JEE Laws of Motion Questions

Let the centre of the ring be $$O$$ and let the centre of the disk be $$C$$. Because the disk rolls without slipping on the inner wall of the ring, the distance $$OC$$ remains constant and is equal to the inner radius available for the centre,

$$a = R - r$$

Choose the small angular displacement $$\theta$$ of the centre $$C$$ from the lowest (vertical) position as the generalised coordinate. Arc‐length travelled by the centre is then

$$s = a\theta \qquad (|\theta| \ll 1)$$

1. Constraint of pure rolling
Pure rolling gives a fixed relation between the translational speed of the centre and the angular speed of the disk about its own axis:

$$v_{\text{cm}} = a\dot\theta = r\dot\phi \qquad -(1)$$

where $$\dot\phi$$ is the angular velocity of the disk about its diameter.

2. Kinetic energy
Translational kinetic energy: $$T_{\text{trans}} = \tfrac12 m v_{\text{cm}}^{2} = \tfrac12 m (a\dot\theta)^{2}$$

Moment of inertia of a solid disk about its centre: $$I = \tfrac12 m r^{2}$$

Using the rolling relation $$(1)$$, rotational kinetic energy is

$$T_{\text{rot}} = \tfrac12 I\dot\phi^{2} = \tfrac12\!\left(\tfrac12 m r^{2}\right)\! \left(\tfrac{a\dot\theta}{r}\right)^{2} = \tfrac14 m a^{2}\dot\theta^{2}$$

Thus total kinetic energy is

$$T = T_{\text{trans}} + T_{\text{rot}} = \left(\tfrac12 + \tfrac14\right)m a^{2}\dot\theta^{2} = \tfrac34 m a^{2}\dot\theta^{2} \qquad -(2)$$

3. Potential energy

(i) Gravitational: For a small angular displacement, the rise in the centre’s height is $$h = a(1-\cos\theta) \approx \tfrac12 a\theta^{2}$$ Therefore

$$U_g = mgh = \tfrac12 m g a\theta^{2} \qquad -(3)$$

(ii) Spring: The spring is attached tangentially along the periphery, so its extension/compression equals the arc length $$s = a\theta$$. Hooke’s law gives

$$U_s = \tfrac12 k s^{2} = \tfrac12 k a^{2}\theta^{2} \qquad -(4)$$

Total potential energy: $$U = U_g + U_s$$

$$U = \tfrac12\left(m g a + k a^{2}\right)\theta^{2} \qquad -(5)$$

4. Small-oscillation equation
Using the Lagrangian $$L = T - U$$ with $$T$$ from $$(2)$$ and $$U$$ from $$(5)$$:

$$L = \tfrac34 m a^{2}\dot\theta^{2} - \tfrac12\left(m g a + k a^{2}\right)\theta^{2}$$

Applying the Euler-Lagrange equation $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot\theta}\right) - \frac{\partial L}{\partial\theta}=0$$ gives

$$\frac{d}{dt}\!\left(\tfrac34\cdot 2 m a^{2}\dot\theta\right) + \left(m g a + k a^{2}\right)\theta = 0$$

$$\tfrac32 m a^{2}\ddot\theta + m g a\theta + k a^{2}\theta = 0 \qquad -(6)$$

Divide by $$\tfrac32 m a^{2}$$:

$$\ddot\theta + \left[\frac{2}{3}\left(\frac{g}{a} + \frac{k}{m}\right)\right]\theta = 0$$

This is the standard SHM form $$\ddot\theta + \omega^{2}\theta = 0$$ with

$$\omega^{2} = \frac{2}{3}\left(\frac{g}{a} + \frac{k}{m}\right) = \frac{2}{3}\left(\frac{g}{R - r} + \frac{k}{m}\right) \qquad -(7)$$

5. Time period

$$T = \frac{2\pi}{\omega}$$ with $$\omega$$ given by $$(7)$$.

Hence the correct expression is

$$\boxed{\displaystyle \omega = \sqrt{\frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}\right)}}$$

Option A perfectly matches this result.

Mass of the projectile: $$m = 200 \text{ g} = 0.2 \text{ kg}$$

The drag force is linear: $$\vec F = -c\vec v,$$ with $$c = 0.1 \text{ kg s}^{-1}.$$
Hence for any component of velocity, $$m\dfrac{dv}{dt} = -cv \; \Rightarrow \; \dfrac{dv}{dt} = -\dfrac{c}{m}\,v.$$

Horizontal component
Initial horizontal speed: $$v_{x0} = u\cos 60^{\circ} = 270 \times 0.5 = 135 \text{ m s}^{-1}.$$

With $$\dfrac{c}{m} = \dfrac{0.1}{0.2} = 0.5 \text{ s}^{-1},$$ the differential equation becomes
$$\dfrac{dv_x}{dt} = -0.5\,v_x.$$

Solution for velocity:
$$v_x(t) = v_{x0}\,e^{-0.5t} = 135\,e^{-0.5t}.$$

Horizontal displacement up to time $$t$$:
$$x(t) = \int_0^{t} v_x(t')\,dt' = 135\int_0^{t} e^{-0.5t'}dt' = 135\left[\dfrac{-1}{0.5}e^{-0.5t'}\right]_0^{t}.$$
This simplifies to
$$x(t) = \dfrac{135}{0.5}\bigl(1-e^{-0.5t}\bigr) = 270\bigl(1-e^{-0.5t}\bigr).$$

The projectile hits the wall at $$t = 2 \text{ s}$$, so
$$x(2) = 270\bigl(1-e^{-1}\bigr).$$

Using the given value $$e = 2.7$$,
$$e^{-1} = \dfrac{1}{2.7} \approx 0.37037,$$
$$1 - e^{-1} \approx 1 - 0.37037 = 0.62963.$$

Therefore
$$x(2) = 270 \times 0.62963 \approx 170 \text{ m}.$$

Hence the vertical wall is about $$170 \text{ m}$$ from the point of projection, which lies within the accepted range 167 - 171 m.

First step is to find out $$v_f$$ which is the final velocity with which both the blocks along with the spring move

Using conservation of momentum:-

so $$m_1v_1=\left(m_1+m_2\right)v_f$$

$$v_f=\frac{m_1v_1}{m_1+m_2}$$

On substituting the given values

$$v_f=\frac{10\times\ 3}{10+5}$$

$$v_f=\frac{30}{15}=2\ \frac{m}{s}$$

Now lets conserve Energy

Initital K.E = $$\frac{1}{2}m_1v_1^2$$

$$=\frac{1}{2}\times\ 10\times\left(\ 3^2\right)$$

$$=5\times9$$

$$K.E_i=45$$

Final K.E is $$\frac{1}{2}\left(m_1+m_2\right)v_f^2$$

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

$$=15\times\ 2$$

$$K.E_f=30$$

Now difference between these is the energy stored in the spring

$$K.E_i-K.E_f$$$$K.E_i-K.E_f\ =\ \ 45-\ 30\ =15$$

Now 

$$\frac{1}{2}kx^2=15$$

$$\frac{1}{2}\times\ 3000\times\ x^2=15$$

$$\ x^2=\frac{30}{3000}$$

$$\ x^2=\frac{1}{100}=10^{-2}$$

$$\ x=10^{-1}=0.1m$$

Given:

$$T_1=\sqrt{\ 3}T_2$$

Horizontal equilibrium

$$T_1\cos\theta_1=T_2\cos\theta_2$$

Substitute:

$$\sqrt{3}T_2\cos\theta_1=T_2\cos\theta_2$$​

Cancel $$T_2$$​:

$$\sqrt{3}\cos\theta_1=\cos\theta_2$$

From the options

We know:

• $$\cos60^{\circ}=\frac{1}{2}$$
• $$\cos30^{\circ}=\frac{\sqrt{3}}{2}$$

Try:

$$\theta_1=60^{\circ\ }$$

Then:

$$\sqrt{3}\cos60^{\circ}=\sqrt{3}\cdot\frac{1}{2}=\frac{\sqrt{3}}{2}$$​​

So:

$$\cos\ \theta\ _2=\frac{\sqrt{\ 3}}{2}$$

$$\theta\ _2=30^{\circ\ }$$

$$T_1\sin\theta_1+T_2\sin\theta_2=mg$$

Substitute:

$$\sqrt{3}T_2\cdot\frac{\sqrt{3}}{2}+T_2\cdot\frac{1}{2}=mg$$

$$\frac{3}{2}T_2+\frac{1}{2}T_2=mg\Rightarrow2T_2=mg$$

$$T_2 = \dfrac{mg}{2}$$

The block slides down a rough plane making an angle $$\theta = 60^{\circ}$$ with the horizontal.

For a body sliding down an inclined plane, the forces along (parallel to) the plane are:
  • Down the plane: component of weight, $$mg\sin\theta$$
  • Up the plane: kinetic friction, $$f_k = \mu_k N$$, where $$N$$ is the normal reaction.

First find the normal reaction. For an incline, $$N = mg\cos\theta$$ because no acceleration is perpendicular to the plane.

Therefore the kinetic friction magnitude is
$$f_k = \mu_k N = \mu_k mg\cos\theta$$.

Apply Newton’s second law along the plane (taking down the plane as positive).
Net force $$= mg\sin\theta - f_k = ma$$.

Substitute $$f_k$$ and the given acceleration $$a = \dfrac{g}{2}$$:
$$mg\sin\theta - \mu_k mg\cos\theta = m\left(\dfrac{g}{2}\right)$$.

Cancel the common factor $$mg$$ from every term:
$$\sin\theta - \mu_k \cos\theta = \dfrac{1}{2}$$ $$-(1)$$

Insert the trigonometric values for $$\theta = 60^{\circ}$$:
$$\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}, \qquad \cos 60^{\circ} = \dfrac{1}{2}$$.

Equation $$(1)$$ becomes
$$\dfrac{\sqrt{3}}{2} - \mu_k \left(\dfrac{1}{2}\right) = \dfrac{1}{2}$$.

Multiply every term by 2 to remove the denominators:
$$\sqrt{3} - \mu_k = 1$$.

Solve for $$\mu_k$$:
$$\mu_k = \sqrt{3} - 1$$.

Hence the coefficient of kinetic friction is $$\sqrt{3} - 1$$, which matches Option A.

Answer: Option A

Initial velocity of the body is given as $$\vec v_{in}=3\hat i+4\hat j$$ ms$$^{-1}$$, so

$$u_x = 3$$ ms$$^{-1},\; u_y = 4$$ ms$$^{-1},\; u_z = 0$$ ms$$^{-1}$$.

The constant force acting on the mass is $$\vec F = 6\hat k$$ N (along +z-axis).

Using Newton’s second law, $$\vec F = m\vec a \Rightarrow \vec a = \frac{\vec F}{m}$$.

The mass of the body is $$m = 2$$ kg, therefore

$$\vec a = \frac{6\hat k}{2} = 3\hat k$$ ms$$^{-2}$$.

The body stays in this field for $$t = \frac{5}{3}$$ s. The kinematics relation $$\vec v = \vec u + \vec a t$$ gives the final velocity:

Along x-axis: $$v_x = u_x + a_x t = 3 + 0 \times t = 3$$ ms$$^{-1}$$ (no force in x).
Along y-axis: $$v_y = u_y + a_y t = 4 + 0 \times t = 4$$ ms$$^{-1}$$ (no force in y).
Along z-axis: $$v_z = u_z + a_z t = 0 + 3 \times \frac{5}{3} = 5$$ ms$$^{-1}$$.

Hence the velocity vector when the body emerges is

$$\vec v_{out}=3\hat i + 4\hat j + 5\hat k$$ ms$$^{-1}$$.

Therefore, the correct option is Option B.

Initially: F_buoy - Mg = Ma, so F_buoy = M(g+a).

After releasing mass m: F_buoy - (M-m)g = (M-m)(3a).

M(g+a) = (M-m)(g+3a), M(g+a) = M(g+3a) - m(g+3a)

m(g+3a) = M(g+3a-g-a) = 2Ma, m = 2Ma/(g+3a).

The correct answer is Option 1.

Here as the system is in equilibrium 

We have to equate horizontal components of both $$T_{1\ }and\ T_2$$ as there is no external Force acting in the horizontal direction

Here horizontal component of $$T_{1\ }is\ T_1\cos60^{\circ\ }$$

Here horizontal component of $$T_{2\ }is\ T_2\cos30^{\circ\ }$$

Equate these 2

$$T_2\cos60^{\circ\ }=\ T_1\cos30^{\circ\ }$$

$$T_2\times\ \frac{\sqrt{\ 3}}{2}=\ T_1\times\ \frac{1}{2}$$

$$T_1\ =\ \sqrt{\ 3}T_2$$

Now coming to the Vertical direction 

There is gravitational force downwards which is Mg and vertical components of $$T_{1\ }and\ T_2$$ add up and are opposing the force due to gravitation

so $$Mg=\ T_1\sin\theta\ \ +\ T_2\sin\ \ \theta\ $$

$$Mg=\ T_1\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$$

Substituting  $$T_1\ =\ \sqrt{\ 3}T_2$$ in that equation

$$Mg=\ \sqrt{\ 3}T_2\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$$

$$Mg=\frac{3T_2}{2}+\frac{T_2}{2}$$
$$Mg=2T_2$$

$$T_2=\frac{g}{2}\ \ \left(as\ M=1\right)$$

$$T_2=5N$$

as $$T_1\ =\ \sqrt{\ 3}T_2$$

$$T_1=5\sqrt{\ 3}N$$

The speed of the object varies with its position as $$v = 4\sqrt{x}$$, where $$x$$ is the coordinate (in metres) measured along the +x-axis.

First convert the mass to SI units: given mass $$m = 500\,$$g $$= 0.5\,$$kg.

To find the force we need the acceleration. For one-dimensional motion, acceleration can be expressed in terms of position as
$$a = v\,\frac{dv}{dx}$$

STEP 1 - Differentiate the speed with respect to $$x$$:
$$v = 4x^{1/2} \; \Rightarrow \; \frac{dv}{dx} = 4 \cdot \frac{1}{2}x^{-1/2} = 2x^{-1/2}$$

STEP 2 - Use $$a = v\,\dfrac{dv}{dx}$$:
Substitute $$v = 4x^{1/2}$$ and $$\dfrac{dv}{dx} = 2x^{-1/2}$$:
$$a = \left(4x^{1/2}\right)\left(2x^{-1/2}\right) = 8 \text{ m\,s}^{-2}$$

The acceleration turns out to be a constant $$8 \text{ m\,s}^{-2}$$ (independent of $$x$$).

STEP 3 - Apply Newton’s second law $$F = ma$$:
$$F = 0.5 \times 8 = 4 \text{ N}$$

Hence the force acting on the object is $$4 \text{ N}$$.

Therefore, Option D is correct.

Two cars P and Q move on a road in the same direction. Car P has an acceleration that increases linearly with time, and car Q has constant acceleration. They cross each other at $$t = 0$$ for the first time, and we need to find the maximum possible number of crossings.

We begin by setting up the position equations. Let both cars be at the same position at $$t = 0$$ and define the relative position $$\Delta x = x_P - x_Q$$.

For car P the acceleration is $$a_P = \alpha t$$ (linearly increasing, where $$\alpha > 0$$ is a constant). Integrating gives the velocity $$v_P = v_{P0} + \frac{\alpha t^2}{2}$$ and the position $$x_P = x_0 + v_{P0}t + \frac{\alpha t^3}{6}$$.

Similarly, for car Q the constant acceleration is $$a_Q = \beta$$. Integrating gives the velocity $$v_Q = v_{Q0} + \beta t$$ and the position $$x_Q = x_0 + v_{Q0}t + \frac{\beta t^2}{2}$$.

Subtracting these positions, the relative position becomes

$$\Delta x = x_P - x_Q = (v_{P0} - v_{Q0})t + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$$

Letting $$c = v_{P0} - v_{Q0}$$ (the initial relative velocity) simplifies this to

$$\Delta x = ct + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$$.

Crossings occur when $$\Delta x = 0$$, so we set

$$t\left(c + \frac{\alpha t^2}{6} - \frac{\beta t}{2}\right) = 0$$

This equation yields the trivial root $$t = 0$$ (the first crossing) and the quadratic

$$\frac{\alpha}{6}t^2 - \frac{\beta}{2}t + c = 0$$.

This quadratic can have at most two real roots. In order for both of these roots to be positive (since $$t > 0$$), three conditions must be satisfied:

- Discriminant $$\left(\frac{\beta}{2}\right)^2 - 4 \cdot \frac{\alpha}{6} \cdot c \geq 0$$, i.e., $$\frac{\beta^2}{4} \geq \frac{2\alpha c}{3}$$

- Sum of roots $$\frac{\beta/2}{\alpha/6} = \frac{3\beta}{\alpha} > 0$$ (true since $$\alpha, \beta > 0$$)

- Product of roots $$\frac{c}{\alpha/6} = \frac{6c}{\alpha} > 0$$, which requires $$c > 0$$

When these conditions are met for appropriate values of $$\alpha$$, $$\beta$$, and $$c$$, the quadratic has two positive real roots, corresponding to two additional crossings beyond the one at $$t = 0$$.

To see that no more crossings are possible, note that $$\Delta x / t$$ is a quadratic function, so after factoring out the root at $$t = 0$$ the remaining quadratic factor can have at most two positive zeros. Therefore, the total number of crossings cannot exceed three.

In conclusion, the maximum possible number of crossings (including the one at $$t = 0$$) is $$\boxed{3}$$.

All three masses move upward with acceleration a=$$2 \text{ ms}^{-2}$$

Total mass:

$$m_{total}=4+6+10=20kg$$

Now treat the three blocks as a single body.

Forces on the system:

• Upward: $$T_1$$
• Downward: total weight $$=20g$$

Apply Newton’s second law (upward positive):

$$T_1-20g=20\times2$$

Substitute $$g = 10 \text{ m/s}^2$$

$$T_1-200=40$$

$$T_1=240N$$

Final answer: 240 N

First lets write the Forces acting in Vertical and Horizontal direction

In the Vertical direction,

Forces are $$F_g=mg\ due\ to\ gravity\ and\ T\sin\theta\ \ \left(\ tension\ of\ the\ rope\right)$$

and in Horizontal , The forces are $$F\left(appplied\ force\right)\ and\ T\cos\theta\ \left(tension\ of\ the\ rope\right)$$

First lets equate the Vertical Forces

$$mg\ =\ T\sin\ \theta\ $$

$$10=\ T\times\ \frac{1}{\sqrt{\ 2}}$$

Now substitute this and equal Horizontal forces

$$F=\ T\cos\theta$$

$$F=\ \sqrt{\ 2}\times\ 10\times\ \cos45^{\circ\ }\ $$

$$F=\ \frac{10}{\sqrt{\ 2}}\times\ \sqrt{\ 2}$$

$$F=\ 10N$$

Lets start writing equations for both the blocks

For $$m_1$$ : -

$$m_1g-T=m_1a$$

For $$m_2$$: - 

$$T-m_2g=m_2a$$

Given that $$a=\frac{g}{8}$$

On substituting a in both equations and adding them we get , 

$$m_1g-\ m_2g\ =\ m_1\left(\frac{g}{8}\right)+m_2\left(\frac{g}{8}\right)$$

$$\left(m_1-\ m_2\right)g\ =\ \left(m_1+m_2\right)\left(\frac{g}{8}\right)$$

$$\frac{\left(m_1-\ m_2\right)}{m_1+m_2}=\frac{1}{8}$$

On dividing Numerator and denominator with $$m_2$$

$$\frac{\left(\frac{m_1}{m_2}-1\right)}{\left(\frac{m_1}{m_2}+1\right)}=\frac{1}{8}$$

consider $$\frac{m_1}{m_2}$$ as x and solve for x

$$\frac{\left(x-1\right)}{\left(x+1\right)}=\frac{1}{8}$$

$$\left(x-1\right)=\frac{1}{8}\left(x+1\right)$$

$$x-\frac{1}{8}x=\frac{1}{8}+1$$

$$\frac{7}{8}x=\frac{9}{8}$$

$$x=\frac{9}{7}$$

For an Atwood machine, the acceleration is given by:

$$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$$

(assuming $$m_2 > m_1$$)

Given $$a = \frac{g}{8}$$:

$$\frac{m_2 - m_1}{m_1 + m_2} = \frac{1}{8}$$

$$8(m_2 - m_1) = m_1 + m_2$$

$$8m_2 - 8m_1 = m_1 + m_2$$

$$7m_2 = 9m_1$$

$$\frac{m_2}{m_1} = \frac{9}{7}$$

The correct answer is Option 4: $$9:7$$.

