JEE Centre of Mass Questions

The positions of the two identical masses just before the collision are

$$x_1(t)=\bigl(x_0+d\bigr)+a\sin\omega t,\qquad x_2(t)=\bigl(x_0-d\bigr)-a\sin\omega t.$$

Differentiating, their velocities are

$$v_1(t)=a\omega\cos\omega t,\qquad v_2(t)=-a\omega\cos\omega t.$$

The collision takes place at $$t_0=0$$, so

$$v_{1i}=v_1(0)=a\omega,\qquad v_{2i}=v_2(0)=-a\omega.$$

Particle 3 (mass $$m$$) approaches from the left with speed

$$u_0=\tfrac{a\omega}{2}$$

towards the right (+ve $$x$$ direction). All three masses are equal, so a one-dimensional perfectly elastic collision simply makes the two colliding particles exchange their velocities.

Hence, immediately after the collision,

$$v_{3f}=-a\omega,\qquad v_{2f}=u_0=\tfrac{a\omega}{2},\qquad v_{1f}=v_{1i}=a\omega \;(\text{particle 1 is untouched}).$$

The centre-of-mass velocity of particles 1 and 2 after the collision is

$$v_{cm}=\frac{v_{1f}+v_{2f}}{2}=\frac{a\omega+\tfrac{a\omega}{2}}{2} =\frac{3a\omega/2}{2}= \frac{3}{4}\,a\omega.$$

Therefore,

$$\frac{v_{cm}}{a\omega}=\frac{3}{4}=0.75.$$

The required value is 0.75.

We have two bodies. The first body has mass $$m_1 = 2\ \text{kg}$$ and is moving with initial speed $$u_1 = 4\ \text{m s}^{-1}$$. The second body has mass $$m_2$$ (unknown for now) and is initially at rest, so $$u_2 = 0$$.

Because the collision is perfectly elastic, two basic conservation laws can be applied:

1. Conservation of linear momentum $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$$

2. Conservation of kinetic energy $$\dfrac12 m_1 u_1^{2} + \dfrac12 m_2 u_2^{2} = \dfrac12 m_1 v_1^{2} + \dfrac12 m_2 v_2^{2}.$$

After the collision, the first body continues in the same direction but with one-fourth of its initial speed. Hence

$$v_1 = \dfrac{u_1}{4} = \dfrac{4}{4} = 1\ \text{m s}^{-1}.$$

Let the speed of the second body after collision be $$v_2$$ (to be found). We now substitute the known numbers into the two conservation equations.

Step 1: Momentum conservation $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$2 \times 4 + m_2 \times 0 = 2 \times 1 + m_2 v_2$$ $$8 = 2 + m_2 v_2.$$

Rearranging gives $$m_2 v_2 = 8 - 2 = 6. \quad -(1)$$

Step 2: Kinetic-energy conservation $$\dfrac12 m_1 u_1^{2} + 0 = \dfrac12 m_1 v_1^{2} + \dfrac12 m_2 v_2^{2}$$ $$\dfrac12 \times 2 \times 4^{2} = \dfrac12 \times 2 \times 1^{2} + \dfrac12 m_2 v_2^{2}$$ $$1 \times 16 = 1 \times 1 + \dfrac12 m_2 v_2^{2}$$ $$16 = 1 + \dfrac12 m_2 v_2^{2}.$$

Subtracting 1 from both sides, we obtain $$\dfrac12 m_2 v_2^{2} = 15$$ $$m_2 v_2^{2} = 30. \quad -(2)$$

Step 3: Solving for $$v_2$$ We divide equation (2) by equation (1): $$\dfrac{m_2 v_2^{2}}{m_2 v_2} = \dfrac{30}{6}$$ $$v_2 = 5\ \text{m s}^{-1}.$$

Step 4: Determining $$m_2$$ (only for completeness) Using equation (1): $$m_2 = \dfrac{6}{v_2} = \dfrac{6}{5} = 1.2\ \text{kg}.$$

