$K_f = 0.05 \times 100 = 5$$ J.

Using $$K_f = \frac{1}{2}mv_f^2$$, we get $$5 = \frac{1}{2}(0.5)v_f^2$$, which gives $$v_f^2 = 20$$, so $$v_f = \sqrt{20} = 2\sqrt{5} \approx 4.47$$ m/s.

This is closest to 4.4 m s$$^{-1}$$, so the correct answer is option 2.

Let the original mass of the single block be $$M$$ and its initial speed be $$u = 40\ \text{m s}^{-1}$$. The initial kinetic energy is given by the formula $$K = \dfrac12 m v^{2}$$, so we have

$$K_i = \dfrac12\,M\,u^{2} = \dfrac12\,M\,(40)^{2} = \dfrac12\,M\,(1600) = 800\,M.$$

The block breaks into two pieces whose masses are in the ratio $$1:2$$. We denote the smaller mass by $$m$$ and the larger mass by $$2m$$. Because the total mass must remain the same,

$$m + 2m = 3m = M \;\;\Longrightarrow\;\; m = \dfrac{M}{3}, \qquad 2m = \dfrac{2M}{3}.$$

The problem states that the smaller piece moves forward with speed $$v_1 = 60\ \text{m s}^{-1}$$. We still have to find the speed $$v_2$$ of the larger piece.

The surface is smooth, so no external horizontal force acts on the system; therefore linear momentum is conserved. The law of conservation of momentum says

$$\text{(total momentum before)} = \text{(total momentum after)}.$$

This gives

$$M\,u = m\,v_1 + (2m)\,v_2.$$

Substituting $$u = 40\ \text{m s}^{-1}$$, $$m = \dfrac{M}{3}$$ and $$v_1 = 60\ \text{m s}^{-1}$$, we obtain

$$M(40) = \dfrac{M}{3}(60) + \dfrac{2M}{3}\,v_2.$$

Dividing every term by $$M$$ to cancel the common factor,

$$40 = \dfrac{60}{3} + \dfrac{2}{3}\,v_2.$$

Since $$\dfrac{60}{3} = 20$$, we have

$$40 = 20 + \dfrac{2}{3}\,v_2 \;\;\Longrightarrow\;\; 40 - 20 = \dfrac{2}{3}\,v_2 \;\;\Longrightarrow\;\; 20 = \dfrac{2}{3}\,v_2.$$

Multiplying both sides by $$\dfrac{3}{2}$$ to isolate $$v_2$$, we find

$$v_2 = 20\left(\dfrac{3}{2}\right) = 30\ \text{m s}^{-1}.$$

Now we calculate the kinetic energy after the split. Using $$K = \dfrac12 m v^{2}$$ for each part,

$$\begin{aligned} K_f &= \dfrac12\,m\,v_1^{2} + \dfrac12\,(2m)\,v_2^{2} \\ &= \dfrac12\left(\dfrac{M}{3}\right)(60)^{2} + \dfrac12\left(\dfrac{2M}{3}\right)(30)^{2}. \end{aligned}$$

Simplifying each term one at a time:

First term:

$$\dfrac12\left(\dfrac{M}{3}\right)(60)^{2} = \dfrac{M}{6}\,(3600) = 600\,M.$$

Second term:

$$\dfrac12\left(\dfrac{2M}{3}\right)(30)^{2} = \dfrac{M}{3}\,(900) = 300\,M.$$

Adding these contributions gives

$$K_f = 600\,M + 300\,M = 900\,M.$$

The change in kinetic energy is

$$\Delta K = K_f - K_i = 900\,M - 800\,M = 100\,M.$$

The fractional change in kinetic energy is the ratio $$\dfrac{\Delta K}{K_i}$$, so

$$\dfrac{\Delta K}{K_i} = \dfrac{100\,M}{800\,M} = \dfrac{1}{8}.$$

Hence, the correct answer is Option D.

We need to determine the initial speed of a bullet just before it collides with and becomes embedded in a suspended wooden block, causing the combined system to swing upward.

1. Identify the System Parameters

From the problem statement, we have:

• Mass of the wooden block ($$M$$) = $$5.99\text{ kg}$$
• Mass of the bullet ($$m$$) = $$10\text{ g} = 0.01\text{ kg}$$
• Combined mass of the system ($$M + m$$) = $$5.99 + 0.01 = 6.00\text{ kg}$$
• Vertical height raised ($$h$$) = $$9.8\text{ cm} = 0.098\text{ m}$$
• Acceleration due to gravity ($$g$$) = $$9.8\text{ m s}^{-2}$$

2. Analyze Phase 2: Swing Upward (Conservation of Mechanical Energy)

Immediately after the collision, the combined block-bullet system moves with a common velocity $$V$$. As it swings upward to its maximum height $$h$$, its kinetic energy is completely converted into gravitational potential energy:

$$\frac{1}{2}(M + m)V^2 = (M + m)gh$$

Canceling the combined mass from both sides and solving for $$V$$:

$$V = \sqrt{2gh}$$

Substitute the given numerical values:

$$V = \sqrt{2 \times 9.8 \times 0.098} = \sqrt{2 \times 9.8 \times \frac{9.8}{100}} = \sqrt{\frac{19.6 \times 9.8}{100}} = \sqrt{\frac{1.9208}{1}} = 1.386\text{ m s}^{-1}$$

Alternatively, written more cleanly: $$V = \sqrt{\frac{2 \times 98 \times 98}{10000}} = \frac{98}{100}\sqrt{2} = 0.98 \times 1.4142 \approx 1.386\text{ m s}^{-1}$$

3. Analyze Phase 1: The Collision (Conservation of Linear Momentum)

Since the collision happens almost instantaneously, external forces like tension do not change the horizontal momentum. Let $$u$$ be the initial speed of the bullet just before impact:

$$m \cdot u = (M + m) \cdot V$$

Isolating the initial bullet speed $$u$$:

$$u = \frac{M + m}{m} \cdot V$$

Substitute the mass values and the calculated velocity $$V$$ into the formula:

$$u = \frac{6.00}{0.01} \times 1.386$$

$$u = 600 \times 1.3859 \approx 831.54\text{ m s}^{-1}$$

Using the unrounded exact fractions yields exactly:

$$u = 600 \times 0.98\sqrt{2} = 588\sqrt{2} \approx 588 \times 1.414213 = 831.55\text{ m s}^{-1}$$

This matches closely with the standard rounded evaluation of $$831.4\text{ m s}^{-1}$$.

Conclusion

The speed of the bullet just before the collision is approximately 831.4 m s⁻¹, which corresponds to Option C.

The porter lowers the suitcase with constant velocity, meaning the net force is zero. We need to find the work done by the porter (i.e., the work done by the force applied by the porter).

The suitcase has mass $$m = 80 \text{ kg}$$ and is lowered by a distance $$d = 80 \text{ cm} = 0.80 \text{ m}$$.

Since the suitcase moves down with constant velocity, the porter applies an upward force equal in magnitude to the weight: $$F_{\text{porter}} = mg = 80 \times 9.8 = 784 \text{ N (upward)}$$

The displacement is downward ($$d = 0.80 \text{ m}$$ downward), while the force is upward. Since force and displacement are in opposite directions, the work done by the porter is negative: $$W_{\text{porter}} = -F \cdot d = -(784)(0.80) = -627.2 \text{ J}$$

The work done by the porter in lowering the suitcase is $$-627.2 \text{ J}$$.

Initially the body is at rest at a height $$h$$ above the ground, so its kinetic energy is zero and its gravitational potential energy is $$mgh$$.

At the moment of release we declare this to be the initial state. Let us write the energy-work relation that includes non-conservative forces (air friction):

$$U_{\text i}+K_{\text i}+W_{\text{air}}=U_{\text f}+K_{\text f}.$$

Here $$U$$ denotes gravitational potential energy, $$K$$ denotes kinetic energy and $$W_{\text{air}}$$ is the work done by air friction (which we need to find).

For the initial state we have

$$U_{\text i}=mgh,\qquad K_{\text i}=0.$$

For the final state (just before touching the ground) the height is zero, so

$$U_{\text f}=0.$$

The speed on reaching the ground is given to be $$0.8\sqrt{gh}$$. Hence the final kinetic energy is

$$K_{\text f}=\dfrac12 m\bigl(0.8\sqrt{gh}\bigr)^2 =\dfrac12 m\,(0.64\,gh) =0.32\,mgh.$$

Substituting every quantity into the energy-work relation gives

$$mgh+0+W_{\text{air}}=0+0.32\,mgh.$$

Now we isolate $$W_{\text{air}}$$:

$$W_{\text{air}}=0.32\,mgh-mgh =-0.68\,mgh.$$

The negative sign tells us that air friction has removed mechanical energy from the system.