A train starts from rest, accelerates uniformly to 80 km/h in time $$t$$, then moves at constant speed for time $$3t$$. We need the average speed.

Calculate the distance during acceleration.

During uniform acceleration from 0 to 80 km/h over time $$t$$, the average speed is $$\frac{0 + 80}{2} = 40$$ km/h.

$$ d_1 = \text{average speed} \times \text{time} = 40t \text{ (in km, with } t \text{ in hours)} $$

Calculate the distance during constant speed.

$$ d_2 = 80 \times 3t = 240t \text{ km} $$

Calculate the average speed.

$$ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{d_1 + d_2}{t + 3t} = \frac{40t + 240t}{4t} = \frac{280t}{4t} = 70 \text{ km/h} $$

The correct answer is Option (4): 70.

You have:

• a 2 kg block on a 30° incline
• connected by a string over a pulley
• then a movable pulley holding a 4 kg mass

Key idea: movable pulley changes both force and motion relation

Step 1: understand tension

Let tension in string = T

For the 4 kg block:
It is supported by two segments of the string
So upward force = 2T

Equation:
$$2T−4g=4a₄.$$

Step 2: relation of acceleration

Very important part

If the 2 kg block moves up the incline by x
the movable pulley (and 4 kg mass) moves down by x/2

So:
$$a₂=2a₄$$

Now lets write equation for the 2 Kg block

$$m_2g\sin\ \theta\ -T=m_2a_2$$

 $$2g\times\ \frac{1}{2}-T=2a_2$$

$$g-T=4a_4$$

$$2T−4g=4a₄.$$

now on multiply and adding both so that T gets cancelled

$$-2g=12a_4$$

$$\ a_4=-\frac{g}{6}\left(negative\ as\ we\ considered\ opposite\ direction\right)$$

so now we know 

$$a₂=2a₄$$

so 

$$a₂=2\times\ \frac{g}{6}$$

$$a₂=\ \frac{g}{3}$$

Let us write equations for all three blocks A,B and C

For A:-

$$F-T_2=2a$$

$$80-T_2=2a$$

For B:-

$$T_2-T_1=3a$$

and For C:-

$$T_1=5a$$

Now on adding all 3 equations

We get $$80\ =\ 10a$$

$$a = 8 \text{ m s}^{-2}$$

now substitute value of a in equation 1

$$80-T_2=2a$$

$$80-T_2=2\times\ 8$$

$$80-16\ =T_2$$

$$T_2=64N$$

Substitute the value of a in equation 3

$$T_1=5a$$

$$T_1=5\times\ 8$$

$$T_1=40N$$

Given:

F₁ = $$5\hat{i} + 8\hat{j} + 7\hat{k}$$
F₂ = $$3\hat{i} - 4\hat{j} - 3\hat{k}$$
mass m = 4 kg

First find net force:

$$F_{net}$$ = F₁ + F₂
= $$(5+3)\hat{i} + (8-4)\hat{j} + (7-3) \hat{k}$$
= $$8\hat{i} + 4\hat{j} + \hat{k}$$

Now acceleration:

a =$$\frac{F_{net}}{m}$$
= $$(8\hat{i} + 4\hat{j} + \hat{k})$$ / 4
= $$2\hat{i} + \hat{j} + \hat{k}$$

So the acceleration is:

$$2\hat{i} + \hat{j} + \hat{k}$$

A wooden block (5 kg) and iron cylinder (25 kg) go down together with acceleration $$a = 0.1$$ m/s$$^2$$.

The total mass is $$m = 5 + 25 = 30$$ kg.

According to Newton's second law, the system accelerates downward at $$0.1$$ m/s$$^2$$. The forces acting on the system are weight (downward) and the normal reaction from the floor (upward).

- Weight (downward): $$mg = 30 \times 9.8 = 294$$ N

- Normal reaction from floor (upward): $$N$$

According to Newton's second law,

$$ mg - N = ma $$

Therefore,

$$ N = mg - ma = m(g - a) = 30(9.8 - 0.1) = 30 \times 9.7 = 291 \text{ N} $$

By Newton's third law, the action force of the system on the floor equals $$N = 291$$ N downward.

The correct answer is Option (2): 291 N.

By conservation of momentum (assuming the extra mass is added with zero velocity in the horizontal direction):

$$m_1 v_1 = (m_1 + m_2)v_f$$

$$1000 \times 6 = (1000 + 200) \times v_f$$

$$6000 = 1200 \times v_f$$

$$v_f = 5 \text{ m s}^{-1}$$

The answer is $$5$$ m/s, which corresponds to Option (4).

Given

initially at rest and then Linearly increasing force is applied

$$F=ma$$

as m is constant

F is directly proportional to a

So even $$a=0$$ at first then starts increasing linearly

so option D is correct

A cricket ball of mass 150 g moving at 20 m/s is caught in 0.1 s. Find the force on the player's hand.

Recall Newton's second law in terms of impulse: $$F = \frac{\Delta p}{\Delta t}$$ where $$\Delta p$$ is the change in momentum and $$\Delta t$$ is the time interval.

The initial momentum is $$p_i = mv = 0.15 \times 20 = 3$$ kg m/s, and the final momentum is $$p_f = 0$$ (ball is brought to rest). Hence, $$|\Delta p| = |p_f - p_i| = 3 \text{ kg m/s}$$.

Substituting into the formula gives $$F = \frac{|\Delta p|}{\Delta t} = \frac{3}{0.1} = 30 \text{ N}$$.

By Newton's third law, the force exerted by the ball on the player's hand equals the force exerted by the hand on the ball.

The correct answer is Option (4): 30 N.

F₂ and F₃ are perpendicular. Their resultant = $$\sqrt{8^2 + 6^2} = \sqrt{100} = 10$$ N = F₁.

When F₁ is removed, the net force = resultant of F₂ and F₃ = 10 N.

$$a = F/m = 10/5 = 2$$ m/s²

Given: $$\text{Height of terrace} = 3h$$, $$e = \frac{1}{\sqrt{3}}$$

From conservation of energy on the slide:

$$mgh = \frac{1}{2}mu_0^2 \implies u_0 = \sqrt{2gh} \implies \vec{u}_0 = \sqrt{2gh}\hat{x}$$

For the projectile motion before impact:

$$v_x = u_0 = \sqrt{2gh}$$

$$v_z^2 = 2g(3h) = 6gh \implies v_z = -\sqrt{6gh}$$

$$\tan\theta = \frac{\vert{}v_z\vert{}}{v_x} = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt{3} \implies \theta = 60^\circ$$

After impact with the ground (elastic along $$\hat{x}$$, partially inelastic along $$\hat{z}$$):

$$v'_x = v_x = \sqrt{2gh}$$

$$v'_z = e\vert{}v_z\vert{} = \frac{1}{\sqrt{3}}\sqrt{6gh} = \sqrt{2gh}$$

$$\vec{v} = \sqrt{2gh}\hat{x} + \sqrt{2gh}\hat{z} = \sqrt{2gh}(\hat{x} + \hat{z})$$

Finding maximum height $$h_1$$ and distance $$d$$:

$$h_1 = \frac{(v'_z)^2}{2g} = \frac{2gh}{2g} = h$$

$$d = u_0 \cdot t_{\text{fall}} = \sqrt{2gh} \cdot \sqrt{\frac{2(3h)}{g}} = \sqrt{2gh} \cdot \sqrt{\frac{6h}{g}} = \sqrt{12}h = 2\sqrt{3}h$$

$$\frac{d}{h_1} = \frac{2\sqrt{3}h}{h} = 2\sqrt{3}$$

Given: mass = 10 kg, force F at 30° to horizontal, $$\mu_s = 0.25$$, $$g = 10$$ m/s².

The normal force: $$N = mg - F\sin 30° = 100 - \frac{F}{2}$$

For the block to just start moving, the horizontal component of F equals the maximum static friction:

$$ F\cos 30° = \mu_s N $$

$$ F \times \frac{\sqrt{3}}{2} = 0.25\left(100 - \frac{F}{2}\right) $$

$$ \frac{\sqrt{3}}{2}F = 25 - \frac{F}{8} $$

$$ F\left(\frac{\sqrt{3}}{2} + \frac{1}{8}\right) = 25 $$

$$ F \times \frac{4\sqrt{3} + 1}{8} = 25 $$

$$ F = \frac{200}{4\sqrt{3} + 1} = \frac{200}{7.928} \approx 25.2 \text{ N} $$

Therefore, $$F \approx 25.2$$ N.

Two forces of magnitudes $$A$$ and $$\frac{A}{2}$$ are perpendicular to each other.

The magnitude of the resultant when two forces are perpendicular is:

$$ R = \sqrt{F_1^2 + F_2^2} = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{\frac{5A^2}{4}} = \frac{\sqrt{5}A}{2} $$

The correct answer is $$\dfrac{\sqrt{5}A}{2}$$.

A car moves in a circular horizontal track and a bob is suspended from the roof by a string. We need to find the angle the string makes with the vertical.

We are given that Speed $$v = 20$$ m/s, radius $$R = 40$$ m, $$g = 10$$ m/s$$^2$$.

To begin,

When the car moves in a circular path, it undergoes centripetal acceleration directed toward the centre of the circle. The bob, being inside the car, must also undergo this centripetal acceleration. The only forces on the bob are its weight ($$mg$$, downward) and the tension in the string ($$T$$, along the string). For the bob to have centripetal acceleration, the string must tilt away from the vertical (outward from the centre), so that the horizontal component of tension provides the centripetal force.

Next,

Let $$\theta$$ be the angle the string makes with the vertical.

Vertical equilibrium (no vertical acceleration):

$$ T\cos\theta = mg \quad \cdots (1) $$

Horizontal direction (centripetal acceleration $$a_c = v^2/R$$):

$$ T\sin\theta = \frac{mv^2}{R} \quad \cdots (2) $$

From here,

$$ \frac{T\sin\theta}{T\cos\theta} = \frac{mv^2/R}{mg} $$

$$ \tan\theta = \frac{v^2}{Rg} $$

Continuing,

$$ \tan\theta = \frac{(20)^2}{40 \times 10} = \frac{400}{400} = 1 $$

$$ \theta = \tan^{-1}(1) = 45° = \frac{\pi}{4} $$

The string makes an angle of $$\dfrac{\pi}{4}$$ with the vertical.

The correct answer is Option 3: $$\dfrac{\pi}{4}$$.

Lets write down equation for both

For 4 Kg

$$Mg\sin\theta_1\ -T=Ma$$

$$40\sin60^{\circ\ }\ -T=4a$$

For 1 Kg

$$T-mg\sin\theta_2\ =ma$$

$$T-10\sin30^{\circ\ }\ =a$$

Now add 2 equations

$$40\sin60^{\circ\ }-10\sin30^{\circ\ }\ =5a$$

$$40\times\ \frac{\sqrt{\ 3}}{2}-10\times\ \frac{1}{2}\ =5a$$

$$20\ \sqrt{\ 3}-5\ =5a$$

$$4\sqrt{\ 3}-1=a$$

Now substitute a in any equation

$$T-10\sin30^{\circ\ }\ =4\sqrt{\ 3}-1$$

$$T-5\ =4\sqrt{\ 3}-1$$

$$T=4\sqrt{\ 3}+4\ =\ 4\left(\sqrt{\ 3}+1\right)$$

A garden roller of mass 70 kg is pushed with a force of 200 N at 30 degrees below the horizontal. We need to find the normal reaction.

To begin,

Three forces act on the roller:

1. Weight $$W = mg = 70 \times 10 = 700$$ N (acting downward)

2. Applied force $$F = 200$$ N at 30 degrees below the horizontal (has both horizontal and vertical components)

3. Normal reaction $$N$$ from the ground (acting upward)

Next,

Since the force is applied at 30 degrees below the horizontal:

Horizontal component: $$F_x = F\cos 30° = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3}$$ N (forward)

Vertical component: $$F_y = F\sin 30° = 200 \times \frac{1}{2} = 100$$ N (downward, since the force is below the horizontal)

From here,

Since the roller moves horizontally (no vertical acceleration), the net vertical force must be zero. The upward force (normal reaction) must balance all downward forces (weight + vertical component of applied force):

$$ N = mg + F\sin 30° $$

$$ N = 700 + 100 = 800\;\text{N} $$

Note: The vertical component of the applied force acts downward because the force is directed below the horizontal. This increases the normal reaction beyond just the weight of the roller. If the force were directed above the horizontal, the vertical component would reduce the normal reaction.

The normal reaction on the roller is 800 N.

The correct answer is Option 3: 800 N.

To compare impulses, remember one key idea:

Impulse = area under the F-t graph

Now evaluate each figure based on its shape and dimensions:

For graph a

This is a triangle with $$base\ 1s\ and\ hieght\ 0.5N$$.
Area (impulse) =

$$\frac{1}{2}\times\ 1\times\ 0.5=0.25N$$

For graph B

This is a rectangle with constant force $$0.5N\ \ for\ a\ time\ of\ 2s$$.
Area (impulse) =

$$0.5\times\ 2\ =1N$$

For graph C

Triangle with base 1 s and height 0.75 N
Area =

$$\frac{1}{2}\times\ 1\times\ 0.75=0.375N$$

For graph D

Triangle with base 2 s and height 0.5 
Area =$$\frac{1}{2}\times\ 2\times\ 0.5=0.5N$$

Therefore from this we can conclude that fig(b) has the highest impulse

The assertion states that an electric fan continues to rotate for some time after the current is switched off, and this is correct because when the current to an electric fan is switched off, the fan does not stop immediately but continues to rotate and gradually slows down over several seconds, a commonly observed phenomenon. According to Newton’s First Law of Motion (the Law of Inertia), an object in motion continues in its state of motion unless acted upon by an external unbalanced force, and for rotational motion the analogous principle states that a rotating body continues to rotate unless acted upon by an external torque. The fan blades, motor rotor, and other rotating parts have mass and therefore possess rotational inertia (moment of inertia $$I$$). When the current is switched off, there is no longer a driving torque from the motor; however, the fan has angular momentum given by $$L = I\omega$$ where $$I$$ is the moment of inertia of the rotating parts and $$\omega$$ is the angular velocity. This angular momentum cannot vanish instantaneously, so the fan gradually decelerates due to frictional forces (air resistance and bearing friction) which provide a retarding torque $$\tau_{friction}$$, resulting in an angular deceleration $$\alpha = \frac{\tau_{friction}}{I}$$. Since the frictional torque is relatively small compared to the moment of inertia, it takes considerable time for the fan to come to rest.

The reason suggests that the fan continues to rotate due to inertia of motion, which is correct since it identifies that the fan’s continued rotation is due to rotational inertia, representing a direct application of Newton’s First Law extended to rotational motion, and thus the reason explains the assertion because the rotating fan tends to maintain its state of rotation and only external torques (friction, air resistance) can bring it to rest.

The correct answer is Option 3: Both A and R are correct and R is the correct explanation of A.

Given , 

A momentum-time graph is given,

p-t

momentum p=mv

as m is constant , graph is directly proportional to v

so the slope of the graph is the acceleration 

$$\ F=ma$$

So higher the slope higher the Force

Observe the graph, order of slopes is 

c>b>a

Therefore maximum force is in c and minimum force is in b

Each ball has mass $$m$$ and moves with speed $$v$$. After striking the wall normally, it bounces back with the same speed.

Change in momentum for each ball:

$$ \Delta p = mv - (-mv) = 2mv $$

Total change in momentum for 100 balls:

$$ \Delta p_{\text{total}} = 100 \times 2mv = 200mv $$

The total force exerted on the wall (by Newton's second law):

$$ F = \frac{\Delta p_{\text{total}}}{t} = \frac{200mv}{t} $$

Find the coefficient of kinetic friction for a block sliding down a $$30°$$ incline with acceleration $$\frac{g}{4}$$.

$$ma = mg\sin 30° - \mu_k mg\cos 30°$$

$$\frac{g}{4} = g\sin 30° - \mu_k g\cos 30°$$

$$\frac{1}{4} = \frac{1}{2} - \mu_k \cdot \frac{\sqrt{3}}{2}$$

$$\mu_k \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

$$\mu_k = \frac{1}{4} \cdot \frac{2}{\sqrt{3}} = \frac{1}{2\sqrt{3}}$$

The correct answer is Option B: $$\frac{1}{2\sqrt{3}}$$.

First lets write the Forces acting in Vertical and Horizontal direction

In the Vertical direction,

Forces are $$F_g=\sqrt{\ 3}g\ due\ to\ gravity\ and\ T\ \left(\ tension\ of\ the\ rope\right)$$

and in Horizontal , The forces are $$F\left(appplied\ force\right)\ and\ T\cos\theta\ \left(tension\ of\ the\ rope\right)$$

First lets equate the Vertical Forces 

$$\sqrt{\ 3}g\ =\ T\cos\ \theta\ $$

$$10\sqrt{\ 3}=\ T\times\ \frac{\sqrt{\ 3}}{2}$$

$$\ T=20N$$

Body of mass 10 kg, initial speed 20 m/s, stops in 5 s due to friction.

Find deceleration.

Find coefficient of friction.

The friction force causes deceleration: $$f = \mu mg = ma$$

The answer is Option D: 0.4.

A body of mass 500 g moves along the x-axis with velocity $$v = 10\sqrt{x}$$ m/s. We need to find the force acting on it.

First, we state the given quantities.

Mass: $$m = 500$$ g $$= 0.5$$ kg

Velocity as a function of displacement: $$v = 10\sqrt{x} = 10x^{1/2}$$ m/s

Next, we find the acceleration using the chain rule.

Since velocity is given as a function of position (not time), we use the relation:

$$ a = v\frac{dv}{dx} $$

This follows from the chain rule: $$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$$.

First, find $$\frac{dv}{dx}$$:

$$ \frac{dv}{dx} = \frac{d}{dx}(10x^{1/2}) = 10 \times \frac{1}{2}x^{-1/2} = \frac{5}{\sqrt{x}} $$

Now compute the acceleration:

$$ a = v \cdot \frac{dv}{dx} = 10\sqrt{x} \times \frac{5}{\sqrt{x}} = 50\;\text{m/s}^2 $$

Notice that the $$\sqrt{x}$$ terms cancel, giving a constant acceleration independent of position.

From this, we apply Newton's second law.

$$ F = ma = 0.5 \times 50 = 25\;\text{N} $$

The force acting on the body is 25 N.

The correct answer is Option 2: 25 N.

We have a bullet of mass $$m = 0.1$$ kg travelling at $$u = 400$$ m/s that gets embedded in a block of mass $$M = 3.9$$ kg. The combined system then slides a distance $$s = 20$$ m before coming to rest.

Since the bullet gets embedded in the block, this is a perfectly inelastic collision. By conservation of momentum:

$$mu = (m + M)v$$

$$0.1 \times 400 = (0.1 + 3.9) \times v$$

$$40 = 4v$$

$$v = 10 \text{ m/s}$$

Now, after the collision, the block-bullet system decelerates due to friction and comes to rest. Using the kinematic equation $$v^2 = u^2 - 2as$$ with final velocity zero:

$$0 = (10)^2 - 2a(20)$$

$$a = \frac{100}{40} = 2.5 \text{ m/s}^2$$

Since friction is the only horizontal force causing this deceleration, we have $$a = \mu g$$. So:

$$2.5 = \mu \times 10$$

$$\mu = 0.25$$

Hence, the correct answer is Option 4.

Using conservation of momentum for a perfectly inelastic collision:

$$mv + 2m(0) = (m + 2m)v'$$

$$mv = 3mv'$$

$$v' = \frac{v}{3}$$

The combined mass moves with velocity $$\frac{v}{3}$$.

The correct answer is Option 1.