Step 5: Velocity of the centre of mass The centre of mass (COM) velocity is constant throughout the motion. It can be found from the initial momenta: $$v_{\text{cm}} = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}.$$

Substituting the values, we get $$v_{\text{cm}} = \dfrac{2 \times 4 + 1.2 \times 0}{2 + 1.2} = \dfrac{8}{3.2}\ \text{m s}^{-1}.$$

Now, $$\dfrac{8}{3.2} = 2.5\ \text{m s}^{-1}.$$

The problem states that the COM speed can be expressed as $$\dfrac{x}{10}\ \text{m s}^{-1}$$. So we set

$$\dfrac{x}{10} = 2.5 \quad\Longrightarrow\quad x = 2.5 \times 10 = 25.$$

So, the answer is $$25$$.

For a uniform quarter disc of radius $$a$$ lying in the first quadrant (from $$\theta = 0$$ to $$\theta = \frac{\pi}{2}$$), we find the center of mass coordinates by integration.

Using polar coordinates, the $$x$$-coordinate of the center of mass is $$\bar{x} = \frac{\int_0^{\pi/2}\int_0^a r\cos\theta \cdot r \, dr \, d\theta}{\int_0^{\pi/2}\int_0^a r \, dr \, d\theta}$$. The denominator (total mass divided by $$\sigma$$) is $$\int_0^{\pi/2}\int_0^a r \, dr \, d\theta = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}$$.

The numerator is $$\int_0^{\pi/2}\int_0^a r^2 \cos\theta \, dr \, d\theta = \frac{a^3}{3} \cdot [\sin\theta]_0^{\pi/2} = \frac{a^3}{3} \cdot 1 = \frac{a^3}{3}$$.

Therefore $$\bar{x} = \frac{a^3/3}{\pi a^2/4} = \frac{4a}{3\pi}$$. By symmetry, $$\bar{y} = \frac{4a}{3\pi}$$ as well.

The center of mass is at $$\left(\frac{4a}{3\pi}, \frac{4a}{3\pi}\right)$$. Comparing with the given form $$\left(\frac{xa}{3\pi}, \frac{xa}{3\pi}\right)$$, we get $$x = 4$$.

We take the solid hemisphere of radius $$R=8\,\text{cm}$$. Because of symmetry, its centre of mass will lie on the central vertical axis passing through the centre of the flat circular face. We measure every distance $$z$$ from this flat face, so $$z=0$$ at the plane of the face and $$z=R$$ at the pole of the hemisphere.

To locate the centre of mass we use the continuous form of the centre-of-mass formula for volume distributions:

$$z_{\text{cm}}=\dfrac{\displaystyle\int z\,dV}{\displaystyle\int dV}.$$

We slice the hemisphere into thin circular discs perpendicular to the axis. A slice at distance $$z$$ from the base has thickness $$dz$$ and radius $$r$$. From the equation of a sphere we have

$$r^{2}+z^{2}=R^{2}\;\Longrightarrow\;r^{2}=R^{2}-z^{2}.$$

The volume of this thin disc is

$$dV=\pi r^{2}\,dz=\pi\bigl(R^{2}-z^{2}\bigr)\,dz.$$

Now we write the numerator of the centre-of-mass expression:

$$\int_{0}^{R} z\,dV=\int_{0}^{R} z\,\pi\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\int_{0}^{R}\bigl(R^{2}z - z^{3}\bigr)\,dz.$$

Integrating term by term,

$$\pi\left[\frac{R^{2}z^{2}}{2}-\frac{z^{4}}{4}\right]_{0}^{R} =\pi\left(\frac{R^{2}R^{2}}{2}-\frac{R^{4}}{4}\right) =\pi\left(\frac{R^{4}}{2}-\frac{R^{4}}{4}\right) =\pi\,\frac{R^{4}}{4}.$$

Next we find the denominator (the total volume of the hemisphere):

$$\int_{0}^{R} dV=\int_{0}^{R}\pi\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\int_{0}^{R}\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\left[R^{2}z-\frac{z^{3}}{3}\right]_{0}^{R} =\pi\left(R^{3}-\frac{R^{3}}{3}\right) =\pi\,\frac{2R^{3}}{3}.$$