Hence, the correct answer is Option A.

We begin by recalling the definition of mechanical power. When a force $$F$$ moves an object with instantaneous speed $$v$$, the power delivered is

$$P = F\,v.$$

For the automobile of mass $$m$$, the only horizontal force producing acceleration is the net engine force. Newton’s second law gives

$$F = m\,a = m\,\frac{dv}{dt}.$$

Substituting this value of $$F$$ in the power expression, we get

$$P = \left(m\frac{dv}{dt}\right)v.$$

So

$$m\,v\,\frac{dv}{dt} = P.$$

To separate the variables, multiply both sides by $$dt$$:

$$m\,v\,dv = P\,dt.$$

Now integrate each side. At $$t = 0$$ the car is at rest, so the lower limit for velocity is $$0$$.

$$\int_{0}^{v} m\,v\,dv = \int_{0}^{t} P\,dt.$$

Carry out the integrations:

$$m\left[\frac{v^{2}}{2}\right]_{0}^{v} = P\,[t]_{0}^{t}.$$

So

$$\frac{m\,v^{2}}{2} = P\,t.$$

Solving for $$v^{2}$$ gives

$$v^{2} = \frac{2P}{m}\,t.$$

Taking the positive square root (speed is non-negative), we obtain the velocity as a function of time:

$$v = \sqrt{\frac{2P}{m}}\;t^{1/2}.$$

Next, we use the relationship between velocity and position, namely

$$v = \frac{dx}{dt}.$$

Substituting the expression for $$v$$ just found, we have

$$\frac{dx}{dt} = \sqrt{\frac{2P}{m}}\;t^{1/2}.$$

Separate the variables:

$$dx = \sqrt{\frac{2P}{m}}\;t^{1/2}\,dt.$$

Integrate again, with the initial condition that the car starts from the origin $$x = 0$$ at $$t = 0$$:

$$\int_{0}^{x} dx = \sqrt{\frac{2P}{m}}\int_{0}^{t} t^{1/2}\,dt.$$

The left integral simply yields $$x$$. For the right side, recall the power rule of integration, $$\int t^{n}\,dt = \frac{t^{\,n+1}}{n+1}$$. Here $$n = \tfrac{1}{2}$$, so

$$\int t^{1/2}\,dt = \frac{t^{3/2}}{3/2} = \frac{2}{3}t^{3/2}.$$

Therefore,

$$x = \sqrt{\frac{2P}{m}}\;\left(\frac{2}{3}\,t^{3/2}\right).$$

Simplify the coefficient. The numerical factor becomes $$\frac{2}{3}$$, and under the radical we can combine constants:

$$\frac{2}{3}\sqrt{\frac{2P}{m}} = \sqrt{\frac{4}{9}}\;\sqrt{\frac{2P}{m}} = \sqrt{\frac{8P}{9m}}.$$

Thus the position as a function of time is finally

$$x(t) = \left(\frac{8P}{9m}\right)^{\frac{1}{2}}\,t^{\frac{3}{2}}.$$

This matches Option D.

Hence, the correct answer is Option 4.

Kinetic energy and momentum are related by $$KE = \frac{p^2}{2m}$$, so $$p = \sqrt{2m \cdot KE}$$, meaning $$p \propto \sqrt{KE}$$.

If the kinetic energy becomes 4 times its initial value, i.e., $$KE_f = 4\,KE_i$$, then $$p_f = \sqrt{4\,KE_i \cdot 2m} = 2\sqrt{2m \cdot KE_i} = 2p_i$$.

The percentage change in momentum is $$\frac{p_f - p_i}{p_i} \times 100 = \frac{2p_i - p_i}{p_i} \times 100 = 100\%$$.

We begin by realising that when water falls from the top of the falls to the bottom, the gravitational potential energy it loses is converted almost entirely into internal (thermal) energy, and this causes its temperature to rise. Hence we equate the loss in potential energy to the heat gained by the water.

The loss of gravitational potential energy for a mass $$m$$ falling through a height $$h$$ is given by the well-known expression

$$\text{Potential energy lost}=mgh,$$

where $$g$$ is the acceleration due to gravity.

We take a convenient mass of water, say $$m=1\ \text{g}$$. To use SI units in the energy calculation we write this as

$$m=1\ \text{g}=0.001\ \text{kg}.$$

The height of Victoria’s Falls is given as

$$h = 63\ \text{m},$$

and we use the standard value

$$g = 9.8\ \text{m s}^{-2}.$$

Substituting these values into the potential-energy formula, we obtain

$$\begin{aligned} \text{Potential energy lost} &= m g h \\ &= (0.001\ \text{kg})(9.8\ \text{m s}^{-2})(63\ \text{m}) \\ &= 0.6174\ \text{J}. \end{aligned}$$

Now, this energy becomes heat $$Q$$ absorbed by the same mass of water. To find the corresponding heat in calories, we recall the conversion factor stated in the question,

$$1\ \text{cal} = 4.2\ \text{J}.$$

Hence

$$\begin{aligned} Q &= \frac{0.6174\ \text{J}}{4.2\ \text{J cal}^{-1}} \\ &= 0.147\ \text{cal}. \end{aligned}$$

Next, we relate this heat to the rise in temperature using the definition of specific heat capacity. For a substance of mass $$m$$, specific heat $$c$$, and temperature change $$\Delta T$$, the heat absorbed is

$$Q = m c \Delta T.$$

The specific heat of water is given as

$$c = 1\text{ cal g}^{-1}\,^{\circ}\text{C}^{-1}$$

and we are still considering $$m = 1\ \text{g}$$ of water. Substituting these values, we get

$$\begin{aligned} 0.147\ \text{cal} &= (1\ \text{g})(1\ \text{cal g}^{-1}\,^{\circ}\text{C}^{-1})\Delta T \\ \Delta T &= 0.147\ ^{\circ}\text{C}. \end{aligned}$$

Thus the water becomes warmer by approximately $$0.147\ ^{\circ}\text{C}$$ between the top and the bottom of the falls.

Hence, the correct answer is Option C.

First, we translate the given horse‐power of the motor into the standard SI unit of power, the watt. The relation stated in the question is $$1\text{ HP}=746\text{ W}.$$ We have a $$60\text{ HP}$$ motor, so the available mechanical power is $$P = 60 \times 746\ \text{W}.$$

Carrying out the multiplication step by step, $$60 \times 700 = 42000,$$ $$60 \times 40 = 2400,$$ $$60 \times 6 = 360,$$ and adding these partial products, $$42000 + 2400 + 360 = 44760.$$ So, $$P = 44760\ \text{W}.$$

Next we examine all the forces that the motor must overcome while lifting the elevator at steady speed. There are two such forces:

1. The gravitational weight of the fully loaded elevator. With a maximum mass $$m = 2000\ \text{kg}$$ and taking the acceleration due to gravity as $$g = 10\ \text{m s}^{-2},$$ the weight is $$W = m g = 2000 \times 10 = 20000\ \text{N}.$$

2. The additional constant frictional force acting downward, given directly as $$F_{\text{fr}} = 4000\ \text{N}.$$

Since both forces oppose the upward motion, the motor must supply enough power to counter their combined effect. Therefore the total opposing or effective force is $$F_{\text{total}} = W + F_{\text{fr}} = 20000 + 4000 = 24000\ \text{N}.$$

For uniform upward motion, the mechanical power needed is related to force and speed by the general formula $$P = F v,$$ where $$v$$ is the (constant) speed of the elevator. Solving this formula for speed gives $$v = \frac{P}{F}.$$

Substituting the numerical values we have determined, $$v = \frac{44760\ \text{W}}{24000\ \text{N}}.$$

Performing the division carefully, we first note that both numerator and denominator have the same factor of 1000, so we can simplify: $$\frac{44760}{24000} = \frac{4476}{2400}.$$ Now dividing, $$\frac{4476}{2400} \approx 1.865.$$ Rounding to two significant figures that match the precision of the data, $$v \approx 1.9\ \text{m s}^{-1}.$$

Hence, the correct answer is Option B.

We begin by noting that the motor has to do two kinds of work while the elevator moves upward with uniform speed:

1. It must balance the gravitational pull on the combined mass of the elevator and all the passengers.
2. It must overcome the constant frictional force that opposes the motion.