Average force exerted by a machine gun equals the rate of change of momentum of the fired bullets.

$$F = n \cdot m \cdot v$$

$$125 = n \cdot (10 \times 10^{-3}) \cdot 250$$

$$125 = n \cdot 2.5 \implies n = \frac{125}{2.5} = 50$$

Let mass $$=m,angle, θ=45°$$

Normal reaction:
$$N=mg\cosθ$$

Friction:
$$f=μN=μmg\cosθ$$

Now case 1: just moving up the incline

Friction acts down the plane

$$F₁=mg\sinθ+μmg\cosθ$$

Now case 2: just preventing sliding down

Friction acts up the plane

$$F₂=mg\sinθ−μmg\cosθ.$$

Given:

$$F_1=2F_2$$

Substitute:

$$mg\sinθ+μmg\cosθ=2(mg\sinθ−μmg\cosθ)$$

Cancel mg:

$$\sinθ+μ\cosθ=2\sinθ−2μ\cosθ$$

Bring terms together:

$$\sinθ−2\sinθ=−2μ\cosθ−μ\cosθ$$
$$−\sinθ=−3μ\cosθ$$

So:

$$μ=\frac{(\sinθ)}{(3\cosθ)}$$
$$μ=\frac{(\tanθ)}{3}$$

Now$$θ=45°,\tan45=1$$

So:

$$μ=\frac{1}{3}$$

Final answer:

$$\frac{1}{3}$$

Solution :

Given :

Mass of body,

$$m = 20\text{ kg}$$

Force acts for,

$$t_1 = 20\text{ s}$$

Distance travelled in next 10s after force ceases :

$$s = 50\text{ m}$$

After the force ceases, body moves with constant velocity.

Therefore,

$$v = \frac{s}{t}$$

$$= \frac{50}{10}$$

$$= 5\text{ m s}^{-1}$$

This is the velocity acquired after 20s.

Using,

$$v = u + at$$

Since body starts from rest,

$$u = 0$$

$$5 = 0 + a(20)$$

$$a = \frac{5}{20}$$

$$= 0.25\text{ m s}^{-2}$$

Using Newton’s second law :

$$F = ma$$

$$= 20 \times 0.25$$

$$= 5\text{ N}$$

Final Answer :

$$5\text{ N}$$

Let the mass of the elevator be $$M$$ and the magnitude of its velocity be $$v$$.

For an entire elevator moving vertically, the only forces acting are:
  • its weight $$W = Mg$$ acting downward,
  • the tension $$T$$ in the supporting cable acting upward.

By Newton’s second law the net force along the vertical is
$$T - Mg = M a$$ $$-(1)$$
where $$a$$ is the elevator’s vertical acceleration (taken positive upward).

Case 1: Elevator moving with uniform speed (either upward or downward).
Uniform speed means $$a = 0$$. Substituting $$a = 0$$ in $$(1)$$ gives
$$T - Mg = 0 \; \Longrightarrow \; T = Mg$$ $$-(2)$$

Equation $$(2)$$ shows that whenever the tension equals the weight, the elevator’s acceleration is zero, so it can remain at rest or travel upward/downward with constant speed. Hence Statement I is true.

Case 2: Person of mass $$m$$ standing on the elevator floor while the elevator goes down with increasing speed.
“Down with increasing speed” means the elevator has an acceleration of magnitude $$a$$ directed downward, i.e. upward acceleration is $$-a$$.

For the person, the forces are:
  • true weight $$mg$$ downward,
  • normal reaction $$N$$ from the floor upward.

Taking upward as positive, Newton’s second law for the person gives
$$N - mg = m(-a)$$ $$-(3)$$

Simplifying $$(3)$$,
$$N = m(g - a)$$ $$-(4)$$

Since the elevator is speeding up downward, $$a \gt 0$$, so from $$(4)$$:
$$g - a \lt g \; \Longrightarrow \; N \lt mg$$

The normal force (apparent weight) is less than the actual weight when the elevator accelerates downward. Therefore Statement II is false.

Combining the results: Statement I is true and Statement II is false. According to the given options, this corresponds to Option B.

Answer: Option B

$$\vec{r} = 10t\hat{i} + 15t^2\hat{j} + 7\hat{k}$$

Velocity: $$\vec{v} = \frac{d\vec{r}}{dt} = 10\hat{i} + 30t\hat{j}$$

Acceleration: $$\vec{a} = \frac{d\vec{v}}{dt} = 30\hat{j}$$

Since $$\vec{F} = m\vec{a}$$, the force is along the positive y-axis direction ($$30m\hat{j}$$).

The direction of net force is along the positive y-axis.

A ball of mass $$200$$ g rests on a vertical post of height $$20$$ m. A bullet of mass $$10$$ g hits it horizontally. After collision, both travel independently.

Both the ball and bullet undergo projectile motion from height $$h = 20$$ m with zero initial vertical velocity.

$$h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{10}} = 2$$ s

Ball lands at 30 m: $$v_{\text{ball}} = \frac{30}{2} = 15$$ m/s

Bullet lands at 120 m: $$v_{\text{bullet}} = \frac{120}{2} = 60$$ m/s

Before collision: only bullet has momentum.

$$m_{\text{bullet}} \times u = m_{\text{ball}} \times v_{\text{ball}} + m_{\text{bullet}} \times v_{\text{bullet}}$$

$$0.01 \times u = 0.2 \times 15 + 0.01 \times 60$$

$$0.01u = 3 + 0.6 = 3.6$$

$$u = 360$$ m/s

The initial velocity of the bullet is $$360$$ m/s.

The correct answer is Option 4: $$360$$ m/s.

3015150

Given: mass $$m = 500$$ g = 0.5 kg, velocity $$\vec{v} = (2t\hat{i} + 3t^2\hat{j})$$ m/s.

Acceleration:

$$ \vec{a} = \frac{d\vec{v}}{dt} = (2\hat{i} + 6t\hat{j})\;\text{m/s}^2 $$

At $$t = 1$$ s: $$\vec{a} = (2\hat{i} + 6\hat{j})$$ m/s².

Force:

$$ \vec{F} = m\vec{a} = 0.5(2\hat{i} + 6\hat{j}) = (\hat{i} + 3\hat{j})\;\text{N} $$

Comparing with $$(\hat{i} + x\hat{j})$$: $$x = 3$$.

The correct answer is $$x = 3$$.

Impulse equals the change in momentum. By Newton's third law, the impulse on the gun equals the impulse on the bullet (in magnitude).

The bullet starts from rest and leaves with velocity $$v = 600$$ m/s. Mass of bullet $$m = 10$$ g $$= 0.01$$ kg.

Impulse on bullet = change in momentum of bullet:

$$J = mv - 0 = 0.01 \times 600 = 6 \text{ N s}$$

By Newton's third law, the impulse supplied to the gun is also $$\mathbf{6}$$ N s (in the opposite direction).

The force acting on the particle is $$\vec{F}= -k\left(x\hat{i}+y\hat{j}\right)$$ with $$k=1$$ and the mass is $$m=1\;\text{kg}$$.

Hence the equations of motion in the $$x$$ and $$y$$ directions are
$$m\frac{d^2x}{dt^2}= -x,\qquad m\frac{d^2y}{dt^2}= -y$$
or, since $$m=1$$,
$$\frac{d^2x}{dt^2}= -x,\qquad \frac{d^2y}{dt^2}= -y.$$

Notice that the force always points along $$-\,(x\hat{i}+y\hat{j})$$, i.e. along the radius vector $$\vec{r}_{xy}=x\hat{i}+y\hat{j}$$ lying in the $$xy$$-plane. Therefore the torque about the origin is

$$\vec{\tau}= \vec{r}\times\vec{F}= (x\hat{i}+y\hat{j})\times\bigl(-(x\hat{i}+y\hat{j})\bigr)=\vec{0},$$

which implies that the angular momentum about the origin is conserved. The component of angular momentum along the $$z$$-axis is

$$L_z = (\vec{r}\times\vec{p})_z = x p_y - y p_x = m(xv_y - yv_x).$$

Because $$m=1\;\text{kg}$$, the conserved quantity simplifies to

$$xv_y - yv_x = \text{constant}.$$

Compute this constant using the initial data at $$t=0$$:
$$x_0 = \frac{1}{\sqrt{2}},\quad y_0 = \sqrt{2},\quad v_{x0} = -\sqrt{2},\quad v_{y0} = \sqrt{2}.$$

$$\begin{aligned} x_0v_{y0} - y_0v_{x0} &= \left(\frac{1}{\sqrt{2}}\right)(\sqrt{2}) - (\sqrt{2})(-\sqrt{2}) \\ &= 1 - (-2) \\ &= 3\;\text{m}^2\text{s}^{-1}. \end{aligned}$$

The motion in the $$z$$-direction is unaffected by the force, so $$z(t) = z_0 + v_{z0}t$$. With $$z_0 = 0$$ and $$v_{z0}= \dfrac{2}{\pi}\;\text{m s}^{-1}$$, the condition $$z=0.5\;\text{m}$$ gives $$t = \dfrac{\pi}{4}\;\text{s}$$, but the conserved quantity $$xv_y - yv_x$$ remains the same at every instant.

Therefore, when $$z = 0.5\;\text{m}$$, the value of $$(xv_y - yv_x)$$ is still

$$3\;\text{m}^2\text{s}^{-1}.$$

Concept:
For an inextensible string,

$$\sum_{ }^{ }T⃗⋅a⃗=0$$

i.e., sum of components of accelerations along tension = 0.

Applying to system (counting segments):

$$-4Ta_1-2Ta_2-Ta_3-Ta_4=0$$

Cancel T:

$$-4a_1-2a_2-a_3-a_4=0$$

Final Answer:

$$-4a_1-2a_2-a_3-a_4=0$$

Given values:

Mass $$M = 100\text{ kg}$$

Mass $$m_1 = 10\text{ kg}$$

Mass $$m_2 = 20\text{ kg}$$

Upward acceleration of mass $$m_2$$, $$a_{2,y} = 2\text{ m s}^{-2}$$

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

1. Motion of $$m_2$$ relative to the wedge $$M$$:

For mass $$m_2$$ to move upward relative to the block/wedge $$M$$ with an acceleration $$a_r = 2\text{ m s}^{-2}$$, mass $$m_1$$ must move to the right relative to $$M$$ with the same relative acceleration $$a_r = 2\text{ m s}^{-2}$$ (since the string is inextensible).

2. Equation for mass $$m_2$$ in the vertical direction:

The vertical forces acting on $$m_2$$ are tension $$T$$ pulling upward and weight $$m_2 g$$ pulling downward:

$$$T - m_2 g = m_2 a_r$$$

$$$T - 20(10) = 20(2) \implies T = 200 + 40 = 240\text{ N}$$$

3. Horizontal motion of $$m_1$$ in the frame of reference of wedge $$M$$:

Let the horizontal acceleration of the whole system (the wedge $$M$$) to the right be $$a_x$$.

In the non-inertial reference frame of the wedge $$M$$ moving rightward with acceleration $$a_x$$, a pseudo force $$m_1 a_x$$ acts to the left on $$m_1$$.

The equation of motion for $$m_1$$ along the horizontal direction relative to the wedge $$M$$ is:

$$$m_1 a_x - T = m_1 a_r$$$

Substitute the known values:

$$$10 a_x - 240 = 10(2)$$$

$$$10 a_x = 240 + 20 = 260 \implies a_x = 26\text{ m s}^{-2}$$$

4. Total Applied Force $$F$$:

Considering the entire $$M\ and\ m_2$$, both having same acceleration in x direction

$$F-T=(M+m_2)a_x$$

The value of $$F$$ is 3360 N 

Let us start by considering the forces on the 4M block

There is an external Force pulling it which is $$F=2Mg$$ and $$T\left(tension\ of\ the\ rope\right)$$

we know that here masses M and 4M have the same acceleration , lets take it as 'a'

now writing equation for the 4M block

$$2Mg\ -\ T\ =4Ma$$

and for the M block 

Forces at $$T\left(tension\ of\ the\ rope\right)$$ and $$F=Mg$$(due to gravity)

now writing equation for M block

$$T-Mg=Ma$$

now on adding these both equations

$$2Mg-Mg=5Ma$$

$$Mg=5Ma$$

$$\frac{g}{5}=a$$

Now substituting value of a in equation 1

$$2Mg\ -\ T\ =4M\times\ \frac{g}{5}$$

$$2Mg\ -\frac{4Mg}{5}\ =T$$

$$\frac{\left(10Mg\ -4Mg\right)}{5}\ =T$$

$$\frac{6Mg}{5}\ =T$$

Therefore x=6

We begin by noting that a mass of $$10$$ kg hangs from a rope of length $$5$$ m, and that a horizontal force of $$30$$ N is applied at the midpoint of the rope. Let $$T_1$$ denote the tension in the upper half and $$T_2$$ the tension in the lower half, while $$\alpha$$ and $$\beta$$ are the angles these halves make with the vertical.

Since there is no horizontal force below the midpoint, the lower half supports only the weight of the mass, so it hangs vertically and thus $$\beta = 0$$. Consequently, $$T_2 = mg = 10 \times 10 = 100$$ N.

At the midpoint, three forces are in equilibrium: the tension $$T_1$$ along the upper rope at an angle $$\alpha$$ to the vertical, the downward tension $$T_2 = 100$$ N, and the applied horizontal force $$F = 30$$ N. From horizontal equilibrium we have $$T_1 \sin\alpha = 30$$ ... (i), and from vertical equilibrium we have $$T_1 \cos\alpha = T_2 = 100$$ ... (ii).

Dividing equation (i) by equation (ii) gives $$\tan\alpha = \frac{30}{100} = \frac{3}{10} = 3 \times 10^{-1}$$, so $$\alpha = \tan^{-1}(3 \times 10^{-1})$$.

Comparing this expression for $$\alpha$$ with the given form $$\alpha = \tan^{-1}(x \times 10^{-1})$$ shows that $$x = 3$$.

The answer is $$\boxed{3}$$.

Given: Mass $$m = 100$$ g $$= 0.1$$ kg, Force $$\vec{F} = (10\hat{i} + 5\hat{j})$$ N, initial velocity = 0 (starts from rest), $$t = 2$$ s.

Find the acceleration: Using Newton's second law $$\vec{F} = m\vec{a}$$:

$$\vec{a} = \frac{\vec{F}}{m} = \frac{10\hat{i} + 5\hat{j}}{0.1} = (100\hat{i} + 50\hat{j}) \text{ m/s}^2$$

Find the position at $$t = 2$$ s: Since the object starts from rest ($$\vec{u} = 0$$), using $$\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2$$:

$$\vec{s} = \frac{1}{2}(100\hat{i} + 50\hat{j})(2)^2$$

$$\vec{s} = \frac{1}{2}(100\hat{i} + 50\hat{j})(4)$$

$$\vec{s} = (200\hat{i} + 100\hat{j}) \text{ m}$$

So $$a = 200$$ and $$b = 100$$.

Find $$\frac{a}{b}$$: $$\frac{a}{b} = \frac{200}{100} = 2$$

The answer is 2.

Step 1: Acceleration of system
When the 6th ball just leaves the table:

• Number of hanging balls = 6
• Driving force = 6mg N
• Total mass of system = 10m

$$a=\frac{\text{net force}}{\text{total mass}}=\ \frac{\ 6mg}{10m}=\ \frac{\ 3g}{5}$$

​Step 2: Tension between 7th and 8th ball

Consider the last 3 balls (8th, 9th, 10th) lying on the table:

• Mass of these 3 balls = 3m
• They move with same acceleration $$a=\frac{3g}{5}$$

$$T=(\text{mass})\times(\text{acceleration})$$

 $$T=3m\cdot\frac{3g}{5}$$

$$substitute\ m=2kg,g=10m/s^2:$$

$$T=3\times2\times\ \frac{\ 3\times\ 10}{5}​=36N$$

Final Answer:

36 N

We need to determine the value of $$x$$ given that four concurrent forces acting at point $$P$$ are in a state of translational equilibrium.

1. Resolve the Forces into Components

For the particle to be in equilibrium, the net force along both the horizontal ($$x$$-axis) and vertical ($$y$$-axis) directions must equal zero. Let's resolve the angled forces ($$2\text{ N}$$ and $$1\text{ N}$$) into their components using $$\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$$:

• $$2\text{ N}$$ Force (Top-Left):
  • Horizontal component: $$2 \cos 45^\circ = \frac{2}{\sqrt{2}} = \sqrt{2}\text{ N}$$ (pointing in the negative $$x$$-direction)
  • Vertical component: $$2 \sin 45^\circ = \frac{2}{\sqrt{2}} = \sqrt{2}\text{ N}$$ (pointing in the positive $$y$$-direction)
• $$1\text{ N}$$ Force (Top-Right):
  • Horizontal component: $$1 \cos 45^\circ = \frac{1}{\sqrt{2}}\text{ N}$$ (pointing in the positive $$x$$-direction)
  • Vertical component: $$1 \sin 45^\circ = \frac{1}{\sqrt{2}}\text{ N}$$ (pointing in the positive $$y$$-direction)

2. Set Up the Equilibrium Equations

• Along the horizontal ($$x$$) direction ($$\Sigma F_x = 0$$):

  $$F_1 + 1 \cos 45^\circ - 2 \cos 45^\circ = 0$$

  $$F_1 + \frac{1}{\sqrt{2}} - \frac{2}{\sqrt{2}} = 0$$

  $$F_1 - \frac{1}{\sqrt{2}} = 0 \implies F_1 = \frac{1}{\sqrt{2}}\text{ N}$$

• Along the vertical ($$y$$) direction ($$\Sigma F_y = 0$$):

  $$2 \sin 45^\circ + 1 \sin 45^\circ - F_2 = 0$$

  $$\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} - F_2 = 0$$

  $$F_2 = \frac{3}{\sqrt{2}}\text{ N}$$

3. Find the Ratio ($$F_1 : F_2$$)

Taking the ratio of the calculated forces:

$$\frac{F_1}{F_2} = \frac{\frac{1}{\sqrt{2}}}{\frac{3}{\sqrt{2}}} = \frac{1}{3}$$

Comparing this result with the given ratio format ($$1 : x$$)

$$\frac{1}{x} = \frac{1}{3} \implies x = 3$$

Conclusion

The value of $$x$$ is 3.

We need to find the impulse imparted to the ball when the batsman hits it back without changing its speed.

Since the mass of the ball is $$m = 0.4$$ kg and its speed is $$v = 15$$ m/s, and the ball is hit back in the opposite direction with the same speed, we define a direction convention to set up our calculations.

Let the initial direction of the ball towards the batsman be negative, so $$v_i = -15$$ m/s. After being hit back, the ball moves in the opposite direction, which we take as positive: $$v_f = +15$$ m/s.

Now the impulse can be calculated as the change in momentum: Impulse = Change in momentum = $$m(v_f - v_i)$$, which gives $$0.4 \times (15 - (-15))$$, or equivalently $$0.4 \times 30$$, resulting in $$12$$ N s.

The impulse imparted to the ball is 12 N s.

Based on the provided image, we can accurately solve for x by balancing the forces acting on the block along the inclined plane.

1. Forces and Components

To keep the block stationary on a smooth incline, the net force along the plane must be zero.

• Component of Weight (mg): Acts down the incline as $$mg \sin(60^\circ)$$.
• Component of Horizontal Force (F): Acts up the incline as $$F \cos(60^\circ)$$.

2. Equilibrium Equation

For the block to remain stationary:

$$F \cos(60^\circ) = mg \sin(60^\circ)$$

$$F = mg \tan(60^\circ)$$

3. Numerical Calculation

Given:

• Mass (m) = $$200\text{ g} = 0.2\text{ kg}$$
• $$g = 10\text{ m/s}^2$$
• Weight (m
• g) = $$0.2 \times 10 = 2\text{ N}$$
• Angle ($$\theta$$) = $$60^\circ$$
• Force ($$F$$) = $$\sqrt{x}\text{ N}$$

Substituting the values:

$$\sqrt{x} = 2 \times \tan(60^\circ)$$

$$\sqrt{x} = 2 \times \sqrt{3}$$

Squaring both sides to find $x$:

$$x = (2\sqrt{3})^2$$$$x = 4 \times 3$$

$$\boxed{x = 12}$$

We apply the principle of conservation of linear momentum to solve this problem. The mass of the man is $$m_1 = 60$$ kg, the mass of the trolley is $$m_2 = 120$$ kg, the initial velocity of the trolley is $$0$$ m/s (stationary), and the final velocity of the (man + trolley) system is $$v_f = 2$$ m/s. Let the velocity of the running man be $$v$$ m/s.

Total momentum before equals total momentum after. Substituting into this gives:

$$m_1 \times v + m_2 \times 0 = (m_1 + m_2) \times v_f$$

$$60 \times v + 120 \times 0 = (60 + 120) \times 2$$

$$60v = 180 \times 2$$

$$60v = 360$$

$$v = \frac{360}{60}$$

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

Hence, the velocity of the running man when he jumps into the car is 6 m/s.

A ball of mass $$0.15 \text{ kg}$$ hits the wall at $$12 \text{ m s}^{-1}$$ and bounces back with the same speed under a force of $$100 \text{ N}$$, so we need to find the contact time.