Substituting these two results into the centre-of-mass formula,

$$z_{\text{cm}}=\frac{\pi\,\dfrac{R^{4}}{4}} {\pi\,\dfrac{2R^{3}}{3}} =\frac{R^{4}}{4}\times\frac{3}{2R^{3}} =R\,\frac{3}{8} =\frac{3R}{8}.$$

Putting $$R=8\,\text{cm}$$, we obtain

$$z_{\text{cm}}=\frac{3\times8}{8}\,\text{cm}=3\,\text{cm}.$$

So, the answer is $$3\,\text{cm}$$.

Let us denote the mass of each body by $$m$$ and the common magnitude of their initial velocities by $$v$$. Thus, at the very beginning each body possesses a momentum vector of magnitude

$$p_0 = m\,v$$

but the two momentum vectors point in different directions within the same plane. Suppose the angle between these two directions is $$\theta$$ (this is precisely the angle we are required to find).

To combine the two individual momentum vectors into a single “net” or “resultant” initial momentum, we treat the vectors exactly the way we add two sides of a parallelogram. For two vectors of equal magnitude $$p_0$$ separated by an angle $$\theta$$, the magnitude of their resultant, by the law of cosines for vectors, is

$$\lvert\vec P_{\text{initial}}\rvert \;=\;\sqrt{p_0^{\,2} + p_0^{\,2} + 2\,p_0\,p_0\,\cos\theta}\;=\;\sqrt{2p_0^{\,2}\,(1+\cos\theta)}$$

Inside the square root we notice the standard trigonometric identity $$1 + \cos\theta = 2\cos^{2}\dfrac{\theta}{2}$$. Substituting this, we obtain

$$\lvert\vec P_{\text{initial}}\rvert \;=\;\sqrt{2p_0^{\,2}\,\bigl(2\cos^{2}\tfrac{\theta}{2}\bigr)} \;=\; \sqrt{4p_0^{\,2}\,\cos^{2}\tfrac{\theta}{2}} \;=\; 2p_0\,\cos\dfrac{\theta}{2}.$$

Replacing $$p_0$$ by $$m\,v$$ gives

$$\lvert\vec P_{\text{initial}}\rvert \;=\;2\,m\,v\,\cos\dfrac{\theta}{2}.$$

Now the collision is stated to be completely inelastic, meaning the two bodies stick together and thereafter behave like a single composite body. The mass of this composite body is simply the sum of the individual masses, namely

$$m_{\text{final}} \;=\; m + m \;=\; 2m.$$

The problem also tells us that the common final speed of this combined body is half of the original speed, that is

$$v_{\text{final}} \;=\; \dfrac{v}{2}.$$

Therefore the magnitude of the final momentum is

$$\lvert\vec P_{\text{final}}\rvert \;=\; m_{\text{final}}\,v_{\text{final}} \;=\; 2m \times \dfrac{v}{2} \;=\; m\,v.$$

According to the principle of conservation of linear momentum, the total momentum of an isolated system remains constant, so we can equate the magnitudes of the initial and final momenta:

$$2\,m\,v\,\cos\dfrac{\theta}{2} \;=\; m\,v.$$

Dividing both sides by the common positive factor $$m\,v$$ leaves us with a very simple trigonometric equation:

$$2\,\cos\dfrac{\theta}{2} \;=\; 1.$$

We now isolate the cosine term:

$$\cos\dfrac{\theta}{2} \;=\; \dfrac{1}{2}.$$

The standard value of the cosine function that gives one-half is $$60^\circ$$, so we can write

$$\dfrac{\theta}{2} \;=\; 60^\circ.$$

Multiplying both sides by 2, we finally obtain

$$\theta \;=\; 120^\circ.$$

So, the answer is $$120^\circ$$.