The useful relation to find the power of the motor is the mechanical power formula

$$P = F \, v$$

where $$P$$ is the power delivered, $$F$$ is the total force that the motor must supply, and $$v$$ is the constant speed of the elevator.

Now we calculate each quantity step by step.

Step 1: Mass of passengers.
The problem states that the elevator is carrying its full capacity of 10 persons, each of average mass 68 kg. Hence

$$m_{\text{persons}} = 10 \times 68 \text{ kg} = 680 \text{ kg}.$$

Step 2: Mass of the empty elevator.
This is given directly as

$$m_{\text{elevator}} = 920 \text{ kg}.$$

Step 3: Total mass being lifted.
Adding the two masses, we obtain

$$m_{\text{total}} = m_{\text{persons}} + m_{\text{elevator}} = 680 \text{ kg} + 920 \text{ kg} = 1600 \text{ kg}.$$

Step 4: Gravitational force on this mass.
Using the standard relation $$\text{Weight} = m g$$ with $$g = 10 \text{ m/s}^2$$, we get

$$F_{\text{gravity}} = m_{\text{total}} \, g = 1600 \text{ kg} \times 10 \text{ m/s}^2 = 16000 \text{ N}.$$

Step 5: Frictional force.
The friction opposing the upward motion is specified as

$$F_{\text{friction}} = 6000 \text{ N}.$$

Step 6: Total opposing force.
Since the motor must counter both gravity and friction simultaneously, we add the two forces:

$$F_{\text{total}} = F_{\text{gravity}} + F_{\text{friction}} = 16000 \text{ N} + 6000 \text{ N} = 22000 \text{ N}.$$

Step 7: Constant speed of the elevator.
The elevator moves upward at

$$v = 3 \text{ m/s}.$$

Step 8: Required power of the motor.
Substituting $$F_{\text{total}}$$ and $$v$$ into $$P = F v$$, we obtain

$$ P = F_{\text{total}} \times v = 22000 \text{ N} \times 3 \text{ m/s} = 66000 \text{ W}. $$

Thus, the motor must supply at least $$66\,000 \text{ watts}$$ of power while lifting the fully-loaded elevator at the stated speed.

Hence, the correct answer is Option D.

We are told that the engine supplies a constant power $$P = 1\,\text{J s}^{-1}$$ to a body of mass $$m = 2\,\text{kg}$$. The body starts from rest, so its initial velocity is $$v_0 = 0$$. Power is defined as the rate at which work is done, and, for translational motion in a straight line, we have the relation

$$P \;=\; \frac{dW}{dt} \;=\; F\,v$$

Because the force responsible for the motion provides the acceleration, we can rewrite the force in terms of mass and acceleration: $$F = m\,a$$. Substituting this into the power expression gives

$$P \;=\; (m\,a)\,v \;=\; m\,a\,v$$

The acceleration $$a$$ is the time derivative of velocity, $$a = \dfrac{dv}{dt}$$. Replacing $$a$$ by $$\dfrac{dv}{dt}$$, we obtain

$$P \;=\; m\,v\,\frac{dv}{dt}$$

All the quantities in this differential equation are separable, so we rearrange to isolate the variables:

$$v\,dv \;=\; \frac{P}{m}\,dt$$

Now we integrate both sides from the initial state (velocity $$0$$ at time $$0$$) to some general state (velocity $$v$$ at time $$t$$):

$$\int_{0}^{v} v\,dv \;=\; \frac{P}{m}\int_{0}^{t} dt$$

Evaluating the integrals, we get

$$\frac{1}{2}v^{2} \;=\; \frac{P}{m}\,t$$

Multiplying by $$2$$ yields an explicit expression for the square of the velocity:

$$v^{2} \;=\; 2\,\frac{P}{m}\,t$$

Taking the positive square root (because the body moves forward), we find the velocity as a function of time:

$$v(t) \;=\; \sqrt{\frac{2P}{m}\,t}$$

The displacement $$s$$ in time $$t$$ is the time integral of the velocity:

$$s \;=\; \int_{0}^{t} v(t')\,dt' \;=\; \int_{0}^{t} \sqrt{\frac{2P}{m}\,t'}\,dt'$$

We factor out the constants to simplify the integral:

$$s \;=\; \sqrt{\frac{2P}{m}}\;\int_{0}^{t} (t')^{1/2}\,dt'$$

We now integrate the power-law function. The integral of $$t'^{1/2}$$ is $$\tfrac{2}{3}t'^{3/2}$$. Substituting the bounds gives

$$s \;=\; \sqrt{\frac{2P}{m}}\;\left[\frac{2}{3}\,t^{3/2}\right]$$

This can be rewritten more compactly as

$$s \;=\; \frac{2}{3}\,\sqrt{\frac{2P}{m}}\;t^{3/2}$$

We now substitute the numerical values $$P = 1\,\text{J s}^{-1}$$, $$m = 2\,\text{kg}$$ and $$t = 9\,\text{s}$$.

First, calculate the square-root factor:

$$\sqrt{\frac{2P}{m}} \;=\; \sqrt{\frac{2 \times 1}{2}} \;=\; \sqrt{1} \;=\; 1$$

With this, the expression for $$s$$ simplifies to

$$s \;=\; \frac{2}{3}\;t^{3/2}$$

Next, evaluate $$t^{3/2}$$ for $$t = 9\,\text{s}$$. We note that $$9^{1/2} = 3$$, so

$$9^{3/2} \;=\; (9^{1/2})^{3} \;=\; 3^{3} \;=\; 27$$

Substituting this value, we find

$$s \;=\; \frac{2}{3}\times 27 \;=\; 18$$

Therefore, after $$9\,\text{s}$$ the body has travelled a distance of $$18\,\text{m}$$.

So, the answer is $$18\,\text{m}$$.

The mass of the cricket ball is given as $$m = 0.15 \text{ kg}$$ and the ball finally rises to a maximum height of $$h = 20 \text{ m}$$ after leaving the machine. We first find the speed with which the ball must leave the machine.

According to the principle of conservation of mechanical energy, the kinetic energy with which the ball leaves the machine is completely converted into gravitational potential energy at the highest point. Writing this idea mathematically, we state the formulae:

Potential energy at height $$h$$: $$U = mgh$$

Kinetic energy at launch: $$K = \dfrac12 m v^2$$

Since all of the kinetic energy changes into potential energy, we have

$$\dfrac12 m v^2 = m g h.$$

Dividing both sides by $$m$$ (because $$m \neq 0$$) gives

$$\dfrac12 v^2 = g h.$$

Multiplying both sides by $$2$$, we obtain

$$v^2 = 2 g h.$$

Now we substitute the numerical values $$g = 10 \text{ m s}^{-2}$$ and $$h = 20 \text{ m}$$:

$$v^2 = 2 \times 10 \times 20 = 400.$$

Taking the square root, we get the speed at launch:

$$v = \sqrt{400} = 20 \text{ m s}^{-1}.$$

Next, we relate this speed to the force that the bowling machine exerts. While the ball is in contact with the machine, it travels a distance $$s = 0.2 \text{ m}$$ under the action of a constant force $$F$$. The work-energy theorem tells us that the work done by this force equals the change in kinetic energy of the ball:

Work done by the force: $$W = F s.$$

Change in kinetic energy (initially the ball is at rest): $$\Delta K = \dfrac12 m v^2 - 0.$$

Equating the two, we write

$$F s = \dfrac12 m v^2.$$

Solving for $$F$$, we have

$$F = \dfrac{\dfrac12 m v^2}{s}.$$

Substituting $$m = 0.15 \text{ kg}$$, $$v = 20 \text{ m s}^{-1}$$, and $$s = 0.2 \text{ m}$$, we get

$$F = \dfrac{\dfrac12 \times 0.15 \times (20)^2}{0.2}.$$

First compute the numerator:

$$\dfrac12 \times 0.15 = 0.075,$$ $$0.075 \times (20)^2 = 0.075 \times 400 = 30.$$

Now divide by the distance $$s$$:

$$F = \dfrac{30}{0.2} = 150 \text{ N}.$$

So, the answer is $$150$$.

We begin by recalling the energy principle for a charged particle moving through an electric potential difference. The work done by the electric field is converted into kinetic energy. Mathematically we state the relation

$$\text{Work done} \;=\; qV \;=\; \tfrac12 m v^{2},$$

where $$q$$ is the charge on the particle, $$V$$ is the potential difference, $$m$$ is the mass of the particle and $$v$$ is the final speed acquired from rest.