Taking the initial direction of motion as positive, the initial momentum is $$p_i = m \times v = 0.15 \times 12 = 1.8 \text{ kg m s}^{-1}$$. Next, the final momentum after bouncing back is $$p_f = m \times (-v) = 0.15 \times (-12) = -1.8 \text{ kg m s}^{-1}$$. From this, the change in momentum is $$\Delta p = p_f - p_i = -1.8 - 1.8 = -3.6 \text{ kg m s}^{-1}$$.

Since $$F \cdot \Delta t = |\Delta p|$$ and the magnitude of the change in momentum is $$|\Delta p| = 3.6 \text{ kg m s}^{-1}$$, substituting gives $$\Delta t = \dfrac{|\Delta p|}{F} = \dfrac{3.6}{100} = 0.036 \text{ s}$$.

The correct answer is Option B: $$0.036 \text{ s}$$.

1. Identify the Forces acting on the Chain

Let the linear mass density of the chain be λ. Since the chain is uniform, it is given by:

At the given point in time:

• The length of the chain hanging on one side is l. The mass of this section is m1 = λl.
• The length of the chain hanging on the other side is L - l. The mass of this section is m2 = λ(L - l).

Assuming the longer side (L - l) moves downward, the net driving force (Fnet) on the chain is the difference between the gravitational forces acting on both sides:

2. Formulate the Acceleration Equation

The total mass being accelerated is the entire mass of the chain, m = λL.

Using Newton's second law (Fnet = m · a):

3. Substitute the Given Values

We are given that when l = L / x, the acceleration of the chain is a = g / 2. Substituting these into our acceleration equation:

Cancel out g from both sides:

Rearranging the terms to solve for x:

```

We need to find the additional force (both magnitude and direction) that must be applied to the body so that its net acceleration becomes zero, meaning the body remains in translational equilibrium.

1. Find the Net Existing Force

From the free body diagram ,let's break down the given forces along the horizontal ($$x$$) and vertical ($$y$$) axes:

• Along the $$x$$-axis:

  $$\Sigma F_x = 5\text{ N} - 6\text{ N} = -1\text{ N}$$

  (This means there is a net force of $$1\text{ N}$$ pointing to the left).
• Along the $$y$$-axis:

  $$\Sigma F_y = 7\text{ N} - 8\text{ N} = -1\text{ N}$$

  (This means there is a net force of $$1\text{ N}$$ pointing downward).

Expressing the net existing force vector ($$\vec{F}_{\text{net}}$$):

$$\vec{F}_{\text{net}} = -1\hat{i} - 1\hat{j}$$

2. Determine the Balancing Force Vector ($$\vec{F}_{\text{add}}$$)

For the net acceleration of the body to be zero, the total vector sum of all forces must be zero:

$$\vec{F}_{\text{net}} + \vec{F}_{\text{add}} = 0 \implies \vec{F}_{\text{add}} = -\vec{F}_{\text{net}}$$

$$\vec{F}_{\text{add}} = -(-1\hat{i} - 1\hat{j}) = 1\hat{i} + 1\hat{j}$$

3. Calculate Magnitude and Direction

• Magnitude:

  $$|\vec{F}_{\text{add}}| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}\text{ N}$$

• Direction ($\theta$ with the positive $$x$$-axis):

  $$\tan \theta = \frac{F_y}{F_x} = \frac{1}{1} = 1$$

  Since both the $$x$$ and $$y$$ components of our additional force are positive, the vector lies in the first quadrant:

  $$\theta = \tan^{-1}(1) = 45^\circ$$

Conclusion

The additional force that must be applied is $$\sqrt{2}\text{ N}$$ at an angle of $$45^\circ$$ with the positive $$x$$-axis (Option A).

A block of mass $$M$$ inside a box descends vertically with acceleration $$a$$. The block exerts a force equal to one-fourth of its weight on the floor.

When the system descends with acceleration $$a$$, the apparent weight (normal force $$N$$) is:

$$N = M(g - a)$$

The block exerts a force $$N = \frac{Mg}{4}$$ on the floor (by Newton's third law, the floor exerts the same force on the block).

$$M(g - a) = \frac{Mg}{4}$$

$$g - a = \frac{g}{4}$$

$$a = g - \frac{g}{4} = \frac{3g}{4}$$

Hence, the correct answer is Option B: $$\dfrac{3g}{4}$$.

An object of mass 5 kg moving at 25 m/s hits a wall and comes to rest. We compare two scenarios in which it stops in 3 s and in 5 s. The impulse imparted to the object is given by the change in momentum: $$m \times \Delta v$$. In both cases the initial momentum is the same and the final momentum is zero, so the impulse is

$$J = m(v_f - v_i) = 5(0 - 25) = -125 \text{ kg m/s}$$

and its magnitude is 125 N·s.

The average force over the stopping time is calculated as

$$\frac{\text{Impulse}}{\Delta t}$$

When the object stops in 3 s, the average force is

$$F_1 = \frac{125}{3} = 41.67$$ N,

whereas stopping in 5 s yields

$$F_2 = \frac{125}{5} = 25$$ N.

Thus, although the impulse is the same in both cases, the average force differs due to the different time intervals. Therefore, the correct answer is Option B: impulse will be the same for both cases but the average force will be different.

Two billiard balls of mass $$m = 0.05 \text{ kg}$$ each, moving in opposite directions at $$10 \text{ m s}^{-1}$$, collide and rebound with the same speed. Contact time $$t = 0.005 \text{ s}$$.

Calculate the change in momentum of one ball.

Taking the initial direction of one ball as positive, its velocity changes from $$+10 \text{ m s}^{-1}$$ to $$-10 \text{ m s}^{-1}$$:

$$\Delta p = m(v_f - v_i) = 0.05 \times (-10 - 10) = 0.05 \times (-20) = -1 \text{ kg m s}^{-1}$$

Calculate the force using the impulse-momentum theorem.

$$F = \frac{|\Delta p|}{t} = \frac{1}{0.005} = 200 \text{ N}$$

The correct answer is Option B: $$200 \text{ N}$$.

We have three forces $$\vec F_1$$, $$\vec F_2$$ and $$\vec F_3$$. The statement tells us that these three vectors are represented by the three sides of a triangle and that they are taken in the same cyclic order, i.e. the head of $$\vec F_1$$ is joined to the tail of $$\vec F_2$$, the head of $$\vec F_2$$ is joined to the tail of $$\vec F_3$$ and finally the head of $$\vec F_3$$ returns to the tail of $$\vec F_1$$, thereby closing the triangle.

Whenever three vectors form the three sides of a closed triangle in the same order, we can write, by simple head-to-tail addition, the vector equation

$$\vec F_1 + \vec F_2 + \vec F_3 = \vec 0.$$

Rearranging this equation, we obtain

$$\vec F_1 + \vec F_2 = -\vec F_3.$$

Now, the condition for equilibrium of concurrent forces is that the resultant force must be zero. This condition is expressed algebraically as

$$\vec F_{\text{net}} = \sum \vec F = \vec 0.$$

Comparing this general equilibrium condition with the relation $$\vec F_1 + \vec F_2 + \vec F_3 = \vec 0,$$ we see that the resultant of the three forces vanishes. Hence there is no unbalanced force, so the system is in equilibrium. Since we can imagine these three forces acting at the same point (because only their magnitudes and directions matter for the triangle law), they are described as concurrent forces. Therefore, Statement I is true.

Next, let us examine Statement II. The condition for translatory (translational) equilibrium is exactly the same: the vector sum of all external forces acting on the body must be zero, i.e.

$$\sum \vec F = \vec 0.$$

We have already established that if three forces can be arranged to form a closed triangle in the same order, then

$$\vec F_1 + \vec F_2 + \vec F_3 = \vec 0.$$

This directly satisfies the translational equilibrium condition. No net force means no linear acceleration of the body. Hence Statement II is also true.

Both statements are individually correct, and Statement II is merely a restatement of the vector equilibrium condition without going into the concurrency detail. Consequently, both statements are true.

Hence, the correct answer is Option A.

We apply the law of conservation of momentum. Initially, both the bullet and gun are at rest, so the total momentum is zero.

The bullet has mass $$m = 4 \text{ g} = 0.004 \text{ kg}$$ and muzzle speed $$v = 50 \text{ ms}^{-1}$$.

The impulse imparted to the gun equals the change in momentum of the bullet (Newton's third law). The magnitude of impulse is: $$J = mv = 0.004 \times 50 = 0.2 \text{ kg m s}^{-1}$$

By conservation of momentum, the gun (mass $$M = 4 \text{ kg}$$) recoils with velocity $$V$$ such that: $$MV = mv$$ $$V = \frac{mv}{M} = \frac{0.004 \times 50}{4} = \frac{0.2}{4} = 0.05 \text{ m s}^{-1}$$

Therefore, the impulse imparted to the gun is $$0.2 \text{ kg m s}^{-1}$$ and the velocity of recoil is $$0.05 \text{ m s}^{-1}$$.

We consider the motion from the point of view of an observer sitting inside the car. Since the car is accelerating, its interior is a non-inertial frame, so in addition to the real forces we must introduce a pseudo (fictitious) force on every mass in the direction opposite to the car’s acceleration.

The given data are

$$a = 10 \text{ m s}^{-2}$$ (acceleration of the car upward along the plane) $$g = 10 \text{ m s}^{-2}$$ (acceleration due to gravity, vertically downward) $$\alpha = 30^\circ \quad$$ (inclination of the plane to the horizontal)

The real force on the bob is its weight $$\vec{W}=m\vec{g},$$ vertically downward.

The pseudo force is $$\vec{F}_p = -m\vec{a},$$ opposite to the car’s acceleration, i.e. down the plane. Because the car’s acceleration is parallel to the plane and up the slope, the pseudo force is parallel to the plane and down the slope, of magnitude $$|\vec{F}_p| = ma = 10m.$$

We now add these two vectors to obtain the effective force (or effective acceleration) that the bob “feels” in the non-inertial frame:

$$\vec{g}_{\text{eff}} = \vec{g} + \left(-\vec{a}\right).$$

To combine them, we express the pseudo force in components relative to the vertical:

• The direction “down the slope” makes an angle of $$\alpha = 30^\circ$$ below the horizontal. The angle between this direction and the vertical is therefore $$90^\circ - 30^\circ = 60^\circ.$$
• Hence the pseudo force of magnitude $$10m$$ has
   vertical component $$10m\cos 60^\circ,$$
   horizontal (transverse) component $$10m\sin 60^\circ.$$

Writing the magnitudes only (the factor m cancels everywhere),

Vertical component of $$\vec{g}_{\text{eff}}$$: $$g_{\text{vert}} = 10 + 10\cos 60^\circ = 10 + 10\left(\dfrac12\right) = 10 + 5 = 15.$$

Horizontal (lateral) component of $$\vec{g}_{\text{eff}}$$: $$g_{\text{horiz}} = 10\sin 60^\circ = 10\left(\dfrac{\sqrt3}{2}\right) = 5\sqrt3.$$

The string aligns itself along the direction of $$\vec{g}_{\text{eff}},$$ so the angle $$\theta$$ that the string makes with the true vertical is determined by

$$\tan\theta = \dfrac{g_{$$ horiz $$}}{g_{$$ vert $$}} = \dfrac{5\sqrt3}{15} = \dfrac{\sqrt3}{3} = \dfrac1{\sqrt3}.$$

We know the standard trigonometric value

$$\tan 30^\circ = \dfrac1{\sqrt3}.$$

Therefore,

$$\theta = 30^\circ.$$

So, the answer is $$30^\circ$$.

We consider the instantaneous motion of the rocket when its mass is still $$m = 1000\text{ kg}$$ and it has to acquire an upward acceleration of $$a = 20\ \text{m s}^{-2}$$. At that very instant the rocket ejects gases backward with a speed, relative to the rocket, of $$u = 500\ \text{m s}^{-1}$$.

According to the variable-mass (rocket) mechanics, the backward ejection of mass produces a forward thrust. The magnitude of this thrust is obtained from the basic relation

$$\text{Thrust} = \left(\dfrac{dm}{dt}\right)u,$$

where $$\dfrac{dm}{dt}$$ is the rate at which mass is being lost by the rocket (this rate is positive in magnitude although the actual change in mass is negative).

The forces acting on the rocket, all taken along the vertical line of motion, are:

1. Upward thrust $$\left(\dfrac{dm}{dt}\right)u$$,

2. Downward weight $$mg$$.

The net upward force must produce the required upward acceleration. Hence Newton’s second law gives

$$\left(\dfrac{dm}{dt}\right)u - mg = ma.$$

Now we substitute the known numerical values. We have

$$m = 1000\ \text{kg}, \qquad g = 10\ \text{m s}^{-2}, \qquad a = 20\ \text{m s}^{-2}, \qquad u = 500\ \text{m s}^{-1}.$$

Putting these into the above equation,

$$\left(\dfrac{dm}{dt}\right)(500) - (1000)(10) = (1000)(20).$$

First evaluate the right-hand side:

$$ (1000)(20) = 20000. $$

So we have

$$500\dfrac{dm}{dt} - 10000 = 20000.$$

Move the weight term to the right:

$$500\dfrac{dm}{dt} = 20000 + 10000 = 30000.$$

Solve for $$\dfrac{dm}{dt}$$ by dividing both sides by 500:

$$\dfrac{dm}{dt} = \dfrac{30000}{500}.$$

Carry out the division:

$$\dfrac{dm}{dt} = 60\ \text{kg s}^{-1}.$$

Thus the fuel must be burnt at a rate of $$60\ \text{kilograms per second}$$ to give the rocket the desired acceleration.

Hence, the correct answer is Option B.

We need to determine the ratio of the magnitude of impulses imparted to ball a and ball b along the X-direction by a rigid wall.

1. Understand the Relationship Between Impulse and Momentum

According to the impulse-momentum theorem, the impulse ($\vec{J}$) imparted to an object is equal to its change in linear momentum ($\Delta \vec{p}$):

$$\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$$

Since the question specifically asks for the impulse along the X-direction, we only need to look at the change in the horizontal component of velocity for each ball ($J_x = m \cdot \Delta v_x$). Let the initial speed of both balls be $u$.

2. Calculate the Impulse for Ball (a)

From the diagram:

• Ball (a) moves horizontally straight toward the wall along the positive X-direction: $v_{ix} = +u$
• It rebounds directly backward with the same speed along the negative X-direction: $v_{fx} = -u$

The change in velocity along the X-axis is:

$$\Delta v_{ax} = v_{fx} - v_{ix} = -u - u = -2u$$

The magnitude of the impulse for ball (a) is:

$$|J_a| = 2mu$$

3. Calculate the Impulse for Ball (b)

From the diagram for ball (b):

• The ball approaches the wall at an angle of $45^\circ$ with the horizontal line ($X'X$). The initial velocity component directed into the wall is $v_{ix} = +u \cos(45^\circ)$.
• It rebounds elastically at an angle of $45^\circ$, moving away from the wall with a horizontal component of $v_{fx} = -u \cos(45^\circ)$.

The change in velocity along the X-axis is:

$$\Delta v_{bx} = v_{fx} - v_{ix} = -u \cos(45^\circ) - u \cos(45^\circ) = -2u \cos(45^\circ)$$

The magnitude of the impulse for ball (b) is:

$$|J_b| = 2mu \cos(45^\circ) = 2mu \left(\frac{1}{\sqrt{2}}\right) = \sqrt{2}mu$$

4. Determine the Ratio

Now, we take the ratio of the two horizontal impulse magnitudes:

$$\frac{|J_a|}{|J_b|} = \frac{2mu}{\sqrt{2}mu} = \frac{2}{\sqrt{2}} = \frac{\sqrt{2}}{1}$$

Conclusion

The ratio of the magnitude of impulses along the X-direction is $$\sqrt{2} : 1$$, which corresponds exactly to Option B.

$$M = m_{\text{block}} + 3 \times m_{\text{cylinder}} = 10 + 3(20) = 70\ \text{kg}$$

$$Mg - R' = Ma$$

$$\implies R' = M(g - a)$$ $$\implies R' = 70(10 - 0.2) = 70 \times 9.8 = 686\ \text{N}$$

Let the initial velocity of $$m_1$$ be $$u$$ and $$m_2$$ is at rest. After collision, both move with equal speeds $$v$$ in opposite directions. So $$m_1$$ moves with speed $$v$$ in the negative direction and $$m_2$$ moves with speed $$v$$ in the positive direction.

Applying conservation of momentum: $$m_1 u = m_2 v - m_1 v$$, which gives $$m_1 u = v(m_2 - m_1)$$.

For the collision to be physically valid, we also use the coefficient of restitution. For elastic collision, $$e = 1$$: $$e = \frac{v - (-v)}{u - 0} = \frac{2v}{u} = 1$$, giving $$u = 2v$$.

Substituting into the momentum equation: $$m_1(2v) = v(m_2 - m_1)$$, so $$2m_1 = m_2 - m_1$$, which gives $$m_2 = 3m_1$$.

Therefore $$m_2 : m_1 = 3 : 1$$.

Let the tension in the string connected to the $$2\text{ kg}$$ mass be $$T$$.

$$\text{Tension on the movable pulley holding the } 8\text{ kg} \text{ mass} = 2T$$

$$\text{Constraint relation: } 2 \cdot a_1 = a_2 \implies a_{2\text{kg}} = 2a,\ a_{8\text{kg}} = a$$

Equations of motion:

$$8g - 2T = 8a \implies 4g - T = 4a$$

$$T - 2g = 2(2a) \implies T - 2g = 4a$$

Adding the equations:  $$2g = 8a \implies a = \frac{g}{4} = \frac{10}{4} = 2.5\text{ m/s}^2$$

Time taken to strike the ground ($$h = 20\text{ cm} = 0.2\text{ m}$$):

$$h = \frac{1}{2}at^2 \implies 0.2 = \frac{1}{2}(2.5)t^2$$

$$t^2 = \frac{0.4}{2.5} = 0.16 \implies t = 0.4\text{ s}$$

We have a constant force $$\vec F = 40\hat i + 10\hat j \;{\rm N}$$ acting on a body whose mass is $$m = 5\;{\rm kg}$$.

According to Newton’s second law, the relation between force and acceleration is first stated as $$\vec F = m\,\vec a.$$

We now solve for the acceleration $$\vec a$$ by dividing the force vector by the mass:

$$\vec a = \frac{\vec F}{m} = \frac{40\hat i + 10\hat j}{5} = \left(\frac{40}{5}\right)\hat i + \left(\frac{10}{5}\right)\hat j = 8\hat i + 2\hat j\;{\rm m\,s^{-2}}.$$

The body starts from rest, so its initial velocity is

$$\vec v_0 = 0\hat i + 0\hat j\;{\rm m\,s^{-1}}.$$

For motion with constant acceleration, the position vector after time $$t$$ is given by the kinematic formula (which we state explicitly):

$$\vec r = \vec r_0 + \vec v_0\,t + \frac12\,\vec a\,t^{2}.$$

The body is released from the origin, so $$\vec r_0 = 0\hat i + 0\hat j$$. Because $$\vec v_0 = 0$$, the second term also vanishes. Therefore, the position vector simplifies to

$$\vec r = \frac12\,\vec a\,t^{2}.$$

Substituting the acceleration $$\vec a = 8\hat i + 2\hat j$$ and the time $$t = 10\;{\rm s}$$, we obtain

$$\vec r = \frac12\,(8\hat i + 2\hat j)\,(10)^{2} = \frac12\,(8\hat i + 2\hat j)\,(100).$$

Carrying out the multiplication, we get component-wise values:

$$\vec r = \left[\frac12 \times 8 \times 100\right]\hat i + \left[\frac12 \times 2 \times 100\right]\hat j = (4 \times 100)\hat i + (1 \times 100)\hat j = 400\hat i + 100\hat j\;{\rm m}.$$

Thus the particle’s position vector after 10 seconds is $$\left(400\hat i + 100\hat j\right)\,{\rm m}$$.

Hence, the correct answer is Option C.

We have a particle of mass $$m$$ moving only in the $$xy$$-plane, so its velocity may be written as $$\vec v = v_x \hat i + v_y \hat j$$ with $$v_x = \dfrac{dx}{dt}$$ and $$v_y = \dfrac{dy}{dt}$$. The force acting on it is given in the question as

$$\vec F = k\bigl(v_y\hat i + v_x\hat j\bigr).$$

By Newton’s second law $$\vec F = m\vec a$$, therefore the acceleration components satisfy

$$\begin{aligned} a_x &= \dfrac{F_x}{m} = \dfrac{k}{m}\,v_y,\\[4pt] a_y &= \dfrac{F_y}{m} = \dfrac{k}{m}\,v_x. \end{aligned}$$

So the two coupled differential equations for the velocity components are

$$\begin{aligned} \frac{dv_x}{dt} &= \frac{k}{m}\,v_y, \quad\quad (1)\\[6pt] \frac{dv_y}{dt} &= \frac{k}{m}\,v_x. \quad\quad (2) \end{aligned}$$

Now we examine each statement one by one, using these relations.

Testing the constancy of $$\vec v\times\vec a$$ (Option A).

Because motion is confined to the plane, both $$\vec v$$ and $$\vec a$$ have zero $$z$$-components, so their cross-product points along $$\hat k$$ and its magnitude equals the scalar expression

$$\vec v \times \vec a = (v_x\hat i + v_y\hat j)\times\bigl(a_x\hat i + a_y\hat j\bigr) = (v_x a_y - v_y a_x)\,\hat k.$$

Substituting $$a_x$$ and $$a_y$$ from above,

$$\begin{aligned} v_x a_y - v_y a_x &= v_x\left(\frac{k}{m}v_x\right) - v_y\left(\frac{k}{m}v_y\right)\\ &= \frac{k}{m}\left(v_x^2 - v_y^2\right). \end{aligned}$$

We differentiate this quantity with respect to time to see whether it changes:

$$\begin{aligned} \frac{d}{dt}\!\left(v_x a_y - v_y a_x\right) &=\frac{k}{m}\,\frac{d}{dt}\!\left(v_x^2 - v_y^2\right)\\[6pt] &=\frac{k}{m}\,\bigl(2v_x\dot v_x - 2v_y\dot v_y\bigr)\\[6pt] &=\frac{2k}{m}\,\Bigl(v_x\frac{dv_x}{dt} - v_y\frac{dv_y}{dt}\Bigr). \end{aligned}$$

Using Eqs. (1) and (2): $$\dot v_x = \dfrac{k}{m}v_y$$ and $$\dot v_y = \dfrac{k}{m}v_x$$, so

$$\begin{aligned} \frac{d}{dt}\!\left(v_x a_y - v_y a_x\right) &=\frac{2k}{m}\Bigl(v_x\left(\frac{k}{m}v_y\right) - v_y\left(\frac{k}{m}v_x\right)\Bigr)\\[6pt] &=\frac{2k}{m}\cdot\frac{k}{m}\,(v_x v_y - v_y v_x)\\[6pt] &=\frac{2k^2}{m^2}\,(0)=0. \end{aligned}$$

The time derivative is zero, so $$v_x a_y - v_y a_x$$, and hence $$\vec v\times\vec a$$, is a constant vector. Therefore statement A is correct.

Testing whether $$\vec F$$ could be a magnetic force (Option B).

A magnetic (Lorentz) force on a charge $$q$$ has the form $$\vec F = q\,\vec v\times\vec B$$, which is always perpendicular to $$\vec v$$, implying $$\vec v\cdot\vec F = 0$$. For the given force, however,

$$\vec v\cdot\vec F = v_x(kv_y) + v_y(kv_x) = 2k\,v_x v_y,$$

which is in general non-zero. Hence the given force cannot come purely from a magnetic field, so statement B is false.

Testing the constancy of kinetic energy (Option C).

Kinetic energy is $$K = \dfrac12 m v^2 = \dfrac12 m(v_x^2 + v_y^2).$$ Its time derivative is

$$\frac{dK}{dt} = m\,\vec v\cdot\vec a.$$

We have just computed $$\vec v\cdot\vec a = 2\frac{k}{m}v_x v_y$$, so

$$\frac{dK}{dt} = m\left(2\frac{k}{m}v_x v_y\right) = 2k\,v_x v_y,$$

which is not generally zero. Thus kinetic energy is not constant, and statement C is false.

Testing whether $$\vec v\cdot\vec a$$ itself is constant (Option D).

From above $$\vec v\cdot\vec a = 2\dfrac{k}{m}v_x v_y.$$ Differentiating using the product rule and Eqs. (1) and (2),

$$\begin{aligned} \frac{d}{dt}\!\bigl(\vec v\cdot\vec a\bigr) &=2\frac{k}{m}\Bigl(v_x\dot v_y + v_y\dot v_x\Bigr)\\[6pt] &=2\frac{k}{m}\Bigl(v_x\cdot\frac{k}{m}v_x + v_y\cdot\frac{k}{m}v_y\Bigr)\\[6pt] &=2\frac{k^2}{m^2}\bigl(v_x^2 + v_y^2\bigr), \end{aligned}$$

which is strictly positive (unless the particle is at rest). Therefore $$\vec v\cdot\vec a$$ is not constant, and statement D is false.

Only Option A stands up to scrutiny.

Hence, the correct answer is Option A.

Let the magnitudes of the two given forces be denoted by $$|\vec P| = P$$ and $$|\vec Q| = Q$$.

Their resultant is $$\vec R = \vec P + \vec Q$$ and we are told that its magnitude equals that of $$\vec P$$, that is

$$|\vec R| = |\vec P|\; \Longrightarrow\; |\vec R| = P.$$

First we express the magnitude of the resultant $$\vec R$$ in terms of $$P$$, $$Q$$ and the angle $$\alpha$$ between $$\vec P$$ and $$\vec Q$$. The law of vector addition (or the triangle law) gives

$$|\vec R|^2 \;=\; |\vec P|^2 + |\vec Q|^2 + 2\,|\vec P|\,|\vec Q|\,\cos\alpha.$$

Substituting $$|\vec R| = P$$ and $$|\vec P|=P,\; |\vec Q|=Q$$ we have

$$P^2 = P^2 + Q^2 + 2PQ\cos\alpha.$$

Subtracting $$P^2$$ from both sides yields