We first note the general expression for the position of the centre of mass of a continuous one-dimensional object:

$$x_{\text{cm}}=\frac{1}{M}\int_{0}^{l}x\,\mathrm d m$$

Here the linear mass density is given as $$\rho(x)=a+b\left(\dfrac{x}{l}\right)^2$$, so for an infinitesimal element of length $$\mathrm d x$$ at position $$x$$ we have

$$\mathrm d m=\rho(x)\,\mathrm d x=\left[a+b\left(\frac{x}{l}\right)^2\right]\mathrm d x.$$

We therefore need two separate integrals: one for the total mass $$M$$ and one for the numerator of $$x_{\text{cm}}$$.

Total mass

$$\begin{aligned} M &=\int_{0}^{l}\rho(x)\,\mathrm d x \\ &=\int_{0}^{l}\left[a+b\left(\frac{x}{l}\right)^2\right]\mathrm d x \\ &=\int_{0}^{l}a\,\mathrm d x+\int_{0}^{l}b\left(\frac{x}{l}\right)^2\mathrm d x. \end{aligned}$$

We integrate term by term:

$$\int_{0}^{l}a\,\mathrm d x=a\bigl[x\bigr]_{0}^{l}=a\,l,$$

and

$$\int_{0}^{l}b\left(\frac{x}{l}\right)^2\mathrm d x =b\int_{0}^{l}\frac{x^{2}}{l^{2}}\mathrm d x =\frac{b}{l^{2}}\left[\frac{x^{3}}{3}\right]_{0}^{l} =\frac{b}{l^{2}}\cdot\frac{l^{3}}{3} =\frac{b\,l}{3}.$$

Adding these contributions,

$$M=a\,l+\frac{b\,l}{3}=l\left(a+\frac{b}{3}\right).$$

Numerator for $$x_{\text{cm}}$$

We next evaluate $$\displaystyle\int_{0}^{l}x\,\rho(x)\,\mathrm d x$$:

$$\begin{aligned} \int_{0}^{l}x\,\rho(x)\,\mathrm d x &=\int_{0}^{l}x\left[a+b\left(\frac{x}{l}\right)^2\right]\mathrm d x \\ &=\int_{0}^{l}a\,x\,\mathrm d x+\int_{0}^{l}b\,x\left(\frac{x}{l}\right)^2\mathrm d x. \end{aligned}$$

For the first part,

$$\int_{0}^{l}a\,x\,\mathrm d x =a\left[\frac{x^{2}}{2}\right]_{0}^{l} =a\cdot\frac{l^{2}}{2}=\frac{a\,l^{2}}{2}.$$

For the second part,

$$\int_{0}^{l}b\,x\left(\frac{x}{l}\right)^2\mathrm d x =b\int_{0}^{l}\frac{x^{3}}{l^{2}}\mathrm d x =\frac{b}{l^{2}}\left[\frac{x^{4}}{4}\right]_{0}^{l} =\frac{b}{l^{2}}\cdot\frac{l^{4}}{4} =\frac{b\,l^{2}}{4}.$$

Hence the numerator is

$$\int_{0}^{l}x\,\rho(x)\,\mathrm d x =\frac{a\,l^{2}}{2}+\frac{b\,l^{2}}{4} =l^{2}\left(\frac{a}{2}+\frac{b}{4}\right) =l^{2}\left(\frac{2a+b}{4}\right).$$

Centre of mass position

Substituting the numerator and the total mass into the formula, we get

$$x_{\text{cm}}=\frac{l^{2}\left(\dfrac{2a+b}{4}\right)}{\,l\left(a+\dfrac{b}{3}\right)}.$$ We simplify step by step:

First cancel one factor of $$l$$:

$$x_{\text{cm}}=l\,\frac{\,2a+b\,}{4\left(a+\dfrac{b}{3}\right)}.$$

Next write the denominator over a common denominator of three:

$$a+\frac{b}{3}=\frac{3a+b}{3}.$$

So

$$x_{\text{cm}}=l\,\frac{2a+b}{4}\cdot\frac{3}{3a+b} =l\,\frac{3(2a+b)}{4(3a+b)}.$$

Finally, pull out the numerical factor:

$$x_{\text{cm}}=\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)l.$$

This matches option B exactly. Hence, the correct answer is Option B.