We now apply this relation separately to the hydrogen ion (proton) and to the singly ionised helium atom.

For the hydrogen ion (symbolically $$\text{H}^{+}$$):

Its charge is that of one proton, so $$q_{\text{H}} = +e$$. We denote its mass by $$m_{\text{H}}$$. Substituting these into the energy equation we have

$$q_{\text{H}} V = \tfrac12 m_{\text{H}} v_{\text{H}}^{2}.$$

Explicitly, that reads

$$e\,V = \tfrac12\,m_{\text{H}}\,v_{\text{H}}^{2}.$$

Solving for the speed $$v_{\text{H}}$$ gives

$$v_{\text{H}}^{2} = \frac{2eV}{m_{\text{H}}}, \qquad\text{so}\qquad v_{\text{H}} = \sqrt{\frac{2eV}{m_{\text{H}}}}.$$

For the singly ionised helium atom (symbolically $$\text{He}^{+}$$):

Although the helium nucleus contains two protons, the phrase “singly ionised” means it has lost only one electron, so its net charge is still just one elementary charge. Hence

$$q_{\text{He}} = +e.$$

The mass of a helium nucleus is essentially four times the proton mass, so we write

$$m_{\text{He}} = 4\,m_{\text{H}}.$$

Inserting these values into the same energy equation gives

$$q_{\text{He}} V = \tfrac12 m_{\text{He}} v_{\text{He}}^{2}$$

$$e\,V = \tfrac12 \left(4m_{\text{H}}\right) v_{\text{He}}^{2}.$$

Rearranging to isolate $$v_{\text{He}}^{2}$$:

$$v_{\text{He}}^{2} = \frac{2eV}{4m_{\text{H}}} = \frac{1}{4}\,\frac{2eV}{m_{\text{H}}},$$

and therefore

$$v_{\text{He}} = \sqrt{\frac{1}{4}\,\frac{2eV}{m_{\text{H}}}} = \frac{1}{2}\,\sqrt{\frac{2eV}{m_{\text{H}}}}.$$

Computing the ratio of the two speeds:

$$\frac{v_{\text{H}}}{v_{\text{He}}} = \frac{\sqrt{\dfrac{2eV}{m_{\text{H}}}}} {\dfrac{1}{2}\sqrt{\dfrac{2eV}{m_{\text{H}}}}} = \frac{\sqrt{\dfrac{2eV}{m_{\text{H}}}}}{\sqrt{\dfrac{2eV}{m_{\text{H}}}}/2} = 2.$$

Thus,

$$v_{\text{H}} : v_{\text{He}} = 2 : 1.$$

Hence, the correct answer is Option C.

We have two particles, each of mass $$m$$. Their initial velocity vectors are given as

$$\vec{u}_1 = u\,\hat{i}$$

and

$$\vec{u}_2 = u\left(\frac{\hat{i}+\hat{j}}{2}\right).$$

It is convenient to write the second velocity in component form. Distributing the factor $$u$$ inside the bracket, we obtain

$$\vec{u}_2 = \frac{u}{2}\,\hat{i} + \frac{u}{2}\,\hat{j}.$$

Because the collision is completely inelastic, the two particles stick together. Momentum is conserved, so we equate the total initial momentum to the total final momentum.

Total initial momentum:

$$\vec{p}_{\text{initial}} = m\vec{u}_1 + m\vec{u}_2 = m\left(u\,\hat{i}\right) + m\left(\frac{u}{2}\,\hat{i} + \frac{u}{2}\,\hat{j}\right) = mu\left(1 + \frac12\right)\hat{i} + mu\left(\frac12\right)\hat{j} = \frac{3}{2}mu\,\hat{i} + \frac{1}{2}mu\,\hat{j}.$$

Let the common final velocity after sticking together be $$\vec{v}$$, and note that the combined mass is $$2m$$. Conservation of linear momentum gives

$$\frac{3}{2}mu\,\hat{i} + \frac{1}{2}mu\,\hat{j} = 2m\,\vec{v}.$$

Dividing both sides by $$2m$$, we obtain the final velocity vector:

$$\vec{v} = \frac{3}{4}u\,\hat{i} + \frac{1}{4}u\,\hat{j}.$$

Next we compute the initial kinetic energy. The formula for translational kinetic energy is $$K = \frac12 m v^2$$, where $$v$$ is the magnitude of the velocity.

For particle 1, the speed is simply $$u$$, so

$$K_{1i} = \frac12 m u^2.$$

For particle 2, the magnitude of its velocity is

$$\left|\vec{u}_2\right| = \sqrt{\left(\frac{u}{2}\right)^2 + \left(\frac{u}{2}\right)^2} = \sqrt{\frac{u^2}{4} + \frac{u^2}{4}} = \sqrt{\frac{u^2}{2}} = \frac{u}{\sqrt{2}}.$$

Therefore

$$K_{2i} = \frac12 m \left(\frac{u}{\sqrt{2}}\right)^2 = \frac12 m \left(\frac{u^2}{2}\right) = \frac14 m u^2.$$

The total initial kinetic energy is then

$$K_i = K_{1i} + K_{2i} = \frac12 m u^2 + \frac14 m u^2 = \frac34 m u^2.$$

We now calculate the final kinetic energy of the combined mass. First we need the magnitude squared of the final velocity:

$$|\vec{v}|^2 = \left(\frac{3u}{4}\right)^2 + \left(\frac{u}{4}\right)^2 = \frac{9u^2}{16} + \frac{u^2}{16} = \frac{10u^2}{16} = \frac{5u^2}{8}.$$

The final kinetic energy (with total mass $$2m$$) is

$$K_f = \frac12 (2m) |\vec{v}|^2 = m\,|\vec{v}|^2 = m\left(\frac{5u^2}{8}\right) = \frac{5}{8} m u^2.$$

The loss in kinetic energy during the collision is the difference between the initial and final values:

$$\Delta K = K_i - K_f = \frac34 m u^2 - \frac58 m u^2.$$ Converting the fractions to a common denominator, $$\frac34 = \frac68,$$ so

$$\Delta K = \frac68 m u^2 - \frac58 m u^2 = \frac{1}{8} m u^2.$$

Thus the energy lost in the completely inelastic collision is $$\frac{1}{8}mu^2$$.

Hence, the correct answer is Option B.

We have a block of mass $$M = 1.9\ \text{kg}$$ initially at rest at the edge of a table of height $$h = 1\ \text{m}$$. A bullet of mass $$m = 0.1\ \text{kg}$$ comes horizontally with speed $$u = 20\ \text{m s}^{-1}$$ and sticks to the block. Because the bullet embeds itself, the collision is perfectly inelastic.

During the very short collision time, external horizontal forces are negligible, so we apply conservation of linear momentum in the horizontal direction. The formula is

$$\text{Initial momentum} = \text{Final momentum}.$$

Initially only the bullet moves, so

$$m u = (m + M)\,v,$$

where $$v$$ is the common horizontal velocity of the combined system immediately after the collision.

Substituting the numbers, we get

$$0.1 \times 20 = (0.1 + 1.9)\,v$$

$$2 = 2.0\,v$$

$$v = 1\ \text{m s}^{-1}.$$

This horizontal speed remains unchanged during the subsequent flight because air resistance is neglected.

Now the block-bullet system moves off the table edge with

horizontal speed $$v_x = 1\ \text{m s}^{-1},$$

vertical speed $$v_y = 0$$ at the moment it leaves the table, and falls through a vertical distance $$h = 1\ \text{m}$$ under gravity.

For the vertical motion, we use the kinematic relation for a body starting from rest:

$$v_y^2 = u_y^2 + 2 g h,$$

where $$u_y = 0$$ (no initial vertical speed). Hence

$$v_y^2 = 0 + 2 \times 10 \times 1 = 20,$$

so

$$v_y = \sqrt{20}\ \text{m s}^{-1} \approx 4.47\ \text{m s}^{-1}.$$

Thus, just before striking the floor, the system has two perpendicular components of velocity:

$$v_x = 1\ \text{m s}^{-1}, \quad v_y = \sqrt{20}\ \text{m s}^{-1}.$$

The magnitude of the resultant speed is obtained from Pythagoras, but for kinetic energy we can add the squares directly. The total kinetic energy is

$$ K = \dfrac{1}{2}\,(m + M)\,\left(v_x^2 + v_y^2\right). $$

Substituting,

$$ K = \dfrac{1}{2}\,(0.1 + 1.9)\,\left(1^2 + (\sqrt{20})^2\right) = \dfrac{1}{2}\times 2.0 \times (1 + 20) = 1 \times 21 = 21\ \text{J}. $$

Hence, the correct answer is Option A.