$$0 = Q^2 + 2PQ\cos\alpha.$$

Dividing by the non-zero quantity $$Q$$ we get

$$Q + 2P\cos\alpha = 0.$$

Hence

$$\cos\alpha = -\dfrac{Q}{2P}.$$

Now we form a new resultant by doubling $$\vec P$$ and adding $$\vec Q$$. Let

$$\vec S = 2\vec P + \vec Q.$$

We wish to find the angle $$\theta$$ between $$\vec S$$ and $$\vec Q$$. By the definition of the dot product,

$$\vec S\cdot\vec Q = |\vec S|\,|\vec Q|\,\cos\theta.$$

Compute the dot product on the left:

$$\vec S\cdot\vec Q = (2\vec P + \vec Q)\cdot\vec Q = 2\,\vec P\cdot\vec Q + \vec Q\cdot\vec Q.$$

The individual dot products are

$$\vec P\cdot\vec Q = |\vec P|\,|\vec Q|\,\cos\alpha = P Q \cos\alpha,$$

$$\vec Q\cdot\vec Q = |\vec Q|^2 = Q^2.$$

Therefore

$$\vec S\cdot\vec Q = 2(P Q \cos\alpha) + Q^2.$$

Substituting the previously obtained value $$\cos\alpha = -\dfrac{Q}{2P}$$, we find

$$\vec S\cdot\vec Q = 2\!\left(P Q \left(-\dfrac{Q}{2P}\right)\right) + Q^2 = 2\!\left(-\dfrac{Q^2}{\,2\,}\right) + Q^2 = -Q^2 + Q^2 = 0.$$

Because $$\vec S\cdot\vec Q = 0$$, the cosine of the angle $$\theta$$ is

$$\cos\theta = \dfrac{\vec S\cdot\vec Q}{|\vec S|\,|\vec Q|} = \dfrac{0}{|\vec S|\,|\vec Q|} = 0.$$

An angle whose cosine is zero is a right angle, namely $$\theta = 90^\circ$$.

Hence, the correct answer is Option D (90°).

Let us denote the ceiling point by $$O$$, the mid-point of the rope where the horizontal force is applied by $$P$$, and the point where the 10 kg mass is attached by $$M$$. The rope is massless and its total length is $$4\ \text{m}$$, so each half measures $$2\ \text{m}$$.

The upper half $$OP$$ of the rope is given to make an angle of $$45^{\circ}$$ with the vertical. Writing the length of this segment and resolving it into vertical and horizontal projections, we have

$$OP = 2\ \text{m},\qquad \angle(OP,\text{vertical}) = 45^{\circ}.$$

Therefore

vertical projection $$= OP\cos 45^{\circ}=2\cos45^{\circ}=2\dfrac{\sqrt2}{2}=\sqrt2\ \text{m},$$
horizontal projection $$= OP\sin45^{\circ}=2\sin45^{\circ}=2\dfrac{\sqrt2}{2}=\sqrt2\ \text{m}.$$

Because the horizontal force is applied only at $$P$$ and the rope is light, the lower half $$PM$$ adjusts until the 10 kg mass remains in equilibrium. For the mass to be at rest it must experience no horizontal force at all; hence the tension in the lower half must act vertically. This immediately shows that the segment $$PM$$ is vertical.

On the mass $$M$$ two forces act: its weight downward $$mg$$ and the tension $$T_2$$ of the lower half upward along the same vertical line. Writing the condition of translational equilibrium,

$$T_2 - mg = 0.$$

Substituting the data $$m = 10\ \text{kg},\ g = 10\ \text{m s}^{-2},$$ we obtain

$$T_2 = mg = 10 \times 10 = 100\ \text{N}.$$

Now we inspect the force balance at the junction point $$P$$. Three forces meet there:

1. The tension $$T_1$$ in the upper half, directed from $$P$$ toward the ceiling along $$PO$$.
2. The tension $$T_2 = 100\ \text{N}$$ in the lower half, directed vertically downward toward the mass.
3. The horizontal force $$F$$ applied to the right (say) at $$P$$.

Because the rope makes $$45^{\circ}$$ with the vertical in the upper half, we resolve $$T_1$$ into its components using

$$T_1^{(\text{vertical})}=T_1\cos45^{\circ},\qquad T_1^{(\text{horizontal})}=T_1\sin45^{\circ}.$$

The equilibrium conditions at $$P$$ are

Vertical: $$T_1\cos45^{\circ}-T_2=0,$$

Horizontal: $$F-T_1\sin45^{\circ}=0.$$

From the vertical balance we solve for $$T_1$$:

$$T_1\cos45^{\circ}=T_2\quad\Longrightarrow\quad T_1=\dfrac{T_2}{\cos45^{\circ}} =\dfrac{100}{\frac{\sqrt2}{2}} =100\cdot\dfrac{2}{\sqrt2} =100\sqrt2\ \text{N}.$$

Substituting this value of $$T_1$$ into the horizontal balance gives

$$F = T_1\sin45^{\circ} =(100\sqrt2)\left(\dfrac{\sqrt2}{2}\right) =100\cdot\dfrac{2}{2} =100\ \text{N}.$$

Hence, the correct answer is Option A.

We consider the spaceship and the interplanetary dust as one isolated system. The important physical fact is that the dust is initially at rest in the chosen inertial frame, so before being swept up it has zero momentum. Because no external force acts on the whole system, the total linear momentum remains conserved at every instant.

The instantaneous linear momentum of the combined system after the dust is collected is simply the momentum of the spacecraft, because the newly captured dust instantly acquires the spacecraft’s velocity. Hence at time $$t$$ the momentum is

$$P(t)=M(t)\,v(t).$$

Momentum conservation in the absence of external force means

$$\frac{dP(t)}{dt}=0.$$

First we expand the derivative by using the product rule. The rule states: if $$P=uv$$ then $$\frac{dP}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}.$$ Taking $$u=M(t)$$ and $$v=v(t)$$, we obtain

$$\frac{d}{dt}\Bigl[M(t)\,v(t)\Bigr]=M(t)\,\frac{dv(t)}{dt}+v(t)\,\frac{dM(t)}{dt}=0.$$

We recognise $$\frac{dv(t)}{dt}$$ as the required instantaneous acceleration, so we denote it by $$a(t)$$. Re-writing,

$$M(t)\,a(t)+v(t)\,\frac{dM(t)}{dt}=0.$$

Solving for $$a(t)$$ gives

$$a(t)=-\frac{v(t)}{M(t)}\,\frac{dM(t)}{dt}.$$

The problem statement gives the mass-accretion rate explicitly:

$$\frac{dM(t)}{dt}=b\,v^{2}(t).$$

We now substitute this expression into the previous equation:

$$a(t)=-\frac{v(t)}{M(t)}\Bigl[b\,v^{2}(t)\Bigr].$$

Multiplying the factors of $$v(t)$$, we find

$$a(t)=-\frac{b\,v^{3}(t)}{M(t)}.$$

This is exactly the form listed in option B.

Hence, the correct answer is Option B.

We have particle A with mass $$m_1$$ and initial velocity $$\vec u_1 = \left(\sqrt{3}\,\hat i + \hat j\right)\,\text{m s}^{-1}$$. Particle B has mass $$m_2$$ and is initially at rest, so $$\vec u_2 = \vec 0$$. After the collision the given velocity of A is $$\vec v_1 = \left(\hat i + \sqrt{3}\,\hat j\right)\,\text{m s}^{-1}$$, while the velocity of B is the unknown vector $$\vec v_2 = a\,\hat i + b\,\hat j$$. The masses are related by $$m_1 = 2m_2$$.

Because the collision takes place in isolation, the law of conservation of linear momentum applies:

$$m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2.$$

Substituting the known quantities and remembering that $$\vec u_2 = \vec 0$$ we get

$$m_1\left(\sqrt{3}\,\hat i + \hat j\right) = m_1\left(\hat i + \sqrt{3}\,\hat j\right) + m_2\left(a\,\hat i + b\,\hat j\right).$$

We now compare the $$\hat i$$ and $$\hat j$$ components separately. First, however, it is convenient to express $$m_2$$ in terms of $$m_1$$. Since $$m_1 = 2m_2$$ we have $$m_2 = \dfrac{m_1}{2}$$.

$$\hat i$$-components:

$$m_1\sqrt{3} = m_1\cdot 1 + \dfrac{m_1}{2}\,a.$$

Dividing by $$m_1$$ gives

$$\sqrt{3} = 1 + \dfrac{a}{2}\;\;\Longrightarrow\;\;\dfrac{a}{2} = \sqrt{3}-1\;\;\Longrightarrow\;\;a = 2\!\left(\sqrt{3}-1\right).$$

$$\hat j$$-components:

$$m_1\cdot 1 = m_1\sqrt{3} + \dfrac{m_1}{2}\,b.$$

Again dividing by $$m_1$$ we obtain

$$1 = \sqrt{3} + \dfrac{b}{2}\;\;\Longrightarrow\;\;\dfrac{b}{2} = 1-\sqrt{3} = -\!\left(\sqrt{3}-1\right)\;\;\Longrightarrow\;\;b = -2\!\left(\sqrt{3}-1\right).$$

Thus the velocity of particle B after the collision is

$$\vec v_2 = a\,\hat i + b\,\hat j = 2\!\left(\sqrt{3}-1\right)\hat i - 2\!\left(\sqrt{3}-1\right)\hat j = 2\!\left(\sqrt{3}-1\right)\left(\hat i - \hat j\right)\,\text{m s}^{-1}.$$

Let us call the angle between $$\vec v_1$$ and $$\vec v_2$$ by the symbol $$\theta$$. To find it we use the definition

$$\cos\theta = \dfrac{\vec v_1\!\cdot\!\vec v_2}{\lvert\vec v_1\rvert\,\lvert\vec v_2\rvert}.$$

Dot product:

$$\vec v_1\!\cdot\!\vec v_2 = \bigl(1\,\hat i + \sqrt{3}\,\hat j\bigr)\!\cdot\! \Bigl[2\bigl(\sqrt{3}-1\bigr)\,(\hat i - \hat j)\Bigr]$$ $$= 2\!\left(\sqrt{3}-1\right)\left[1\cdot 1 + \sqrt{3}\,(-1)\right] = 2\!\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right) = -2\!\left(\sqrt{3}-1\right)^{\!2}.$$

Magnitudes:

$$\lvert\vec v_1\rvert = \sqrt{1^{2} + (\sqrt{3})^{2}} = \sqrt{1+3} = 2,$$ $$\lvert\vec v_2\rvert = 2\!\left(\sqrt{3}-1\right)\sqrt{\,(\hat i - \hat j)\!\cdot\!(\hat i - \hat j)} = 2\!\left(\sqrt{3}-1\right)\sqrt{1+1} = 2\!\left(\sqrt{3}-1\right)\sqrt{2}.$$

Cosine of the angle:

$$\cos\theta = \dfrac{-2\!\left(\sqrt{3}-1\right)^{2}} {2 \times 2\!\left(\sqrt{3}-1\right)\sqrt{2}} = \dfrac{-\left(\sqrt{3}-1\right)}{2\sqrt{2}} \approx \dfrac{-0.732}{2.828} \approx -0.259.$$

The value $$\cos\theta = -0.259$$ corresponds to an angle of approximately $$105^\circ$$ (since $$\cos105^\circ \approx -0.2588$$).

Hence, the correct answer is Option D.

We are given the displacement of a particle executing simple harmonic motion as

$$y \;=\; 5\bigl(\sin 3\pi t \;+\; \sqrt{3}\,\cos 3\pi t\bigr)\ \text{cm}.$$

To identify the amplitude and the time period, we rewrite the expression in the standard SHM form

$$y \;=\; A\sin(\omega t + \phi),$$

where $$A$$ is the amplitude and $$\omega$$ is the angular frequency. We note that the given expression is a linear combination of $$\sin 3\pi t$$ and $$\cos 3\pi t$$ with the same angular frequency $$3\pi\ \text{rad s}^{-1}$$. A standard trigonometric result states that for

$$y \;=\; a\sin\theta \;+\; b\cos\theta,$$

the equivalent single-sine form is

$$y \;=\; \sqrt{a^{2}+b^{2}}\;\sin\!\bigl(\theta+\phi\bigr),$$

where $$\sqrt{a^{2}+b^{2}}$$ equals the amplitude. Here

$$a \;=\; 5,\qquad b \;=\; 5\sqrt{3}.$$

Therefore, the amplitude $$A$$ is

$$A \;=\; \sqrt{a^{2}+b^{2}} \;=\; \sqrt{5^{2} + (5\sqrt{3})^{2}} \;=\; \sqrt{25 + 75} \;=\; \sqrt{100} \;=\; 10\ \text{cm}.$$

Next we determine the time period. The angular frequency present in both sine and cosine terms is

$$\omega \;=\; 3\pi\ \text{rad s}^{-1}.$$

For simple harmonic motion, the relation between the angular frequency $$\omega$$ and the time period $$T$$ is

$$\omega \;=\; \frac{2\pi}{T}.$$

Solving for $$T$$, we obtain

$$T \;=\; \frac{2\pi}{\omega} \;=\; \frac{2\pi}{3\pi} \;=\; \frac{2}{3}\ \text{s}.$$

So the motion has an amplitude of $$10\ \text{cm}$$ and a time period of $$\dfrac{2}{3}\ \text{s}$$.

Hence, the correct answer is Option B.

For a simple pendulum oscillating in air, the time period is given by the standard formula

$$T = 2\pi\sqrt{\dfrac{L}{g}}$$

where $$L$$ is the effective length of the pendulum and $$g$$ is the acceleration due to gravity.

When the bob is completely immersed in a non-viscous liquid, it experiences an upward buoyant force in addition to its weight. Because of this buoyancy, the restoring force acting on the bob is reduced, which effectively reduces the value of gravity appearing in the formula. We now calculate that reduction step by step.

Let $$\rho_b$$ be the density of the material of the bob and $$\rho_l$$ be the density of the liquid. The problem states that

$$\rho_l=\dfrac{1}{16}\,\rho_b.$$

If $$V$$ is the volume of the bob, then

mass of the bob, $$m=\rho_b V,$$

weight of the bob, $$W=mg=\rho_b V g,$$

buoyant force on the bob, $$B=\rho_l V g.$$

The effective weight acting downward while the bob is fully immersed is therefore

$$W_{\text{eff}} = W - B = (\rho_b V g) - (\rho_l V g)= (\rho_b-\rho_l)V g.$$

The motion of the pendulum depends on this effective weight, so we define an effective acceleration $$g_{\text{eff}}$$ by equating $$m g_{\text{eff}}$$ with the effective weight:

$$m g_{\text{eff}} = (\rho_b-\rho_l)V g.$$

Because $$m=\rho_b V,$$ we have

$$\rho_b V g_{\text{eff}} = (\rho_b-\rho_l)V g.$$

Canceling the common factor $$V$$ gives

$$g_{\text{eff}} = g\left(1-\dfrac{\rho_l}{\rho_b}\right).$$

Substituting the given ratio $$\rho_l/\rho_b = 1/16,$$ we find

$$g_{\text{eff}} = g\left(1-\dfrac{1}{16}\right) = g\left(\dfrac{15}{16}\right)=\dfrac{15g}{16}.$$

Now we use the time-period formula again, but with $$g_{\text{eff}}$$ in place of $$g$$:

$$T_{\text{liquid}} = 2\pi\sqrt{\dfrac{L}{g_{\text{eff}}}}.$$

Substituting $$g_{\text{eff}}=\dfrac{15g}{16},$$ we obtain

$$T_{\text{liquid}} = 2\pi\sqrt{\dfrac{L}{\dfrac{15g}{16}}}=2\pi\sqrt{\dfrac{16L}{15g}}.$$

We now express this in terms of the original period $$T$$. Because $$T = 2\pi\sqrt{\dfrac{L}{g}},$$ we write

$$T_{\text{liquid}} = 2\pi\sqrt{\dfrac{L}{g}}\;\sqrt{\dfrac{16}{15}} = T\sqrt{\dfrac{16}{15}}.$$

Recognizing that $$\sqrt{16}=4,$$ the result simplifies to

$$T_{\text{liquid}} = T\cdot\dfrac{4}{\sqrt{15}} = 4T\sqrt{\dfrac{1}{15}}.$$

This is exactly the expression given in Option C.

Hence, the correct answer is Option C.

We have a cylindrical plastic bottle whose own mass is negligible but which is filled with water of volume 310 ml. Remember that

$$1\;\text{ml}=1\;\text{cm}^3=1\times10^{-6}\;\text{m}^3,$$

so the volume of water inside the bottle is

$$V = 310\;\text{ml}=310\times10^{-6}\;\text{m}^3 = 3.10\times10^{-4}\;\text{m}^3.$$

The density of water is given as $$\rho = 10^{3}\;\text{kg\,m}^{-3},$$ therefore the mass of the water (and hence of the whole bottle-water system, because the bottle itself is negligibly light) is

$$m = \rho V = (10^{3}\;\text{kg\,m}^{-3})(3.10\times10^{-4}\;\text{m}^3)=0.310\;\text{kg}.$$

The bottle floats upright in the pond. When it is pushed slightly downward through a small vertical distance $$x$$ and then released, the only net restoring force is the extra buoyant force that arises because an extra volume of water has been displaced.

By Archimedes’ principle, the upward buoyant force equals the weight of the displaced water. If the cross-sectional area of the bottle is $$A,$$ pushing the bottle downward by $$x$$ submerges an additional volume $$A x$$ of water. The mass of this extra displaced water is $$\rho A x,$$ so its weight (and hence the restoring force) is

$$F_{\text{buoyant}} = \rho g (A x).$$

This force acts upward, opposite to the downward displacement, so the net restoring force is

$$F = -\,\rho g A\,x.$$

This is of the same mathematical form as Hooke’s law $$F=-k x$$ with an effective spring constant

$$k = \rho g A.$$

The equation of motion for the vertical oscillation is therefore

$$m\,\frac{d^{2}x}{dt^{2}} + \rho g A\,x = 0.$$

The standard form of simple harmonic motion is $$m\,\ddot x + k x = 0,$$ whose angular frequency is

$$\omega = \sqrt{\frac{k}{m}}.$$

Substituting $$k=\rho g A$$ gives

$$\omega = \sqrt{\frac{\rho g A}{m}}.$$

Now we compute the cross-sectional area. The radius of the bottle is

$$r = 2.5\;\text{cm} = 0.025\;\text{m},$$

so

$$A = \pi r^{2} = \pi (0.025\;\text{m})^{2} = \pi (6.25\times10^{-4}\;\text{m}^{2}) = 1.9635\times10^{-3}\;\text{m}^{2}.$$

Next, calculate the effective spring constant:

$$k = \rho g A = (10^{3}\;\text{kg\,m}^{-3})(9.8\;\text{m\,s}^{-2})(1.9635\times10^{-3}\;\text{m}^{2}).$$

First multiply $$9.8$$ by $$1.9635\times10^{-3},$$ obtaining

$$9.8 \times 1.9635\times10^{-3} = 1.9246\times10^{-2}.$$

Then multiply by $$10^{3}$$:

$$k = 10^{3}\times1.9246\times10^{-2}\;\text{N\,m}^{-1} = 19.246\;\text{N\,m}^{-1}.$$

Finally, substitute $$k$$ and $$m=0.310\;\text{kg}$$ into the formula for $$\omega$$:

$$\omega = \sqrt{\frac{19.246\;\text{N\,m}^{-1}}{0.310\;\text{kg}}} = \sqrt{62.086\;\text{s}^{-2}} = 7.88\;\text{rad\,s}^{-1}.$$

The computed value $$7.88\;\text{rad\,s}^{-1}$$ is very close to $$7.9\;\text{rad\,s}^{-1}$$ listed among the options.

Hence, the correct answer is Option C.

Problem Statement

A ball is thrown vertically up (taken as $$+z\text{-axis}$$) from the ground. The correct momentum-height ($$p\text{-}h$$) diagram is:

• (1) Diagram 1
• (2) Diagram 2
• (3) Diagram 3
• (4) Diagram 4
• Initial velocity = $$u$$ (positive, upwards)
• Acceleration due to gravity = $$-g$$ (downwards)
• Displacement $$s$$ = height $$h$$
• The ball starts at the ground ($$h = 0$$) with maximum upward velocity ($$+u$$). Thus, momentum is at its maximum positive value ($$+p_{\text{max}}$$).
• As height $$h$$ increases, the ball slows down until it reaches maximum height ($$h_{\text{max}}$$), where velocity and momentum become zero ($$p = 0$$).
• Trajectory: The path moves from the positive $$p\text{-axis}$$ downward and rightward toward the $$h\text{-axis}$$ (indicated by an arrow pointing down and right).
• From maximum height ($$h_{\text{max}}$$), the ball begins falling back down. Its height decreases back toward zero.
• Since the motion is downward (opposite to $$+z$$), the velocity and momentum become increasingly negative (moving into the $$-p$$ region).
• Trajectory: The path moves from the $$h\text{-axis}$$ downward and leftward toward the negative $$p\text{-axis}$$ (indicated by an arrow pointing down and left).
• Option (3) correctly shows the arrows continuously pointing along the chronological flow of time (first moving towards maximum height, then returning downwards with negative momentum).

Step-by-Step Derivation

To find the correct relation between momentum ($$p$$) and height ($$h$$), we use kinematic equations.

1. Mathematical Relationship

From the third equation of motion:

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

Given parameters for a ball thrown vertically upward:

Substituting these gives:

$$v^2 = u^2 - 2gh$$

Since momentum is defined as $$p = mv \implies v = \frac{p}{m}$$, we substitute this into the equation:

$$\left(\frac{p}{m}\right)^2 = u^2 - 2gh$$

$$p^2 = m^2(u^2 - 2gh)$$

$$p^2 = m^2u^2 - 2m^2gh$$

This equation is of the form $$p^2 = \text{constant} - kh$$, which mathematically represents a parabola symmetric about the $$h\text{-axis}$$ and opening to the left. This eliminates options (1) and (2) which are straight lines.

2. Direction Analysis (Sign of Momentum)

Let's break down the motion into two halves to determine the correct direction arrows:

First Half: Upward Journey

Second Half: Downward Journey

Conclusion

Combining both halves results in a smooth parabolic curve starting on the positive vertical axis, looping down through $$h_{\text{max}}$$, and finishing on the negative vertical axis.

Looking closely at the directional arrows:

Therefore, the correct option is (3).

For the motion of a projectile close to the surface of the earth (neglecting air resistance) the standard Cartesian equation of the trajectory is first written. We have the well-known result

$$y = x\tan\theta_0 - \frac{g\,x^2}{2\,v_0^{\;2}\cos^{2}\theta_0}.$$

In this formula $$\theta_0$$ is the angle of projection with the horizontal, $$v_0$$ is the initial speed and $$g$$ is the magnitude of the acceleration due to gravity.

The question gives the actual path followed by the projectile as

$$y = 2x - 9x^{2}.$$

Because two different expressions describe the same curve they must be identical term by term in powers of $$x$$. Therefore we compare the coefficient of $$x$$ first:

$$\tan\theta_0 = 2.$$

Now we convert this tangent value into the corresponding sine and cosine, because they will be needed later. Starting with

$$\tan\theta_0 = \frac{\sin\theta_0}{\cos\theta_0} = 2,$$

write $$\sin\theta_0 = 2\cos\theta_0$$ and impose the identity $$\sin^{2}\theta_0 + \cos^{2}\theta_0 = 1.$$ Substituting gives

$$4\cos^{2}\theta_0 + \cos^{2}\theta_0 = 1 \;\Longrightarrow\; 5\cos^{2}\theta_0 = 1 \;\Longrightarrow\; \cos^{2}\theta_0 = \frac{1}{5}.$$

Taking the positive square-root (because the angle of projection is acute in projectile motion) we get

$$\cos\theta_0 = \frac{1}{\sqrt{5}}.$$

From $$\sin\theta_0 = 2\cos\theta_0$$ we immediately have

$$\sin\theta_0 = \frac{2}{\sqrt{5}}.$$

Hence one may write the angle either as $$\theta_0 = \cos^{-1}\!\left(\dfrac{1}{\sqrt{5}}\right)$$ or as $$\theta_0 = \sin^{-1}\!\left(\dfrac{2}{\sqrt{5}}\right).$$ Both representations describe the same direction because they come from the same sine and cosine pair.

Next compare the coefficient of $$x^{2}$$ in the two trajectory equations. From the standard formula the coefficient of $$x^{2}$$ is

$$-\frac{g}{2\,v_0^{\;2}\cos^{2}\theta_0},$$

whereas in the given curve it is $$-9$$. Setting them equal produces

$$-\frac{g}{2\,v_0^{\;2}\cos^{2}\theta_0} = -9 \;\Longrightarrow\; \frac{g}{2\,v_0^{\;2}\cos^{2}\theta_0} = 9.$$

Substituting $$g = 10\text{ m s}^{-2}$$ and $$\cos^{2}\theta_0 = \dfrac{1}{5}$$, we get

$$\frac{10}{2\,v_0^{\;2}\left(\dfrac{1}{5}\right)} = 9 \;\Longrightarrow\; \frac{10}{\dfrac{2v_0^{\;2}}{5}} = 9 \;\Longrightarrow\; \frac{10 \times 5}{2v_0^{\;2}} = 9 \;\Longrightarrow\; \frac{50}{2v_0^{\;2}} = 9.$$

Now multiply both sides by $$2v_0^{\;2}$$ to isolate $$v_0$$:

$$50 = 18v_0^{\;2} \;\Longrightarrow\; v_0^{\;2} = \frac{50}{18} = \frac{25}{9}.$$

Taking the positive square-root gives the initial speed

$$v_0 = \frac{5}{3}\text{ m s}^{-1}.$$

We have therefore found

$$\theta_0 = \cos^{-1}\!\left(\frac{1}{\sqrt{5}}\right), \qquad v_0 = \frac{5}{3}\text{ m s}^{-1}.$$

These values coincide exactly with those specified in Option A.