Let us denote the uniform mass-density of the material by $$\rho$$.

We first write the masses of the two spheres explicitly. For the big sphere (centre at $$C$$, radius $$R$$)

$$M_{\text{big}}=\dfrac{4\pi}{3}\rho R^{3}$$

and for the cavity (centre at $$O$$, radius $$1$$)

$$M_{\text{cav}}=\dfrac{4\pi}{3}\rho \;1^{3}= \dfrac{4\pi}{3}\rho.$$

The mass of the remaining (shaded) solid is therefore

$$M_{\text{rem}}=M_{\text{big}}-M_{\text{cav}}=\dfrac{4\pi}{3}\rho\left(R^{3}-1\right).$$

Because the situation is completely symmetrical about the straight line $$CO$$, the centre of mass of the remainder must lie somewhere on this very line. We choose the origin of the one-dimensional coordinate axis at $$O$$ and measure every distance along $$OC$$ as positive. With this choice

$$x_{O}=0,\qquad x_{C}=d \;(=\;OC).$$

The point $$G$$, by statement of the problem, is situated on the surface of the cavity, that is exactly one unit away from $$O$$ along the same line, so

$$x_{G}=1.$$

Now we invoke the centre-of-mass formula for a composite body:

$$M_{\text{rem}}\;x_{G}=M_{\text{big}}\;x_{C}-M_{\text{cav}}\;x_{O}.$$

Substituting the known quantities, we have

$$\left[\dfrac{4\pi}{3}\rho\left(R^{3}-1\right)\right](1)=\left[\dfrac{4\pi}{3}\rho R^{3}\right]\,d-\left[\dfrac{4\pi}{3}\rho\right](0).$$

All common factors $$\dfrac{4\pi}{3}\rho$$ cancel immediately, leaving the simple relation

$$R^{3}-1=R^{3}\,d\qquad\Longrightarrow\qquad d=\dfrac{R^{3}-1}{R^{3}}=1-\dfrac{1}{R^{3}}.$$

Up to this point, every value of $$R>1$$ would merely give a different permissible separation $$d$$ of the two centres. However, the drawing that accompanies the question shows the cavity touching the outer surface of the large sphere at a single point. For two spheres of radii $$R$$ (big) and $$1$$ (cavity) this “tangential” condition means quite simply

$$OC+1=R\qquad\Longrightarrow\qquad d=R-1.$$

Putting this geometric condition together with the mass-balance condition just obtained we write

$$1-\dfrac{1}{R^{3}}=R-1.$$

Rearranging,

$$2-R-\dfrac{1}{R^{3}}=0\qquad\Longrightarrow\qquad 1=R^{3}(2-R).$$

We still have the factor $$R^{3}$$ on the right. It is convenient to eliminate it by first observing from an earlier step that

$$R^{3}-1=R^{3}(R-1).$$

Dividing both members of this identity by $$R-1$$ (with $$R\neq1$$) we get the well-known algebraic result

$$R^{3}=R^{2}+R+1.$$

Substituting this expression for $$R^{3}$$ in the equation $$1=R^{3}(2-R)$$ leads directly to

$$1=(R^{2}+R+1)(2-R).$$

Exactly the same algebra appears, term for term, in option A, viz.

$$(R^{2}+R+1)(2-R)=1.$$

Hence, the correct answer is Option A.

We have a solid, right-circular cone of uniform (constant) density. Its geometrical dimensions are the height $$h$$ and the radius of the base $$R$$. We want the distance $$z_0$$ of the centre of mass from the vertex (the sharp tip). By symmetry, the centre of mass must lie on the axis of the cone, so we only need the coordinate along this axis.

To find $$z_0$$ we will use the definition of the centre of mass for a continuous body:

$$z_0=\dfrac{\displaystyle\int z \, dm}{\displaystyle\int dm}$$

Here the element of mass $$dm$$ is the mass of a thin disc cut perpendicular to the axis at a distance $$z$$ from the vertex. Every algebraic step follows.