We first recall the definition of mechanical work. The work $$W$$ done by a variable or constant force acting along the line of motion is given by the formula

$$W = \int F\,\mathrm{d}x,$$

where $$F$$ is the instantaneous force and $$x$$ is the displacement. When the force remains constant over a segment, the integral reduces to the simple product of the constant force and the distance covered, $$W = F \times x$$. When the force varies linearly, the integral is numerically equal to the product of the average force and the distance.

According to the statement, the motion of the box can be divided into two successive parts:

(i) A first stretch of $$15\ \text{m}$$ during which the person keeps pushing with a constant force of $$200\ \text{N}$$.

(ii) A second and equal stretch of $$15\ \text{m}$$ during which the force decreases uniformly (linearly) from $$200\ \text{N}$$ down to $$100\ \text{N}$$ because the person is getting tired.

We now compute the work done in each part and then add the two contributions.

Work during the first 15 m

The force is constant, so we apply $$W = F \times x$$ directly:

$$W_1 = 200\ \text{N} \times 15\ \text{m} = 3000\ \text{J}.$$

Work during the next 15 m

Here the force varies linearly from $$200\ \text{N}$$ at the start of the segment to $$100\ \text{N}$$ at the end. For a linear variation, the average (mean) force $$F_{\text{avg}}$$ is simply the arithmetic mean:

$$F_{\text{avg}} = \dfrac{F_{\text{initial}} + F_{\text{final}}}{2} = \dfrac{200\ \text{N} + 100\ \text{N}}{2} = 150\ \text{N}.$$

Using again the constant-force formula with this average value gives the work for that segment:

$$W_2 = F_{\text{avg}} \times x = 150\ \text{N} \times 15\ \text{m} = 2250\ \text{J}.$$

Total work

Now we add the two pieces together:

$$W_{\text{total}} = W_1 + W_2 = 3000\ \text{J} + 2250\ \text{J} = 5250\ \text{J}.$$

Hence, the correct answer is Option D.

First, we translate every given quantity into standard SI units. The mass of the bullet is 20 g, and since $$1\ \text{kg}=1000\ \text{g}$$, we have

$$m = 20\ \text{g} = 20 \times 10^{-3}\ \text{kg}=0.02\ \text{kg}.$$

The initial speed of the bullet, just before entering the wall, is already given in metres per second:

$$u = 1\ \text{m s}^{-1}.$$

The thickness of the mud wall is 20 cm. Converting centimetres to metres (because $$1\ \text{m}=100\ \text{cm}$$):

$$s = 20\ \text{cm}=20 \times 10^{-2}\ \text{m}=0.20\ \text{m}.$$

The wall offers a mean resistive force of $$2.5 \times 10^{-2}\ \text{N}$$, so

$$F = 2.5 \times 10^{-2}\ \text{N}.$$

Now we invoke the Work-Energy Theorem, which states:

$$\text{Work done by all forces} = \text{Change in kinetic energy}.$$

The resistive force does negative work (it removes energy from the bullet). The magnitude of the work done by this constant resistive force while the bullet travels the full thickness of the wall is

$$W = F \, s.$$

Substituting the known values, we obtain

$$W = \left(2.5 \times 10^{-2}\ \text{N}\right)\left(0.20\ \text{m}\right) = 0.5 \times 10^{-2}\ \text{J} = 5.0 \times 10^{-3}\ \text{J}.$$

The bullet’s initial kinetic energy is found from the standard formula $$K = \tfrac12 m u^2$$:

$$K_{\text{initial}} = \frac12 (0.02\ \text{kg}) (1\ \text{m s}^{-1})^{2} = \frac12 (0.02)\,(1) = 0.01\ \text{J}.$$

Because the wall’s resistive force removes energy, the final kinetic energy of the bullet after it just emerges is

$$K_{\text{final}} = K_{\text{initial}} - W.$$

Substituting the numerical results,

$$K_{\text{final}} = 0.01\ \text{J} - 5.0 \times 10^{-3}\ \text{J} = 5.0 \times 10^{-3}\ \text{J}.$$

Let $$v$$ denote the bullet’s speed on emerging. Using the kinetic-energy relation once more,

$$K_{\text{final}} = \frac12 m v^{2}.$$

Therefore,

$$\frac12 m v^{2} = 5.0 \times 10^{-3}\ \text{J}.$$

Solving explicitly for $$v^{2}$$:

$$v^{2} = \frac{2 \times 5.0 \times 10^{-3}\ \text{J}}{0.02\ \text{kg}} = \frac{1.0 \times 10^{-2}}{0.02} = 0.50.$$

Taking the positive square root (speed is positive by definition),

$$v = \sqrt{0.50}\ \text{m s}^{-1} \approx 0.707\ \text{m s}^{-1}.$$

This value is most nearly $$0.7\ \text{m s}^{-1}$$ among the listed choices.

Hence, the correct answer is Option A.

According to the work-energy theorem,

$$\text{Work done}=\Delta KE$$

$$W=KE_{\text{final}}-KE_{\text{initial}}$$

The particle starts from rest, therefore

$$KE_{\text{initial}}=0$$

Work done is equal to the area under the F-x graph.

From the graph:

Rectangle area:

$$2\times2=4$$

Trapezium area:

$$\frac{1}{2}(2+3)\times1=2.5$$

Therefore,

$$W=4+2.5$$

$$W=6.5\ \text{J}$$

Hence,

$$KE_{\text{final}}=6.5\ \text{J}$$

$$\boxed{6.5\ \text{J}}$$

We begin with the bob of the simple pendulum. It is released from the angle $$\theta_0$$ and swings down to the lowest point. At the lowest point all the gravitational potential energy that the bob lost has been converted into kinetic energy.

The change in height of the bob, measured vertically from the lowest point, is obtained from the length $$l$$ of the pendulum. For a small angle $$\theta$$ we use the approximation $$\cos\theta \simeq 1-\dfrac{\theta^2}{2}$$, so

$$h = l - l\cos\theta \;\; \Longrightarrow \;\; h \simeq l\left(1-\Bigl[1-\frac{\theta^2}{2}\Bigr]\right)=\frac{l\theta^2}{2}.$$

Hence the speed of the bob just before impact is found from conservation of mechanical energy:

$$\frac{1}{2}m v_0^2 = m g h \;\; \Longrightarrow \;\; \frac{1}{2}m v_0^2 = m g \Bigl(\frac{l\theta_0^2}{2}\Bigr)$$

which simplifies to

$$v_0 = \sqrt{g l}\;\theta_0.$$ (Note that $$\sqrt{g l}$$ is the natural speed scale of a simple pendulum.)

The bob collides elastically with a block of mass $$M$$ that is initially at rest on a horizontal surface. For a perfectly elastic, one-dimensional collision we invoke the standard results (stated first):

For masses $$m$$ and $$M$$ with initial velocities $$u_m$$ and $$u_M$$, the velocities just after collision are $$ v_m = \frac{m-M}{m+M}\;u_m + \frac{2M}{m+M}\;u_M , \qquad v_M = \frac{2m}{m+M}\;u_m + \frac{M-m}{m+M}\;u_M . $$

Because the block is initially at rest, we put $$u_m = v_0$$ and $$u_M = 0$$, giving

$$ v_1 = \frac{m-M}{m+M}\;v_0, \qquad V = \frac{2m}{m+M}\;v_0, $$

where $$v_1$$ is the speed of the bob just after the collision (now moving upward) and $$V$$ is the speed of the block.

After the bounce the bob rises to an angle $$\theta_1$$ before momentarily coming to rest. Using the same energy argument as before, the kinetic energy at the lowest point converts back into gravitational potential energy at that maximum angle:

$$\frac{1}{2}m v_1^2 = m g \Bigl(\frac{l\theta_1^2}{2}\Bigr).$$

Solving for $$v_1$$ yields

$$v_1 = \sqrt{g l}\;\theta_1.$$

We now have two expressions for $$v_1$$, so we equate them:

$$\sqrt{g l}\;\theta_1 = \frac{m-M}{m+M}\;\Bigl(\sqrt{g l}\;\theta_0\Bigr).$$

Dividing both sides by $$\sqrt{g l}$$ gives a purely algebraic relation:

$$\theta_1 = \frac{m-M}{m+M}\;\theta_0.$$ Multiplying both sides by $$(m+M)$$ we obtain $$m\theta_1 + M\theta_1 = m\theta_0 - M\theta_0.$$

Gathering the terms that contain $$M$$ on the left and those that contain $$m$$ on the right:

$$M\theta_1 + M\theta_0 = m\theta_0 - m\theta_1.$$ Factoring the common symbols, $$M(\theta_1+\theta_0) = m(\theta_0-\theta_1).$$

Finally, solving for $$M$$ we divide both sides by $$(\theta_1+\theta_0)$$:

$$ M = m\;\frac{\theta_0 - \theta_1}{\theta_0 + \theta_1}. $$

This matches Option A.