Hence, the correct answer is Option A.

We are told that the two vectors $$\vec A$$ and $$\vec B$$ have the same magnitude. Let us denote this common magnitude by $$a$$, so that we can write $$|\vec A| = |\vec B| = a$$.

The angle between the two vectors will be denoted by $$\theta$$. Our objective is to find an expression for this angle in terms of the given constant $$n$$.

First, we recall the standard vector identity for the magnitude of the sum of two vectors:

$$|\vec A + \vec B|^2 = |\vec A|^2 + |\vec B|^2 + 2|\vec A|\,|\vec B|\cos\theta.$$

Substituting $$|\vec A| = |\vec B| = a$$ into this formula gives

$$|\vec A + \vec B|^2 \;=\; a^2 + a^2 + 2a\cdot a\cos\theta \;=\; 2a^2(1+\cos\theta).$$

In exactly the same way we use the identity for the magnitude of the difference of two vectors:

$$|\vec A - \vec B|^2 = |\vec A|^2 + |\vec B|^2 - 2|\vec A|\,|\vec B|\cos\theta.$$

Again substituting $$a$$ for the magnitudes, we obtain

$$|\vec A - \vec B|^2 \;=\; a^2 + a^2 - 2a\cdot a\cos\theta \;=\; 2a^2(1-\cos\theta).$$

The statement of the problem says that the magnitude of $$\vec A + \vec B$$ is $$n$$ times the magnitude of $$\vec A - \vec B$$, that is,

$$|\vec A + \vec B| = n\,|\vec A - \vec B|.$$

We square both sides to eliminate the square roots:

$$|\vec A + \vec B|^2 = n^2\,|\vec A - \vec B|^2.$$

Now we substitute the expressions we have already derived:

$$2a^2(1+\cos\theta) = n^2\,[\,2a^2(1-\cos\theta)\,].$$

Because the factor $$2a^2$$ is present on both sides, we can divide it out immediately, giving

$$(1+\cos\theta) = n^2(1-\cos\theta).$$

Let us bring all the terms involving $$\cos\theta$$ to one side so that we can solve for $$\cos\theta$$ explicitly:

$$(1+\cos\theta) = n^2 - n^2\cos\theta,$$

$$\cos\theta + n^2\cos\theta = n^2 - 1,$$

$$\cos\theta\,(1 + n^2) = n^2 - 1,$$

$$\cos\theta = \dfrac{n^2 - 1}{n^2 + 1}.$$

Thus the angle $$\theta$$ between the two vectors is given by the inverse cosine of the right-hand side:

$$\theta = \cos^{-1}\!\left(\dfrac{n^2 - 1}{n^2 + 1}\right).$$

This expression exactly matches Option A.

Hence, the correct answer is Option A.

$$N = mg\cos30^\circ$$

$$f_{\text{max}} = \mu N = \mu mg\cos30^\circ$$

Case 1: Maximum force down the incline before slipping ($$F_1 = 2\text{ N}$$), friction acts up the incline:

$$F_1 + mg\sin30^\circ = f_{\text{max}}$$

$$2 + mg\sin30^\circ = \mu mg\cos30^\circ \quad \text{--- (1)}$$

Case 2: Maximum force up the incline before slipping ($$F_2 = 10\text{ N}$$), friction acts down the incline:

$$F_2 = mg\sin30^\circ + f_{\text{max}}$$

$$10 = mg\sin30^\circ + \mu mg\cos30^\circ \quad \text{--- (2)}$$

Subtracting (1) from (2): $$10 - 2 = 2mg\sin30^\circ \implies 8 = 2mg\left(\frac{1}{2}\right) \implies mg = 8\text{ N}$$

$$2 + 8\left(\frac{1}{2}\right) = \mu (8)\left(\frac{\sqrt{3}}{2}\right)$$

$$2 + 4 = 4\sqrt{3}\mu \implies 6 = 4\sqrt{3}\mu$$

$$\mu = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$$

Case (A) - Push:

$$N_A = mg + F\sin\theta = 5(10) + 20\sin30^\circ = 50 + 10 = 60\text{ N}$$

$$F_{\text{net}, A} = F\cos\theta - \mu N_A = m a_A$$

Case (B) - Pull:

$$N_B = mg - F\sin\theta = 5(10) - 20\sin30^\circ = 50 - 10 = 40\text{ N}$$

$$F_{\text{net}, B} = F\cos\theta - \mu N_B = m a_B$$

$$m(a_B - a_A) = \mu(N_A - N_B)$$

$$5(a_B - a_A) = 0.2(60 - 40) = 0.2 \times 20 = 4$$

$$a_B - a_A = \frac{4}{5} = 0.8\text{ m/s}^{-2}$$

We begin by recalling the relation between force and the change in momentum. The total impulse imparted to a particle in a given time interval equals the integral of the force with respect to time over that interval. Stated mathematically, the impulse $$J$$ over the interval from $$t = 0$$ to $$t = T$$ is given by the formula

$$J \;=\; \int_{0}^{T} F \, dt.$$

The problem tells us that the force varies with time according to $$F = k t,$$ where $$k$$ is a constant and $$t$$ is the time measured from zero. Substituting this expression for $$F$$ in the impulse formula, we have

$$J \;=\; \int_{0}^{T} k t \, dt.$$

Because $$k$$ is a constant, it can be taken outside the integral sign:

$$J \;=\; k \int_{0}^{T} t \, dt.$$

Next, we evaluate the elementary integral of $$t$$ with respect to $$t$$. We use the standard result

$$\int t \, dt \;=\; \frac{t^{2}}{2} + C,$$

where $$C$$ is the constant of integration. Applying the definite-integral limits from $$0$$ to $$T,$$ the constant $$C$$ cancels out, giving

$$\int_{0}^{T} t \, dt \;=\; \left[\frac{t^{2}}{2}\right]_{0}^{T} \;=\; \frac{T^{2}}{2} - \frac{0^{2}}{2} \;=\; \frac{T^{2}}{2}.$$

Substituting this result back into the impulse expression, we obtain

$$J \;=\; k \left(\frac{T^{2}}{2}\right) \;=\; \frac{k T^{2}}{2}.$$

Now we turn to the definition of impulse in terms of momentum. The impulse delivered to the particle equals the change in its momentum. Initially the momentum is given as $$p,$$ and after the force has acted for the time interval $$T,$$ the momentum becomes $$3p.$$ Therefore, the change in momentum, denoted $$\Delta p,$$ is

$$\Delta p \;=\; 3p - p \;=\; 2p.$$

Setting the impulse equal to this change in momentum gives

$$\frac{k T^{2}}{2} \;=\; 2p.$$

We wish to solve this equation for the time $$T.$$ First, multiply both sides by $$2$$ to eliminate the denominator:

$$k T^{2} \;=\; 4p.$$

Next, divide both sides by $$k$$ to isolate $$T^{2}$$:

$$T^{2} \;=\; \frac{4p}{k}.$$

Finally, take the positive square root of both sides (time is positive) to obtain $$T$$:

$$T \;=\; 2 \sqrt{\frac{p}{k}}.$$

Comparing this expression with the options supplied, we see that it exactly matches Option B:

$$2\sqrt{\frac{p}{k}}.$$

Hence, the correct answer is Option B.

We consider the usual equations for the motion of a projectile launched from ground level.

For an initial speed $$u$$ and an angle of projection $$\theta$$ with the horizontal, the time of flight is given by the well-known formula

$$t \;=\; \frac{2u\sin\theta}{g},$$

because the vertical component of velocity is $$u\sin\theta$$ and the projectile returns to the same vertical level after rising for $$u\sin\theta/g$$ seconds and falling for the same duration.

The horizontal range reached in that flight is, by the standard range formula,

$$R \;=\; \frac{u^{2}\sin2\theta}{g}.$$

We are told that the shell actually lands at the given distance $$R$$ from the gun, and that two different angles of projection are possible. Let those angles be $$\theta_{1}$$ and $$\theta_{2}$$, with the corresponding times of flight $$t_{1}$$ and $$t_{2}$$.

Since both angles produce the same range, we must have

$$\sin2\theta_{1} \;=\; \sin2\theta_{2} \;=\; \frac{gR}{u^{2}}.$$

Because $$\sin(180^{\circ}-\alpha)=\sin\alpha$$, the two values of $$2\theta$$ that give the same sine add up to $$180^{\circ}$$, so

$$2\theta_{1} + 2\theta_{2} = 180^{\circ} \;\;\Longrightarrow\;\; \theta_{1} + \theta_{2} = 90^{\circ}.$$

Now their times of flight, from the first formula, are

$$t_{1} = \frac{2u\sin\theta_{1}}{g}, \quad t_{2} = \frac{2u\sin\theta_{2}}{g}.$$

We multiply these two expressions to obtain the required product:

$$ t_{1}t_{2} = \left(\frac{2u\sin\theta_{1}}{g}\right)\!\!\left(\frac{2u\sin\theta_{2}}{g}\right) = \frac{4u^{2}}{g^{2}}\,\sin\theta_{1}\sin\theta_{2}. $$

Because $$\theta_{1}+\theta_{2}=90^{\circ}$$, we may write $$\sin\theta_{2}=\cos\theta_{1}$$, giving

$$\sin\theta_{1}\sin\theta_{2} = \sin\theta_{1}\cos\theta_{1} = \tfrac12\sin2\theta_{1}.$$

But from the range equality stated earlier,

$$\sin2\theta_{1} = \frac{gR}{u^{2}}.$$

Substituting this into the product of sines gives

$$\sin\theta_{1}\sin\theta_{2} = \tfrac12 \left(\frac{gR}{u^{2}}\right) = \frac{gR}{2u^{2}}.$$

Putting this back into the expression for $$t_{1}t_{2}$$, we have

$$ t_{1}t_{2} = \frac{4u^{2}}{g^{2}} \times \frac{gR}{2u^{2}} = \frac{4}{g^{2}}\times\frac{gR}{2} = \frac{2R}{g}. $$

Thus

$$t_{1}t_{2} = \frac{2R}{g}.$$

Hence, the correct answer is Option C.

We have a river whose water is flowing in a straight line with a velocity $$\vec v_r = 2\;{\rm km\,h^{-1}}$$. We may imagine this flow to be along the positive $$x$$-axis.

The swimmer is able to propel himself through still water with a speed $$\lvert \vec v_s\rvert = 4\;{\rm km\,h^{-1}}$$. He can choose any direction; let him make an angle $$\theta$$ with the direction of flow of the river. Thus, measured from the positive $$x$$-axis (down-stream direction), his velocity relative to the water can be written in component form as

$$ \vec v_s = 4\cos\theta\,\hat i + 4\sin\theta\,\hat j, $$

where $$\hat i$$ is the unit vector along the flow (east), and $$\hat j$$ is the unit vector pointing straight across the river towards the opposite bank (north).

The actual or ground velocity of the swimmer, which we shall denote by $$\vec v_{g}$$, is obtained by vector addition of the river velocity and the swimmer’s velocity relative to water:

$$ \vec v_{g} \;=\; \vec v_r + \vec v_s \;=\; (2\,\hat i) + \bigl(4\cos\theta\,\hat i + 4\sin\theta\,\hat j\bigr) \;=\; (2 + 4\cos\theta)\,\hat i + 4\sin\theta\,\hat j. $$

To cross the river straight—that is, to reach the point on the opposite bank directly opposite to his starting point—the swimmer’s resultant velocity must point purely across the river. In mathematical terms, the $$x$$-component of $$\vec v_{g}$$ must vanish:

$$ 2 + 4\cos\theta \;=\; 0. $$

Solving for $$\cos\theta$$ gives

$$ \cos\theta \;=\; -\,\frac{2}{4} \;=\; -\frac12. $$

We recall the standard trigonometric result that $$\cos120^\circ = -\dfrac12$$. Hence,

$$ \theta \;=\; 120^\circ. $$

This angle is measured from the direction of the river’s flow and points upstream (because the cosine is negative), exactly the orientation needed to cancel the down-stream drift.

So the swimmer must head at an angle of $$120^\circ$$ with respect to the flow of the river.

Hence, the correct answer is Option C.

Let us denote the two original forces as $$P$$ and $$Q$$, with magnitudes

$$P = 2F \quad\text{and}\quad Q = 3F.$$

The magnitude of the resultant of two forces that act at an angle $$\theta$$ is given by the law of vector addition (parallelogram law):

$$R \;=\; \sqrt{P^{2} + Q^{2} + 2PQ\cos\theta}.$$

Substituting the given values of $$P$$ and $$Q$$, we obtain

$$R \;=\; \sqrt{(2F)^{2} \;+\; (3F)^{2} \;+\; 2\,(2F)(3F)\cos\theta}.$$

Carrying out the squares and the product,

$$R \;=\; \sqrt{4F^{2} \;+\; 9F^{2} \;+\; 12F^{2}\cos\theta}.$$

Adding the first two terms inside the square root,

$$R \;=\; \sqrt{13F^{2} \;+\; 12F^{2}\cos\theta}.$$

So we have

$$R^{2} \;=\; 13F^{2} + 12F^{2}\cos\theta. \quad -(1)$$

Now, the problem states that the force $$Q$$ is doubled. Therefore, the new magnitude of that force becomes

$$Q' = 2Q = 2(3F) = 6F.$$

The magnitude $$R'$$ of the new resultant (with the same angle $$\theta$$ between $$P$$ and the new $$Q'$$) is then

$$R' \;=\; \sqrt{P^{2} + Q'^{2} + 2PQ'\cos\theta}.$$

Substituting $$P = 2F$$ and $$Q' = 6F$$ gives

$$R' \;=\; \sqrt{(2F)^{2} \;+\; (6F)^{2} \;+\; 2\,(2F)(6F)\cos\theta}.$$

This becomes

$$R' \;=\; \sqrt{4F^{2} \;+\; 36F^{2} \;+\; 24F^{2}\cos\theta}.$$

So,

$$R'^{2} \;=\; 40F^{2} + 24F^{2}\cos\theta. \quad -(2)$$

The crucial information given is that the new resultant is twice the original resultant, i.e.

$$R' = 2R.$$

Squaring this relation, we get

$$R'^{2} = 4R^{2}. \quad -(3)$$

Now we substitute the expressions from (1) and (2) into equation (3):

$$40F^{2} + 24F^{2}\cos\theta \;=\; 4\bigl(13F^{2} + 12F^{2}\cos\theta\bigr).$$

Expanding the right-hand side,

$$40F^{2} + 24F^{2}\cos\theta \;=\; 52F^{2} + 48F^{2}\cos\theta.$$

Next, we collect like terms on one side. Subtract the right-hand side from the left-hand side:

$$\bigl(40F^{2} - 52F^{2}\bigr) + \bigl(24F^{2}\cos\theta - 48F^{2}\cos\theta\bigr) = 0.$$

This simplifies to

$$-12F^{2} \;-\; 24F^{2}\cos\theta = 0.$$

Dividing every term by the non-zero quantity $$-12F^{2}$$, we obtain

$$1 + 2\cos\theta = 0.$$

Hence,

$$2\cos\theta = -1 \quad\Longrightarrow\quad \cos\theta = -\dfrac{1}{2}.$$

The cosine of $$120^{\circ}$$ is $$-1/2$$, so

$$\theta = 120^{\circ}.$$

Hence, the correct answer is Option A.

In a two-dimensional elastic collision, the total linear momentum along both the $$x$$-axis and $$y$$-axis is independently conserved.

$$M \times 10 \cos 30^\circ + 2M \times 5 \cos 45^\circ$$

$$= 2M \times V_1 \cos 30^\circ + MV_2 \cos 45^\circ$$

$$5\sqrt{3} + 5\sqrt{2} = 2v_1 \frac{\sqrt{3}}{2} + \frac{v_2}{\sqrt{2}}$$

$$10 \times M \sin 30^\circ - 2M \times 5 \sin 45^\circ$$

$$= Mv_2 \sin 45^\circ - 2Mv_1 \sin 30^\circ$$

$$5 - 5\sqrt{2} = \frac{v_2}{\sqrt{2}} - v_1$$

$$\text{Solving } v_1 = \frac{17.5}{2.7} \simeq 6.5\text{ m/s}$$

$$v_2 \approx 6.3\text{ m/s}$$

We begin by noting that the surface is perfectly smooth, so there is no external horizontal force acting on the combined system of the man and his son. Because of this, the total linear momentum of the system in the horizontal direction must remain constant. This is simply an application of the Law of Conservation of Linear Momentum, which states:

$$\text{Total initial momentum} \;=\;\text{Total final momentum}.$$

Initially, both the man and the son are at rest, so their initial momenta are zero. Hence, the total initial momentum of the system is

$$0 + 0 = 0.$$

Now the man pushes the son. Let us choose the direction in which the son eventually moves as the positive direction. We define:

$$v_s = \text{speed of the son with respect to the ground}$$ $$v_m = \text{speed of the man with respect to the ground}.$$

Because they move apart while facing each other, the man travels in the opposite (negative) direction to the son. Therefore, in our chosen sign convention, the man's final momentum is negative. Applying the conservation of momentum, we write

$$\bigl(\text{mass of son}\bigr)\,(+v_s) \;+\;\bigl(\text{mass of man}\bigr)\,(-v_m) \;=\;0.$$

Substituting their masses, we have

$$20\,v_s \;-\;50\,v_m \;=\;0.$$

Solving this linear equation for $$v_m$$ gives

$$50\,v_m = 20\,v_s,$$ $$v_m = \frac{20}{50}\,v_s,$$ $$v_m = 0.4\,v_s.$$

Next, we use the information about their relative speed. The son moves away from the man at a speed of $$0.70\;\text{m s}^{-1}$$ with respect to the man. The relative speed of two bodies moving in opposite directions equals the sum of their individual speeds with respect to the ground. Therefore, we write

$$v_{\text{relative}} = v_s + v_m.$$

Substituting the given relative speed and the relationship $$v_m = 0.4\,v_s$$ obtained above, we get

$$0.70 = v_s + v_m = v_s + 0.4\,v_s = 1.4\,v_s.$$

Now we solve for $$v_s$$:

$$v_s = \frac{0.70}{1.4} = 0.50\;\text{m s}^{-1}.$$

Substituting this back into $$v_m = 0.4\,v_s$$ gives

$$v_m = 0.4 \times 0.50 = 0.20\;\text{m s}^{-1}.$$

Because we assigned the man's motion opposite to the son's as the negative direction, the magnitude of his speed with respect to the surface is simply $$0.20\;\text{m s}^{-1}.$$

Hence, the correct answer is Option A.

Let the mass be $$m = 10 \text{ kg}$$. Its weight acts vertically downward and is given by the well-known relation $$W = mg$$.

Substituting $$g = 10 \text{ m s}^{-2}$$, we get

$$W = m g = 10 \times 10 = 100 \text{ N}.$$

The horizontal pull applied somewhere on the rope makes the upper part of the rope tilt so that, at the roof, the rope now forms an angle $$\theta = 45^{\circ}$$ with the vertical. The whole system is stated to be in equilibrium; therefore the net force on the suspended mass is zero.

The only forces acting on the mass itself are

(i) its weight $$W = 100 \text{ N}$$ downward, and

(ii) the tension in the rope, $$T$$, directed along the rope, i.e. at $$45^{\circ}$$ to the vertical.

We resolve the tension into mutually perpendicular components, remembering the standard trigonometric relations

$$T_{\text{vertical}} = T \cos\theta, \qquad T_{\text{horizontal}} = T \sin\theta.$$

Because the mass neither rises nor falls, the vertical components must balance:

$$T\cos\theta = W.$$

Substituting $$\theta = 45^{\circ}$$ and $$W = 100 \text{ N}$$, we have

$$T \cos45^{\circ} = 100.$$

Using $$\cos45^{\circ} = \dfrac{\sqrt2}{2},$$ we obtain

$$T \left(\dfrac{\sqrt2}{2}\right) = 100 \;\;\Longrightarrow\;\; T = 100 \times \dfrac{2}{\sqrt2} = 100\sqrt2 \text{ N}.$$

Now we look at the horizontal balance. In equilibrium the horizontal component of the tension must be cancelled by the externally applied horizontal force, which we call $$F$$. Hence

$$F = T\sin\theta.$$

Again inserting $$\theta = 45^{\circ}$$ and $$T = 100\sqrt2 \text{ N},$$ we write

$$F = (100\sqrt2)\sin45^{\circ} = (100\sqrt2)\left(\dfrac{\sqrt2}{2}\right).$$

Simplifying the product of the radicals,

$$F = 100 \left(\dfrac{\sqrt2 \times \sqrt2}{2}\right) = 100 \left(\dfrac{2}{2}\right) = 100 \text{ N}.$$

Hence, the correct answer is Option A.

$$\text{Area}_{0 \rightarrow 2} = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2$$

$$\text{Area}_{2 \rightarrow 4} = \text{Area of rectangle} = \text{width} \times \text{height} = (4 - 2) \times 2 = 4$$

$$\text{Area}_{4 \rightarrow 5} = \text{Area of rectangle} = \text{width} \times \text{height} = (5 - 4) \times 3 = 3$$

$$x(5) = \text{Area}_{0 \rightarrow 2} + \text{Area}_{2 \rightarrow 4} + \text{Area}_{4 \rightarrow 5} = 2 + 4 + 3 = 9\text{ m}$$

We have two bodies moving along the same straight line (take this as the positive x-axis). Initially, the momentum of the system is the vector sum of the individual momenta.

The first body has mass $$m$$ and speed $$2v$$ in the +x direction, so its momentum is $$ p_{1i}=m\,(2v)=2mv. $$

The second body has mass $$2m$$ and speed $$v$$, also in the +x direction, so its momentum is $$ p_{2i}=2m\,(v)=2mv. $$

Hence the total initial momentum is $$ p_{\text{initial}} = p_{1i}+p_{2i}=2mv+2mv = 4mv. $$

After the collision the first mass stops completely, so its final momentum is zero. The second body breaks into two equal parts, each of mass $$m$$. Let the speed of each fragment be $$u$$ and let both fragments make an angle $$45^\circ$$ with the original +x-axis, one above and one below, as described.

Because the angles are symmetric, the components of their momenta perpendicular to the +x-axis cancel each other, leaving only the components along the +x-direction.

For one fragment, the momentum component along +x is $$ p_{x} = m\,u\cos 45^\circ. $$

Since $$\cos 45^\circ=\dfrac{1}{\sqrt 2}$$, this becomes $$ p_{x}=m\,u\frac{1}{\sqrt 2}. $$

There are two such fragments, so the total final momentum along the +x-direction is $$ p_{\text{final}} = 2\left(m\,u\frac{1}{\sqrt 2}\right)=\frac{2mu}{\sqrt 2}=\sqrt 2\,m\,u. $$

By the law of conservation of linear momentum, the initial and final momenta along the +x-axis must be equal (no external force acts horizontally), hence $$ p_{\text{initial}} = p_{\text{final}}. $$

Substituting the expressions obtained: $$ 4mv = \sqrt 2\,m\,u. $$

Now divide both sides by $$m$$ to remove the common factor: $$ 4v = \sqrt 2\,u. $$

Finally, isolate $$u$$ by dividing both sides by $$\sqrt 2$$: $$ u = \frac{4v}{\sqrt 2} = 2\sqrt 2\,v. $$

So, the speed of each fragment is $$2\sqrt 2\,v$$, which corresponds to Option C.

Hence, the correct answer is Option C.

For any collision in a straight line, the total linear momentum before impact equals the total linear momentum after impact. We therefore start with the conservation of momentum equation:

$$m_1 v_1 + m_2 v_2 = m_1 v_3 + m_2 v_4.$$

The data given in the question are

$$m_2 = 0.5\,m_1 \quad\text{and}\quad v_3 = 0.5\,v_1.$$

Substituting these two relations into the momentum equation, we obtain

$$m_1 v_1 + \bigl(0.5\,m_1\bigr) v_2 = m_1 \bigl(0.5\,v_1\bigr) + \bigl(0.5\,m_1\bigr) v_4.$$

Now every term contains the common factor $$m_1,$$ so we divide the entire equation by $$m_1$$ to simplify:

$$v_1 + 0.5\,v_2 = 0.5\,v_1 + 0.5\,v_4.$$

We next collect like terms. First, move the term $$0.5\,v_1$$ on the right side to the left:

$$v_1 - 0.5\,v_1 + 0.5\,v_2 = 0.5\,v_4.$$

Simplifying the coefficients of $$v_1$$ gives

$$0.5\,v_1 + 0.5\,v_2 = 0.5\,v_4.$$

To clear the factor $$0.5$$ from every term, multiply the entire equation by $$2$$:

$$v_1 + v_2 = v_4.$$

Finally, solving for $$v_1$$ yields

$$v_1 = v_4 - v_2.$$

Comparing this expression with the options supplied, we see that it exactly matches Option D.

Hence, the correct answer is Option D.

We begin with the motion of the block when it is allowed to slide down the rough incline without any external pull. Let the mass be $$m = 2\ \text{kg}$$, the angle of the incline be $$\theta = 30^{\circ}$$ and the downward acceleration be $$a = 3\ \text{m s}^{-2}$$. The acceleration due to gravity is given to be $$g = 10\ \text{m s}^{-2}$$.