Let the constant volume density be $$\rho$$ (mass per unit volume). First we relate the radius $$r$$ of a disc taken at distance $$z$$ to the fixed dimensions $$R$$ and $$h$$. Because the cone narrows linearly, the ratio of any disc’s radius to its distance from the vertex equals the ratio of the base radius to the full height:

$$\dfrac{r}{z}=\dfrac{R}{h}\qquad\Longrightarrow\qquad r=\dfrac{R}{h}\,z.$$

The disc has thickness $$dz$$, so its volume is the area $$\pi r^{2}$$ multiplied by $$dz$$. Substituting the expression for $$r$$:

$$dV = \pi r^{2}\,dz = \pi\left(\dfrac{R}{h}z\right)^{2}dz = \pi\dfrac{R^{2}}{h^{2}}\,z^{2}\,dz.$$

Because density is uniform, the mass element is

$$dm = \rho\,dV = \rho\,\pi\dfrac{R^{2}}{h^{2}}\,z^{2}\,dz.$$

Next, we calculate the total mass $$M$$ of the cone by integrating $$dm$$ from the vertex $$z=0$$ to the base $$z=h$$:

$$M = \int_{0}^{h} dm = \rho\pi\dfrac{R^{2}}{h^{2}}\int_{0}^{h} z^{2}\,dz.$$

We evaluate the integral $$\int z^{2}dz$$ first:

$$\int_{0}^{h} z^{2}\,dz = \left[\dfrac{z^{3}}{3}\right]_{0}^{h} = \dfrac{h^{3}}{3}.$$

Substituting this back, we get

$$M = \rho\pi\dfrac{R^{2}}{h^{2}}\cdot\dfrac{h^{3}}{3} = \rho\pi R^{2}\dfrac{h}{3}.$$

Now we need the numerator of the centre-of-mass formula, namely $$\int z\,dm$$:

$$\int_{0}^{h} z\,dm = \rho\pi\dfrac{R^{2}}{h^{2}}\int_{0}^{h} z^{3}\,dz.$$

The integral $$\int z^{3}dz$$ is

$$\int_{0}^{h} z^{3}\,dz = \left[\dfrac{z^{4}}{4}\right]_{0}^{h} = \dfrac{h^{4}}{4}.$$

Hence

$$\int z\,dm = \rho\pi\dfrac{R^{2}}{h^{2}}\cdot\dfrac{h^{4}}{4} = \rho\pi R^{2}\dfrac{h^{2}}{4}.$$

We now form the ratio for $$z_0$$:

$$z_0 = \dfrac{\displaystyle\rho\pi R^{2}\dfrac{h^{2}}{4}} {\displaystyle\rho\pi R^{2}\dfrac{h}{3}} = \dfrac{h^{2}}{4}\,\cdot\,\dfrac{3}{h} = \dfrac{3h}{4}.$$

The density $$\rho$$, the factor $$\pi R^{2}$$, and all other common terms cancel in the ratio, leaving a simple fraction of the height.

So, the centre of mass lies on the axis at a distance $$\dfrac{3h}{4}$$ from the vertex.

Hence, the correct answer is Option C.

We have three identical spheres each of mass $$2M$$ at points O(0,0), A(4,0) and B(0,4).

Formula for centre of mass coordinates: $$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}\quad -(1)$$ and $$y_{cm} = \frac{\sum m_i y_i}{\sum m_i}\quad -(2)$$.

Total mass = $$2M + 2M + 2M = 6M$$.

From $$(1)$$:
$$x_{cm} = \frac{2M\cdot0 + 2M\cdot4 + 2M\cdot0}{6M} = \frac{8M}{6M} = \frac{4}{3}$$.

From $$(2)$$:
$$y_{cm} = \frac{2M\cdot0 + 2M\cdot0 + 2M\cdot4}{6M} = \frac{8M}{6M} = \frac{4}{3}$$.