Hence, the correct answer is Option A.

We have a body of mass $$m_1 = 1\;\text{kg}$$ that is released from rest at a height $$h = 100\;\text{m}$$ above a platform of mass $$m_2 = 3\;\text{kg}$$. The platform is fixed to the upper end of a vertical spring whose force constant is $$k = 1.25 \times 10^{6}\;\text{N\,m}^{-1}$$. During the motion the body falls, strikes the platform, sticks to it (perfectly inelastic impact) and then the combined system moves downward, compressing the spring.

The body of mass $$m_1$$ possesses an initial gravitational potential energy (taking the top of the spring as the reference level) equal to

$$U_{\text{g,\,initial}} \;=\; m_1 g h \;=\; 1 \times 10 \times 100 \;=\; 1000\;\text{J}. $$

After impact the two masses move together. Let $$x$$ be the maximum compression of the spring measured from its natural (unstretched) length. While the spring is being compressed, the centre of mass of the combined load $$M = m_1 + m_2 = 4\;\text{kg}$$ moves downward through the same distance $$x$$, so its gravitational potential energy decreases by

$$\Delta U_{\text{g,\,down}} \;=\; Mgx \;=\; 4 \times 10 \times x \;=\; 40x\;\text{J}. $$

At the instant of maximum compression the kinetic energy of the system has become zero, and the entire mechanical energy that was available has been stored as elastic potential energy of the spring. The elastic potential energy of a compressed spring is given by the well-known formula

$$U_{\text{spring}} \;=\; \frac12 k x^{2}. $$

Applying conservation of mechanical energy between the moment just before compression starts and the moment of maximum compression, we have

$$\underbrace{m_1 g h}_{1000} \;+\; \underbrace{Mgx}_{40x} \;=\; \underbrace{\frac12 k x^{2}}_{ \tfrac12 (1.25 \times 10^{6}) x^{2} }. $$

Substituting the numerical values,

$$1000 \;+\; 40x \;=\; \frac{1}{2}\bigl(1.25 \times 10^{6}\bigr)x^{2}. $$

Simplifying the right-hand side first,

$$\frac{1}{2}\bigl(1.25 \times 10^{6}\bigr) \;=\; 6.25 \times 10^{5},$$

so the equation becomes

$$6.25 \times 10^{5}\,x^{2} \;-\; 40x \;-\; 1000 \;=\; 0.$$

To solve the quadratic equation we divide every term by $$6.25 \times 10^{5}$$:

$$x^{2} \;-\; \frac{40}{6.25 \times 10^{5}}\,x \;-\; \frac{1000}{6.25 \times 10^{5}} \;=\; 0,$$

and evaluate the small fractions:

$$\frac{40}{6.25 \times 10^{5}} \;=\; 6.4 \times 10^{-5},\qquad \frac{1000}{6.25 \times 10^{5}} \;=\; 1.6 \times 10^{-3}.$$

Thus the quadratic is

$$x^{2} \;-\; 6.4 \times 10^{-5}x \;-\; 1.6 \times 10^{-3} \;=\; 0.$$

Using the quadratic-formula $$x = \dfrac{-b + \sqrt{\,b^{2} - 4ac\,}}{2a}$$ with $$a = 1,\; b = -6.4 \times 10^{-5},\; c = -1.6 \times 10^{-3},$$ we get

$$x \;=\; \frac{6.4 \times 10^{-5} \;+\; \sqrt{\bigl(6.4 \times 10^{-5}\bigr)^{2} + 4 \times 1.6 \times 10^{-3}}} {2}.$$

The term $$(6.4 \times 10^{-5})^{2}$$ is negligibly small, so we have approximately

$$x \;\approx\; \frac{6.4 \times 10^{-5} \;+\; \sqrt{0.0064}}{2} \;=\; \frac{6.4 \times 10^{-5} \;+\; 0.08}{2} \;=\; \frac{0.080064}{2} \;\approx\; 0.040\;\text{m}.$$

Converting metres to centimetres,

$$x \;\approx\; 0.040\;\text{m} = 4.0\;\text{cm}.$$

Hence, the correct answer is Option B.

Let us begin by noting that the cable is uniform, so its mass per unit length (linear mass density) is constant. We denote this density by $$\lambda$$ and write

$$\lambda \;=\;\frac{M}{L},$$

because the total mass $$M$$ is uniformly distributed along the total length $$L$$.

According to the statement, a fraction $$\left(\frac{1}{n}\right)^{\!th}$$ of the cable hangs below the horizontal surface. Hence the vertical length that is hanging is

$$\ell_h \;=\;\frac{L}{n}.$$

We choose the horizontal surface as the reference level $$y = 0$$. Positive $$y$$ is taken upward, so the hanging portion originally lies at negative values of $$y$$. Every small element of cable located at a distance $$y$$ below the surface (that is, at $$y$$ between $$-\,\frac{L}{n}$$ and $$0$$) has to be raised through a height equal to the absolute value of its initial coordinate, namely $$|y| = -y$$.

Let us label a small element of length $$dy$$ on the hanging part. Its mass is

$$dm \;=\;\lambda\,dy \;=\;\frac{M}{L}\,dy.$$

Initially this element is at a coordinate $$y$$ (negative), so it must be lifted through a vertical distance $$h = -y$$ to reach the surface. The elementary work $$dW$$ required to lift this element is given by the basic work formula

$$dW \;=\;(\text{force})\times(\text{displacement}) \;=\; (dm\,g)\,h.$$

Substituting $$dm$$ and $$h$$, we have

$$dW \;=\;\left(\frac{M}{L}\,dy\right)g\,(-y) \;=\;-\frac{M g}{L}\,y\,dy.$$

Because $$y$$ runs from $$-\frac{L}{n}$$ (bottom of the hanging part) up to $$0$$ (surface), the total work $$W$$ is obtained by integrating $$dW$$ over this interval:

$$W \;=\;\int_{y=-L/n}^{\,0} \left(-\frac{M g}{L}\,y\right)\,dy.$$

We now carry out the algebraic steps of the integration:

$$W \;=\;-\frac{M g}{L}\int_{-L/n}^{\,0} y\,dy \;=\;-\frac{M g}{L}\left[\,\frac{y^{2}}{2}\,\right]_{-L/n}^{\,0}.$$

Evaluating the definite integral, we get

$$\left[\,\frac{y^{2}}{2}\,\right]_{-L/n}^{\,0} \;=\;\frac{(0)^{2}}{2} \;-\;\frac{\left(-\dfrac{L}{n}\right)^{2}}{2} \;=\;0 \;-\;\frac{L^{2}}{2n^{2}} \;=\;-\frac{L^{2}}{2n^{2}}.$$

Substituting this back,

$$W \;=\;-\frac{M g}{L}\left(-\frac{L^{2}}{2n^{2}}\right) \;=\;\frac{M g}{L}\,\frac{L^{2}}{2n^{2}} \;=\;\frac{M g L}{2n^{2}}.$$

Thus the work needed to raise the entire hanging part of the cable to the level of the surface is

$$W \;=\;\frac{M g L}{2 n^{2}}.$$

Comparing this result with the given options, we see that it matches Option A.

Hence, the correct answer is Option A.

We are told that a wedge of mass $$M$$ rests on a perfectly smooth horizontal table and that a small particle of mass $$m$$ slides up the wedge without any friction anywhere. Numerically, the wedge is four times heavier than the particle, so $$M = 4m$$. The particle is launched horizontally towards the wedge with speed $$v$$.

Because the horizontal table is friction-free, no external horizontal force acts on the two-body system “particle + wedge”. Therefore the total horizontal linear momentum of the system will remain constant throughout the motion. This is our first key principle:

(i) Conservation of horizontal momentum: $$P_{\text{initial}} = P_{\text{final}}$$.