The component of the weight along the plane is obtained from the trigonometric resolution

$$W_{\parallel}= mg\sin\theta.$$

Substituting the numerical values,

$$W_{\parallel}= 2 \times 10 \times \sin 30^{\circ}= 20 \times \dfrac12 = 10\ \text{N}.$$

The normal reaction is

$$N = mg\cos\theta = 2 \times 10 \times \cos 30^{\circ} = 20 \times \dfrac{\sqrt3}{2}=10\sqrt3\ \text{N}\approx 17.32\ \text{N}.$$

Let the coefficient of kinetic friction be $$\mu_k$$. The frictional force (opposing the motion) is therefore

$$f = \mu_k N = \mu_k \; mg \cos\theta.$$

Along the line of the slope, the downward component of weight drives the motion while friction opposes it. Newton’s second law in the downward direction gives

$$mg\sin\theta - f = ma.$$

Substituting $$f = \mu_k mg\cos\theta$$ we obtain

$$mg\sin\theta - \mu_k mg\cos\theta = ma.$$

Dividing every term by $$m$$ to simplify,

$$g\sin\theta - \mu_k g\cos\theta = a,$$

and therefore

$$\mu_k g\cos\theta = g\sin\theta - a.$$

Now substitute the numbers:

$$\mu_k \;(10)\cos 30^{\circ}= 10\sin 30^{\circ}-3.$$

$$\mu_k \;(10)\times\dfrac{\sqrt3}{2}= 10\times\dfrac12 -3.$$

$$5\sqrt3\,\mu_k = 5-3 = 2.$$

$$\mu_k = \dfrac{2}{5\sqrt3}=\dfrac{2\sqrt3}{15}\approx 0.231.$$

Knowing $$\mu_k$$ makes it possible to find the actual frictional force:

$$f = \mu_k N = \mu_k mg\cos\theta = \left(\dfrac{2}{5\sqrt3}\right)\bigl(2\times10\cos30^{\circ}\bigr) = \dfrac{2}{5\sqrt3}\bigl(20\times\dfrac{\sqrt3}{2}\bigr)=4\ \text{N}.$$

Next we tackle the second part of the problem. Now the block has to move up the same incline with an upward acceleration of the same magnitude, viz. $$a = 3\ \text{m s}^{-2}$$. We apply an external force $$F$$ parallel to the plane and directed upward. Choosing the upward direction as positive, we write Newton’s second law once again.

The forces acting along the plane while the block is accelerating upward are

• External pull $$F$$ upward (positive).
• Component of weight $$mg\sin\theta = 10\ \text{N}$$ downward (negative).
• Friction, which always opposes the motion, is now downward. Its magnitude is already known to be $$4\ \text{N}$$, so it is taken as $$-4\ \text{N}$$ in the upward sign convention.

Applying $$\sum F = ma$$ with the upward direction taken as positive, we therefore have

$$F - mg\sin\theta - f = m a.$$

Substituting the numerical values,

$$F - 10 - 4 = 2 \times 3.$$

$$F - 14 = 6.$$

$$F = 6 + 14 = 20\ \text{N}.$$

Thus an external force of $$20\ \text{N}$$ parallel to the incline and directed upward is necessary to make the block climb the plane with an acceleration of $$3\ \text{m s}^{-2}$$.

Hence, the correct answer is Option D.

We consider the same object sliding the same distance $$s$$ along two different 45$$^\circ$$ planes – one perfectly smooth (no friction) and one rough (kinetic friction coefficient $$\mu_k=\mu$$). Because the initial speed is zero in both cases, the usual constant-acceleration kinematics apply.

First we recall the basic translational kinematic relation for motion with constant acceleration starting from rest:

$$s=\frac12\,a\,t^{2}\;.$$

Here $$a$$ is the magnitude of acceleration along the plane and $$t$$ is the time taken to cover the distance $$s$$.

Smooth 45$$^\circ$$ incline
The component of gravitational acceleration along a smooth plane is

$$a_{\text{smooth}}=g\sin45^\circ=\frac{g}{\sqrt2}\;.$$

Denote the corresponding time of slide by $$t_s$$. Substituting this acceleration into the displacement formula, we have

$$s=\frac12\,a_{\text{smooth}}\,t_s^{2} =\frac12\left(\frac{g}{\sqrt2}\right)t_s^{2}\;.$$

Rough 45$$^\circ$$ incline
With kinetic friction, the frictional force equals $$\mu N$$, where $$N$$ is the normal reaction. For a 45$$^\circ$$ plane, $$N=mg\cos45^\circ=\frac{mg}{\sqrt2}$$. Hence the frictional force magnitude is $$\mu\,\dfrac{mg}{\sqrt2}$$, acting upward along the plane. The net force along the plane therefore equals

$$mg\sin45^\circ-\mu mg\cos45^\circ =\frac{mg}{\sqrt2}-\mu\left(\frac{mg}{\sqrt2}\right) =\frac{mg}{\sqrt2}(1-\mu)\;.$$

Dividing by the mass $$m$$ gives the net acceleration down the plane:

$$a_{\text{rough}}=\frac{g}{\sqrt2}(1-\mu)\;.$$

Let $$t_r$$ be the time of slide on the rough plane. Using the same displacement formula for this case, we write

$$s=\frac12\,a_{\text{rough}}\,t_r^{2} =\frac12\left[\frac{g}{\sqrt2}(1-\mu)\right]t_r^{2}\;.$$

Given relation between the two times
The problem states that the rough slide takes $$n$$ times longer than the smooth slide, so

$$t_r=n\,t_s\;.$$

Equating the two expressions for the same distance $$s$$

$$\frac12\left(\frac{g}{\sqrt2}\right)t_s^{2} =\frac12\left[\frac{g}{\sqrt2}(1-\mu)\right]\!\!\left(n\,t_s\right)^{2}\;.$$

The common factors $$\dfrac12$$, $$\dfrac{g}{\sqrt2}$$, and $$t_s^{2}$$ cancel immediately, leaving

$$1=(1-\mu)\,n^{2}\;.$$

Solving this simple linear equation for $$\mu$$, we proceed step by step:

$$1 = n^{2} - n^{2}\mu$$

$$n^{2}\mu = n^{2} - 1$$

$$\mu = \frac{n^{2}-1}{n^{2}}$$

$$\mu = 1-\frac{1}{n^{2}}\;.$$

This matches Option 2 in the given list.

Hence, the correct answer is Option 2.

Let us examine the situation from the frame of reference of the rocket itself. Because the rocket is accelerating upward with an acceleration $$2g$$, an observer inside the rocket has to introduce a pseudo-force to apply Newton’s laws in the usual way.

Real weight acts downward: $$\vec W = m\vec g\,.$$ Pseudo-force (opposite to the rocket’s upward acceleration) is also downward and equals $$m(2g)\,.$$

Therefore the total effective force that behaves like “weight” for the mass inside the rocket is the vector sum of the two forces, both pointing downward:

$$\vec W_{\text{eff}} = m\vec g + m(2g)\,\hat j = m(3g)\,\hat j\,.$$

In other words, the block feels an effective gravitational acceleration

$$g_{\text{eff}} = 3g\,.$$

The inclined plane makes an angle $$\theta$$ with the horizontal. For such an incline the usual resolution of the weight into components gives:

Component of the effective weight perpendicular to the plane: $$N = mg_{\text{eff}}\cos\theta = m(3g)\cos\theta = 3mg\cos\theta\,.$$

Component of the effective weight parallel to (down) the plane: $$F_{\parallel} = mg_{\text{eff}}\sin\theta = m(3g)\sin\theta = 3mg\sin\theta\,.$$

The static friction force that can oppose this tendency to slide has a maximum value given by the formula

$$f_{\text{max}} = \mu_s N = \mu_s(3mg\cos\theta)\,.$$

For the mass to remain at rest relative to the inclined surface, the frictional force must be at least as large as the downslope component of the effective weight. Hence we demand

$$\mu_s(3mg\cos\theta) \ge 3mg\sin\theta\,.$$

Cancel the common factor $$3mg$$ from both sides:

$$\mu_s\cos\theta \ge \sin\theta\,.$$

Solving for $$\mu_s$$ gives

$$\mu_s \ge \frac{\sin\theta}{\cos\theta} = \tan\theta\,.$$

Thus the least (minimum) value of the coefficient of friction that will just prevent motion is

$$\mu_{\text{min}} = \tan\theta\,.$$

Hence, the correct answer is Option B.

$$F_{net} = mg - kv$$

$$ma = mg - kv \implies a = g - \frac{k}{m}v$$

1. Behaviour of acceleration

At $$t = 0$$: The mass is just starting to fall, so its speed $$v = 0$$. This means the initial acceleration is at its maximum, i.e., $$a = g$$.

As $$t$$ increases: The body speeds up. As $$v$$ increases, the term $$\frac{k}{m}v$$ also increases, which causes the net acceleration $$a$$ to decrease.

As $$t \to \infty$$: Eventually, the drag force equals the gravitational force ($$kv = mg$$). At this point, the acceleration becomes zero, and the body reaches a constant speed called terminal velocity ($$v_T$$).

The acceleration curve must start at a high value and decay exponentially toward zero.

2. Behaviour of velocity

At $$t = 0$$: The body starts from rest ($$v = 0$$).

As $$t$$ increases: Since the acceleration is positive, the speed increases. However, because the acceleration is decreasing over time, the rate of increase of the speed also slows down.

As $$t \to \infty$$: The speed approaches the terminal velocity $$v_T = \frac{mg}{k}$$ asymptotically.

The speed curve must start at the origin and rise asymptotically toward a horizontal line.

Option C is the only graph where the acceleration ($$a$$) starts at a peak and decays to zero, while the speed ($$v$$) starts at zero and levels off at a terminal value.

$$W_{total} = W_A + W_B$$

$$W_{total} = 20\ \text{N} + 100\ \text{N} = 120\ \text{N}$$

The only external force acting upwards to prevent the system from falling is the frictional force ($$f_w$$) exerted by the wall on block B.

Since the system is in equilibrium, the net vertical force is zero, i.e., $$\sum F_y = 0$$

$$f_w - W_{total} = 0$$

$$f_w = W_{total}$$

$$f_w = 120\ \text{N}$$

In elastic collisions between identical masses ($$m$$), the two bodies simply exchange velocities.

Since the beads have a finite radius $$r$$, they occupy space on the rod. Diameter of one bead $$= 2r$$. Total length occupied by $$n$$ beads $$= 2nr$$.

The actual free distance the momentum has to travel between the two supports is $$L_{eff} = L - 2nr$$

The time interval between two consecutive collisions with the same support is $$T = \frac{2 \times L_{eff}}{v} = \frac{2(L - 2nr)}{v}$$

change in momentum ($$\Delta p$$) during one collision is $$\Delta p = mv - (-mv) = 2mv$$

$$F_{avg} = \frac{\Delta p}{T}$$

$$F_{avg} = \frac{2mv}{\frac{2(L - 2nr)}{v}}$$

$$F_{avg} = \frac{mv^2}{L - 2nr}$$

A heavy box is dragged along a rough horizontal floor. The coefficient of friction is given as $$\mu = \frac{\sqrt{3}}{5}$$. Person A pushes the box at an angle of 30° from the horizontal with a minimum force $$F_A$$, and person B pulls it at an angle of 60° from the horizontal with a minimum force $$F_B$$. We need to find the ratio $$\frac{F_A}{F_B}$$.

First, consider the forces acting on the box. The weight of the box is $$mg$$, where $$m$$ is the mass and $$g$$ is the acceleration due to gravity. The normal force $$N$$ is the force exerted by the floor perpendicular to the surface. The friction force opposes motion and is given by $$f = \mu N$$. The minimum force required to move the box occurs when the horizontal component of the applied force equals the maximum static friction force.

Start with person A, who pushes the box at 30° to the horizontal. When pushing, the vertical component of the force acts downward, increasing the normal force. Resolve $$F_A$$ into components:

• Horizontal component: $$F_A \cos 30^\circ$$
• Vertical component: $$F_A \sin 30^\circ$$ (downward)

The normal force $$N_A$$ is:

$$ N_A = mg + F_A \sin 30^\circ $$

The friction force is:

$$ f_A = \mu N_A = \mu (mg + F_A \sin 30^\circ) $$

At the point of motion, the horizontal component equals the friction force:

$$ F_A \cos 30^\circ = \mu (mg + F_A \sin 30^\circ) $$

Substitute the trigonometric values: $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$ and $$\sin 30^\circ = \frac{1}{2}$$:

$$ F_A \cdot \frac{\sqrt{3}}{2} = \mu \left( mg + F_A \cdot \frac{1}{2} \right) $$

Given $$\mu = \frac{\sqrt{3}}{5}$$, substitute:

$$ F_A \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5} \left( mg + \frac{F_A}{2} \right) $$

Divide both sides by $$\sqrt{3}$$ (since $$\sqrt{3} \neq 0$$):

$$ \frac{F_A}{2} = \frac{1}{5} \left( mg + \frac{F_A}{2} \right) $$

Multiply both sides by 5 to eliminate the denominator:

$$ 5 \cdot \frac{F_A}{2} = mg + \frac{F_A}{2} $$

$$ \frac{5F_A}{2} = mg + \frac{F_A}{2} $$

Subtract $$\frac{F_A}{2}$$ from both sides:

$$ \frac{5F_A}{2} - \frac{F_A}{2} = mg $$

$$ \frac{4F_A}{2} = mg $$

$$ 2F_A = mg $$

$$ F_A = \frac{mg}{2} $$

Now, consider person B, who pulls the box at 60° to the horizontal. When pulling, the vertical component acts upward, reducing the normal force. Resolve $$F_B$$ into components:

• Horizontal component: $$F_B \cos 60^\circ$$
• Vertical component: $$F_B \sin 60^\circ$$ (upward)

The normal force $$N_B$$ is:

$$ N_B = mg - F_B \sin 60^\circ $$

The friction force is:

$$ f_B = \mu N_B = \mu (mg - F_B \sin 60^\circ) $$

At the point of motion, the horizontal component equals the friction force:

$$ F_B \cos 60^\circ = \mu (mg - F_B \sin 60^\circ) $$

Substitute the trigonometric values: $$\cos 60^\circ = \frac{1}{2}$$ and $$\sin 60^\circ = \frac{\sqrt{3}}{2}$$:

$$ F_B \cdot \frac{1}{2} = \mu \left( mg - F_B \cdot \frac{\sqrt{3}}{2} \right) $$

Given $$\mu = \frac{\sqrt{3}}{5}$$, substitute:

$$ \frac{F_B}{2} = \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3} F_B}{2} \right) $$

Multiply both sides by 10 to eliminate denominators:

$$ 10 \cdot \frac{F_B}{2} = 10 \cdot \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3} F_B}{2} \right) $$

$$ 5F_B = 2\sqrt{3} \left( mg - \frac{\sqrt{3} F_B}{2} \right) $$

Expand the right side:

$$ 5F_B = 2\sqrt{3} \cdot mg - 2\sqrt{3} \cdot \frac{\sqrt{3} F_B}{2} $$

Simplify the second term:

$$ 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} F_B = \sqrt{3} \cdot \sqrt{3} F_B = 3F_B $$

So:

$$ 5F_B = 2\sqrt{3} \, mg - 3F_B $$

Add $$3F_B$$ to both sides:

$$ 5F_B + 3F_B = 2\sqrt{3} \, mg $$

$$ 8F_B = 2\sqrt{3} \, mg $$

Divide both sides by 2:

$$ 4F_B = \sqrt{3} \, mg $$

$$ F_B = \frac{\sqrt{3} \, mg}{4} $$

Now, find the ratio $$\frac{F_A}{F_B}$$:

$$ \frac{F_A}{F_B} = \frac{\frac{mg}{2}}{\frac{\sqrt{3} \, mg}{4}} = \frac{mg}{2} \cdot \frac{4}{\sqrt{3} \, mg} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} $$

The ratio $$\frac{F_A}{F_B} = \frac{2}{\sqrt{3}}$$, which matches option B.

Hence, the correct answer is Option B.

We are given the mass of the body, $$ m = 5 $$ kg, the initial velocity at $$ t = 0 $$ s as $$ \vec{u} = (6\hat{i} - 2\hat{j}) $$ m/s, and the final velocity at $$ t = 10 $$ s as $$ \vec{v} = +6\hat{j} $$ m/s. The force $$ \vec{F} = F_x\hat{i} + F_y\hat{j} $$ is constant. We need to find $$ \vec{F} $$.

According to Newton's second law, the force is given by $$ \vec{F} = m\vec{a} $$, where $$ \vec{a} $$ is the acceleration. Acceleration is the rate of change of velocity, so $$ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} $$.

First, calculate the change in velocity $$ \Delta \vec{v} $$:

$$ \Delta \vec{v} = \vec{v} - \vec{u} = (6\hat{j}) - (6\hat{i} - 2\hat{j}) $$

Distribute the negative sign:

$$ \Delta \vec{v} = 6\hat{j} - 6\hat{i} + 2\hat{j} $$

Combine like terms:

$$ \Delta \vec{v} = -6\hat{i} + (6 + 2)\hat{j} = -6\hat{i} + 8\hat{j} \text{ m/s} $$

The time interval $$ \Delta t = t_{\text{final}} - t_{\text{initial}} = 10 - 0 = 10 $$ s.

Now, find the acceleration $$ \vec{a} $$:

$$ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} = \frac{-6\hat{i} + 8\hat{j}}{10} $$

Separate the components:

$$ \vec{a} = \left( \frac{-6}{10} \right) \hat{i} + \left( \frac{8}{10} \right) \hat{j} $$

Simplify the fractions:

$$ \vec{a} = -\frac{6}{10}\hat{i} + \frac{8}{10}\hat{j} = -\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \text{ m/s}^2 $$

Now, use Newton's second law to find the force:

$$ \vec{F} = m \vec{a} = 5 \times \left( -\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \right) $$

Distribute the mass to each component:

$$ \vec{F} = 5 \times \left( -\frac{3}{5} \right) \hat{i} + 5 \times \left( \frac{4}{5} \right) \hat{j} $$

Simplify each term:

$$ \vec{F} = \left( 5 \times -\frac{3}{5} \right) \hat{i} + \left( 5 \times \frac{4}{5} \right) \hat{j} $$

$$ \vec{F} = \left( -3 \right) \hat{i} + \left( 4 \right) \hat{j} $$

So, $$ \vec{F} = (-3\hat{i} + 4\hat{j}) $$ N.

Comparing with the options:

• A. $$ (-3\hat{i} + 4\hat{j}) $$ N
• B. $$ \left( -\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \right) $$ N
• C. $$ (3\hat{i} - 4\hat{j}) $$ N
• D. $$ \left( \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j} \right) $$ N

Our result matches option A.

Hence, the correct answer is Option A.

First, we have a bullet of mass 4 g fired horizontally at 300 m/s into a wooden block of mass 0.8 kg at rest. The bullet embeds into the block, making this an inelastic collision. We use conservation of momentum to find the velocity of the block-bullet system right after the collision.

Convert the bullet's mass to kilograms: mass of bullet, $$ m_b = 4 $$ g = $$ 0.004 $$ kg. Mass of block, $$ m_{bl} = 0.8 $$ kg. Initial velocity of bullet, $$ u_b = 300 $$ m/s. Initial velocity of block, $$ u_{bl} = 0 $$ m/s. Let $$ v $$ be the common velocity after collision.

Conservation of momentum:

$$ m_b u_b + m_{bl} u_{bl} = (m_b + m_{bl}) v $$

Substitute the values:

$$ (0.004 \times 300) + (0.8 \times 0) = (0.004 + 0.8) v $$

$$ 1.2 + 0 = 0.804 v $$

$$ 1.2 = 0.804 v $$

Solve for $$ v $$:

$$ v = \frac{1.2}{0.804} = \frac{1200}{804} $$

Simplify the fraction by dividing numerator and denominator by 12:

$$ \frac{1200 \div 12}{804 \div 12} = \frac{100}{67} $$

So, $$ v = \frac{100}{67} $$ m/s ≈ 1.4925 m/s.

Now, the block-bullet system slides on the table with initial velocity $$ v = \frac{100}{67} $$ m/s and comes to a stop due to friction. The coefficient of friction is $$ \mu = 0.3 $$. The friction force opposes the motion.

The normal force $$ N $$ equals the weight of the system since the table is horizontal. Total mass, $$ m = m_b + m_{bl} = 0.004 + 0.8 = 0.804 $$ kg. Using $$ g = 9.8 $$ m/s², weight = $$ m g = 0.804 \times 9.8 $$ N.

Friction force, $$ f = \mu N = \mu m g = 0.3 \times 0.804 \times 9.8 $$.

By Newton's second law, friction causes deceleration $$ a $$:

$$ f = m a $$

$$ 0.3 \times 0.804 \times 9.8 = 0.804 a $$

The mass $$ 0.804 $$ cancels out:

$$ a = 0.3 \times 9.8 = 2.94 \text{ m/s}^2 $$

So, deceleration $$ a = -2.94 $$ m/s² (negative indicates opposition to motion).

We need to find the distance $$ s $$ the block slides before stopping. Initial velocity $$ u = v = \frac{100}{67} $$ m/s, final velocity $$ v_f = 0 $$ m/s, acceleration $$ a = -2.94 $$ m/s².

Use the equation of motion:

$$ v_f^2 = u^2 + 2 a s $$

Substitute the values:

$$ 0^2 = \left( \frac{100}{67} \right)^2 + 2 \times (-2.94) \times s $$

$$ 0 = \frac{10000}{4489} - 5.88 s $$

Solve for $$ s $$:

$$ 5.88 s = \frac{10000}{4489} $$

$$ s = \frac{10000}{4489 \times 5.88} $$

Convert $$ 5.88 $$ to a fraction: $$ 5.88 = \frac{588}{100} = \frac{147}{25} $$.

$$ s = \frac{10000}{4489} \times \frac{1}{5.88} = \frac{10000}{4489} \times \frac{25}{147} $$

$$ s = \frac{10000 \times 25}{4489 \times 147} $$

Calculate numerator: $$ 10000 \times 25 = 250000 $$.

Calculate denominator: $$ 4489 \times 147 $$. Break it down:

$$ 4489 \times 100 = 448900 $$

$$ 4489 \times 40 = 179560 $$

$$ 4489 \times 7 = 31423 $$

Add them: $$ 448900 + 179560 = 628460 $$, then $$ 628460 + 31423 = 659883 $$.

So,

$$ s = \frac{250000}{659883} \approx 0.37878 \text{ m} $$

Approximately, $$ s \approx 0.379 $$ m.

Comparing with the options, 0.379 m corresponds to Option B.

Hence, the correct answer is Option B.

Mass $$m$$: $$\vec{v}_1 = 3u \hat{i}$$

Mass $$2m$$: $$\vec{v}_2 = 2u(-\cos 60^\circ \hat{i} - \sin 60^\circ \hat{j}) = -u \hat{i} - u\sqrt{3} \hat{j}$$

Mass $$3m$$: $$\vec{v}_3 = u(-\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = -\frac{u}{2} \hat{i} + \frac{u\sqrt{3}}{2} \hat{j}$$

$$P_x = m(3u) + 2m(-u) + 3m\left(-\frac{u}{2}\right) = 3mu - 2mu - 1.5mu = -0.5mu$$

$$P_y = m(0) + 2m(-u\sqrt{3}) + 3m\left(\frac{u\sqrt{3}}{2}\right) = -2mu\sqrt{3} + 1.5mu\sqrt{3} = -0.5mu\sqrt{3}$$

$$\vec{P}_{initial} = -\frac{mu}{2} \hat{i} - \frac{mu\sqrt{3}}{2} \hat{j}$$

$$\vec{P}_{final} = \vec{P}_{initial}$$

$$6m \cdot \vec{V} = -\frac{mu}{2} \hat{i} - \frac{mu\sqrt{3}}{2} \hat{j}$$

$$\vec{V} = \frac{u}{12} (-\hat{i} - \sqrt{3}\hat{j})$$

Initially, the applied force increases from zero ($$F = kt$$). The block remains stationary as long as the force is less than or equal to the maximum limiting static friction ($$f_{s,max} = \mu_s mg$$). Acceleration: $$a = 0$$ for $$t \leq \frac{\mu_s mg}{k}$$.

Once motion begins, the friction coefficient suddenly drops from static ($$\mu_s$$) to kinetic ($$\mu_k$$, where $$\mu_k < \mu_s$$). This sudden reduction in friction causes an instantaneous net force, forcing the acceleration to jump vertically from zero to a finite value.

Once the block is moving, kinetic friction remains constant ($$f_k = \mu_k mg$$). The equation of motion becomes:

$$kt - \mu_k mg = ma \implies a = \left(\frac{k}{m}\right)t - \mu_k g$$

Since this follows a linear form ($$y = mx + c$$), the acceleration increases along a straight line with a positive slope.

The correct graph showing a flat line at zero, a vertical jump, and a linear incline is Option B.