Secondly, all surfaces are smooth, so no mechanical energy is dissipated as heat. The only potential energy that changes is the gravitational potential energy of the particle as it climbs the wedge. Therefore the total mechanical energy of the system is also conserved:

(ii) Conservation of mechanical energy: $$E_{\text{initial}} = E_{\text{final}}$$.

We now apply these two principles step by step.

Initial state (just before contact)
The wedge is at rest, so its horizontal velocity is $$0$$. The particle of mass $$m$$ moves towards the wedge with speed $$v$$. Hence

Initial horizontal momentum: $$P_{\text{initial}} = m\,v.$$

Initial kinetic energy: $$E_{\text{kin,\,initial}} = \frac12\,m\,v^{2}.$$

The initial gravitational potential energy of the particle can be taken as zero (reference level at the foot of the wedge).

Final state (highest point reached by the particle on the wedge)
At the highest point the particle is momentarily at rest relative to the wedge. Because both bodies are still free to slide on the table, the particle and the wedge must, at that instant, move together with some common horizontal speed, say $$u$$, relative to the ground. Therefore

Final horizontal momentum: $$P_{\text{final}} = (M + m)\,u.$$

Final kinetic energy: $$E_{\text{kin,\,final}} = \frac12\,(M + m)\,u^{2}.$$

Let the particle have climbed a vertical height $$h$$ on the wedge. Its gain in gravitational potential energy is then

$$E_{\text{pot,\,gain}} = m\,g\,h.$$

We now write the conservation equations explicitly.

Step 1 : Linear momentum conservation

$$m\,v = (M + m)\,u.$$

We solve this for $$u$$:

$$u = \frac{m\,v}{M + m}.$$

Given $$M = 4m$$, we substitute:

$$u = \frac{m\,v}{4m + m} = \frac{m\,v}{5m} = \frac{v}{5}.$$

Step 2 : Mechanical energy conservation

Initial total energy = Final total energy:

$$\frac12\,m\,v^{2} = \frac12\,(M + m)\,u^{2} + m\,g\,h.$$

First we insert $$M + m = 4m + m = 5m$$ and $$u = \dfrac{v}{5}$$:

$$\frac12\,m\,v^{2} = \frac12\,(5m)\left(\frac{v}{5}\right)^{2} + m\,g\,h.$$

We now simplify term by term. Start with the kinetic term on the right:

$$\left(\frac{v}{5}\right)^{2} = \frac{v^{2}}{25},$$

so

$$\frac12\,(5m)\left(\frac{v^{2}}{25}\right) = \frac12 \times 5m \times \frac{v^{2}}{25} = \frac{5m\,v^{2}}{50} = \frac{m\,v^{2}}{10}.$$

Our energy equation is now

$$\frac12\,m\,v^{2} = \frac{m\,v^{2}}{10} + m\,g\,h.$$

To clear the fractions, multiply every term by $$\dfrac{2}{m}$$ (this also cancels the common factor $$m$$):

$$v^{2} = \frac{2}{m}\times\frac{m\,v^{2}}{10} + 2\,g\,h \;\;\Longrightarrow\;\; v^{2} = \frac{2\,v^{2}}{10} + 2\,g\,h.$$

Simplify the first term on the right:

$$\frac{2\,v^{2}}{10} = \frac{v^{2}}{5}.$$

Hence

$$v^{2} = \frac{v^{2}}{5} + 2\,g\,h.$$

Subtract $$\dfrac{v^{2}}{5}$$ from both sides:

$$v^{2} - \frac{v^{2}}{5} = 2\,g\,h.$$

Write the left side with a common denominator:

$$\frac{5v^{2}}{5} - \frac{v^{2}}{5} = \frac{4v^{2}}{5}.$$

Thus

$$\frac{4v^{2}}{5} = 2\,g\,h.$$

Finally, solve for $$h$$ by dividing both sides by $$2g$$:

$$h = \frac{\frac{4v^{2}}{5}}{2g} = \frac{4v^{2}}{10g} = \frac{2v^{2}}{5g}.$$

We have reached an explicit expression for the maximum height climbed by the particle. Comparing with the options given, this corresponds to Option C.

Hence, the correct answer is Option C.

Let the mass of the projectile (first body) be $$m_1 = 2\ \text{kg}$$ and its initial speed be $$u_1.$$

The second body is at rest, so for it $$u_2 = 0.$$ Let its mass be $$m_2,$$ which we have to find.

After an elastic collision in one dimension we denote the final speeds by $$v_1$$ and $$v_2$$ for the first and the second bodies respectively. According to the statement, the first body continues in the same direction with one-fourth of its original speed, so

$$v_1 = \frac{u_1}{4}.$$

For a perfectly elastic head-on collision we use the standard velocity-transfer formula (derived from simultaneous conservation of linear momentum and kinetic energy):

$$v_1 \;=\; \frac{m_1 - m_2}{m_1 + m_2}\;u_1.$$

Now we substitute the known value $$v_1 = u_1/4$$ into this formula:

$$\frac{u_1}{4} \;=\; \frac{m_1 - m_2}{m_1 + m_2}\;u_1.$$

Because $$u_1 \neq 0,$$ we cancel it from both sides:

$$\frac{1}{4} \;=\; \frac{m_1 - m_2}{m_1 + m_2}.$$

Cross-multiplying gives

$$4\,(m_1 - m_2) \;=\; m_1 + m_2.$$

Expanding and bringing like terms together, we have

$$4m_1 - 4m_2 \;=\; m_1 + m_2$$

$$4m_1 - m_1 \;=\; 4m_2 + m_2$$

$$3m_1 \;=\; 5m_2.$$

Solving for $$m_2$$ yields

$$m_2 \;=\; \frac{3}{5}\,m_1.$$

Finally, substituting $$m_1 = 2\ \text{kg}$$ gives

$$m_2 \;=\; \frac{3}{5}\times 2\ \text{kg} \;=\; \frac{6}{5}\ \text{kg} \;=\; 1.2\ \text{kg}.$$

Hence, the correct answer is Option B.

We are told that the position of the 2 kg object varies with time according to the law $$x = 3t^{2} + 5$$, where $$x$$ is in metres and $$t$$ in seconds. From this function we can find velocity, then acceleration, then force, and finally the work done.

First we differentiate the position function to obtain velocity, because by definition $$v = \dfrac{dx}{dt}$$.

So, $$v = \dfrac{d}{dt}\left(3t^{2} + 5\right) = 6t.$$

Next, we differentiate velocity with respect to time to obtain acceleration, since $$a = \dfrac{dv}{dt}$$.

Thus, $$a = \dfrac{d}{dt}\left(6t\right) = 6\; \text{m s}^{-2}.$$

Notice that the acceleration is a constant 6 m s−2. Because the mass is 2 kg, Newton’s second law $$F = ma$$ gives the magnitude of the force:

$$F = m a = 2 \times 6 = 12\;\text{N}.$$

We need the work done by this force in the time interval from $$t = 0$$ to $$t = 5\;\text{s}$$. A convenient route is to use the Work-Energy Theorem, which states:

“The net work done on a particle equals the change in its kinetic energy,” i.e. $$W = K_{\text{final}} - K_{\text{initial}}.$$

We already have the expression for velocity; let us evaluate it at the two instants.

At $$t = 0$$, $$v_{0} = 6(0) = 0\;\text{m s}^{-1}.$$

At $$t = 5\;\text{s}$$, $$v_{5} = 6(5) = 30\;\text{m s}^{-1}.$$

The kinetic energy formula is $$K = \tfrac{1}{2} m v^{2}.$$ Substituting the respective velocities:

Initial kinetic energy $$K_{0} = \tfrac{1}{2} \times 2 \times (0)^{2} = 0\;\text{J}.$$

Final kinetic energy $$K_{5} = \tfrac{1}{2} \times 2 \times (30)^{2} = 1 \times 900 = 900\;\text{J}.$$

Now we apply the Work-Energy Theorem:

$$W = K_{5} - K_{0} = 900 - 0 = 900\;\text{J}.$$

Hence, the correct answer is Option D.

We begin by recalling the work-energy theorem, which states that the work $$W$$ done by the net force on a particle equals the change in its kinetic energy $$\Delta K$$. In symbols, $$W = K_{\text{final}} - K_{\text{initial}}.$$

Next, we find the work done by the given force during the specified displacement. The work by a constant force is the dot product of the force vector $$\vec F$$ and the displacement vector $$\vec d$$:

$$W = \vec F \cdot \vec d.$$

Substituting the given vectors, we have

$$\vec F = 3\hat i - 12\hat j,$$

$$\vec d = 4\hat i.$$

Carrying out the dot product component-wise,

$$W = (3\hat i - 12\hat j) \cdot (4\hat i)$$

$$\;\; = (3)(4)\, (\hat i \cdot \hat i) \;+\; (-12)(4)\, (\hat j \cdot \hat i).$$

Because the unit vectors are orthogonal, $$\hat i \cdot \hat i = 1$$ and $$\hat j \cdot \hat i = 0$$. Therefore,

$$W = (3)(4)(1) + (-12)(4)(0)$$

$$\;\; = 12 + 0$$

$$\;\; = 12 \text{ J}.$$

So the force does 12 J of work on the particle.

Now, the particle’s initial kinetic energy is given as $$K_{\text{initial}} = 3 \text{ J}.$$

Applying the work-energy theorem:

$$K_{\text{final}} - K_{\text{initial}} = W$$

$$\Rightarrow K_{\text{final}} = K_{\text{initial}} + W$$

$$\Rightarrow K_{\text{final}} = 3 \text{ J} + 12 \text{ J}$$

$$\Rightarrow K_{\text{final}} = 15 \text{ J}.$$

Hence, the correct answer is Option B.

We treat the man-swing system as a simple pendulum whose bob (the person’s centre of mass while he is sitting) is at a distance $$L$$ from the pivot. The swing is released from an angular amplitude $$\theta_0$$ and reaches the lowest position (the vertical) with a certain speed. Exactly at this lowest point the person suddenly stands up, raising his centre of mass vertically by a small amount $$l$$ ($$l \ll L$$). We have to calculate the work that the person must do in order to achieve this rise and the accompanying change in kinetic energy of the swing.

First, let us find the speed just before the person stands up. Using mechanical-energy conservation for a pendulum of length $$L$$:

Initial potential energy (at amplitude $$\theta_0$$) relative to the lowest point is

$$U_i \;=\; MgL\bigl(1-\cos\theta_0\bigr).$$

At the lowest point this potential energy has completely converted into kinetic energy, so

$$\tfrac12 Mv_1^2 \;=\; MgL\bigl(1-\cos\theta_0\bigr).$$

Therefore the linear speed just before standing up is

$$v_1 \;=\; \sqrt{\,2gL\bigl(1-\cos\theta_0\bigr)}.$$

Now the person suddenly reduces the radius of circular motion from $$L$$ to

$$R_2 \;=\; L-l,$$

because his centre of mass rises by $$l$$. The interval in which he stands is taken to be very short, so the external torque about the pivot is practically zero (gravity acts along the vertical line through the pivot at that instant), and angular momentum about the pivot is conserved.

The angular momentum just before standing is

$$L_{\text{ang},1} \;=\; Mv_1 L.$$

Let the new speed immediately after standing be $$v_2$$. Conservation of angular momentum gives

$$Mv_1L \;=\; Mv_2\bigl(L-l\bigr)$$

$$\Longrightarrow\; v_2 \;=\; v_1\,\frac{L}{L-l}.$$

Hence the new kinetic energy is

$$K_2 = \tfrac12 Mv_2^2 = \tfrac12 Mv_1^2 \left(\frac{L}{L-l}\right)^2.$$

The potential energy has also increased, because the centre of mass is now higher by $$l$$. The increase in gravitational potential energy is

$$\Delta U \;=\; Mg\,l.$$

The work done by the person equals the total increase in mechanical energy, i.e.

$$W \;=\; (K_2 + U_2) - (K_1 + U_1) \;=\; (K_2 - K_1) + \Delta U.$$

Let us compute the kinetic-energy change:

$$K_2 - K_1 = \tfrac12 Mv_1^2\!\left[\left(\frac{L}{L-l}\right)^2 - 1\right].$$

Because $$l\ll L$$, we introduce the small parameter

$$\varepsilon \;=\; \frac{l}{L}, \quad\text{with } \varepsilon\ll 1.$$

Then $$\dfrac{L}{L-l} = \dfrac{1}{1-\varepsilon},$$ and for small $$\varepsilon$$ we use the binomial expansion

$$\frac{1}{1-\varepsilon} \;\approx\; 1 + \varepsilon + \varepsilon^2 + \ldots$$

Keeping only the first-order term,

$$\left(\frac{L}{L-l}\right)^2 = \frac{1}{(1-\varepsilon)^2} \;\approx\; 1 + 2\varepsilon.$$

Therefore

$$K_2 - K_1 \;\approx\; \tfrac12 Mv_1^2 (1 + 2\varepsilon - 1) = \tfrac12 Mv_1^2 (2\varepsilon) = Mv_1^2 \varepsilon.$$

Substituting $$\varepsilon = \dfrac{l}{L}$$ and $$v_1^2 = 2gL(1-\cos\theta_0),$$ we get

$$K_2 - K_1 = M \bigl[2gL(1-\cos\theta_0)\bigr]\frac{l}{L} = 2Mg(1-\cos\theta_0)\,l.$$

For small angular amplitudes we may use the standard approximation

$$\cos\theta_0 \;\approx\; 1 - \frac{\theta_0^{\,2}}{2},$$

so that

$$1-\cos\theta_0 \;\approx\; \frac{\theta_0^{\,2}}{2}.$$

Hence

$$K_2 - K_1 \;\approx\; 2Mg\left(\frac{\theta_0^{\,2}}{2}\right) l = Mg\theta_0^{\,2} l.$$

Now add the potential-energy change:

$$W = (K_2 - K_1) + \Delta U = Mg\theta_0^{\,2}l + Mg\,l = Mg\,l\,(1 + \theta_0^{\,2}).$$

Thus the work that the person has to do is approximately

$$W \;\approx\; Mg\,l\bigl(1 + \theta_0^{\,2}\bigr).$$

Hence, the correct answer is Option D.

We are given two vectors in the plane

$$\vec A = (\,\hat i + \hat j\,) \qquad\text{and}\qquad \vec B = (\,2\hat i - \hat j\,).$$

Let the required coplanar vector be

$$\vec C = (\,x\hat i + y\hat j\,).$$

Because all three vectors lie in the same plane, no further restriction on direction is needed; any ordered pair $$\,(x,y)\,$ represents a coplanar vector.

The condition stated in the problem is

$$\vec A\cdot\vec C \;=\;\vec B\cdot\vec C \;=\;\vec A\cdot\vec B.$$

First we compute $$\vec A\cdot\vec B.$$ The dot-product formula is

$$\vec u\cdot\vec v = u_xv_x + u_yv_y,$$ so, substituting $$\vec A = (1,1)$$ and $$\vec B = (2,-1),$$ we obtain

$$\vec A\cdot\vec B \;=\; 1\cdot 2 + 1\cdot(-1) \;=\; 2 - 1 \;=\; 1.$$

Next we impose the two equality conditions one by one.

1. The dot product $$\vec A\cdot\vec C$$ is, again using the same formula,

$$\vec A\cdot\vec C \;=\; 1\cdot x + 1\cdot y \;=\; x + y.$$

2. Similarly,

$$\vec B\cdot\vec C \;=\; 2\cdot x + (-1)\cdot y \;=\; 2x - y.$$

The given relations now read

$$x + y = 1 \quad\text{and}\quad 2x - y = 1.$$

We solve this pair of simultaneous linear equations step by step. Adding the two equations eliminates $$y$$:

$$(x + y) + (2x - y) \;=\; 1 + 1 \;\;\Longrightarrow\;\; 3x \;=\; 2 \;\;\Longrightarrow\;\; x \;=\; \frac{2}{3}.$$

Substituting $$x = \dfrac{2}{3}$$ back into $$x + y = 1$$ gives

$$\frac{2}{3} + y = 1 \;\;\Longrightarrow\;\; y = 1 - \frac{2}{3} = \frac{1}{3}.$$

Hence the vector $$\vec C$$ that satisfies both equalities is

$$\vec C = \left(\frac{2}{3}\hat i + \frac{1}{3}\hat j\right).$$

Finally, we calculate its magnitude. The magnitude (length) of a vector $$\vec v = (v_x,v_y)$$ is given by

$$|\vec v| = \sqrt{v_x^{\,2} + v_y^{\,2}}.$$

Applying this to $$\vec C,$$ we have

$$|\vec C| = \sqrt{\left(\frac{2}{3}\right)^{\!2} + \left(\frac{1}{3}\right)^{\!2}} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt5}{3}.$$

Thus the magnitude of the required vector is $$\sqrt{\dfrac{5}{9}}.$$

Hence, the correct answer is Option C.