Position vector magnitude of centre of mass:
Formula: $$r = \sqrt{x_{cm}^2 + y_{cm}^2}\quad -(3)$$.

Substituting:
$$r = \sqrt{\Bigl(\frac{4}{3}\Bigr)^2 + \Bigl(\frac{4}{3}\Bigr)^2} = \frac{4}{3}\sqrt{2} = \frac{4\sqrt{2}}{3}$$.

Comparing with $$\frac{4\sqrt{2}}{x}$$ gives $$x = 3$$.

Answer: 3.

Three identical spheres, each of mass $$M$$, are placed at the corners of a right-angled triangle with mutually perpendicular sides of $$3 \text{ m}$$ each. Taking the point of intersection of the perpendicular sides as the origin, the three vertices are at:

$$A = (0, 0)$$, $$B = (3, 0)$$, $$C = (0, 3)$$

The coordinates of the centre of mass are:

$$x_{cm} = \dfrac{M(0) + M(3) + M(0)}{3M} = \dfrac{3}{3} = 1 \text{ m}$$

$$y_{cm} = \dfrac{M(0) + M(0) + M(3)}{3M} = \dfrac{3}{3} = 1 \text{ m}$$

The magnitude of the position vector of the centre of mass is:

$$r = \sqrt{x_{cm}^2 + y_{cm}^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ m}$$

Comparing with $$\sqrt{x}$$, we get $$x = 2$$.

Therefore, the answer is $$\boxed{2}$$.

We have a one-dimensional rod AB of length $$L$$ with mass density $$\rho = \rho_0\left(1 - \frac{x^2}{L^2}\right)$$ kg/m, where $$x$$ is measured from end A. We need to find the distance of the centre of mass from end A.

The centre of mass is given by $$x_{\text{cm}} = \frac{\int_0^L x \, \rho \, dx}{\int_0^L \rho \, dx}$$.

First, let us compute the denominator (total mass). We have $$\int_0^L \rho_0\left(1 - \frac{x^2}{L^2}\right) dx = \rho_0\left[x - \frac{x^3}{3L^2}\right]_0^L = \rho_0\left(L - \frac{L}{3}\right) = \frac{2\rho_0 L}{3}$$.

Now for the numerator, $$\int_0^L x \cdot \rho_0\left(1 - \frac{x^2}{L^2}\right) dx = \rho_0\int_0^L \left(x - \frac{x^3}{L^2}\right) dx = \rho_0\left[\frac{x^2}{2} - \frac{x^4}{4L^2}\right]_0^L = \rho_0\left(\frac{L^2}{2} - \frac{L^2}{4}\right) = \frac{\rho_0 L^2}{4}$$.

So the centre of mass position is $$x_{\text{cm}} = \frac{\rho_0 L^2 / 4}{2\rho_0 L / 3} = \frac{L^2}{4} \times \frac{3}{2L} = \frac{3L}{8}$$.

Comparing with the given expression $$\frac{3L}{\alpha}$$, we get $$\alpha = 8$$.

Hence, the value of $$\alpha$$ is $$\textbf{8}$$.

A body of mass $$M$$ at rest explodes into three pieces with mass ratio 1 : 1 : 2. The two smaller pieces fly off perpendicular to each other with velocities 30 m s$$^{-1}$$ and 40 m s$$^{-1}$$.

Using conservation of momentum, let the masses be $$\frac{M}{4}$$, $$\frac{M}{4}$$, and $$\frac{M}{2}$$.

Initial momentum = 0, so final momentum must also be zero:

$$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$$

Calculating the momentum of the two smaller pieces gives

$$p_1 = \frac{M}{4} \times 30 = \frac{30M}{4}$$

$$p_2 = \frac{M}{4} \times 40 = \frac{40M}{4}$$

Since the two pieces move perpendicular to each other:

$$|\vec{p}_1 + \vec{p}_2| = \sqrt{p_1^2 + p_2^2} = \frac{M}{4}\sqrt{30^2 + 40^2} = \frac{M}{4}\sqrt{900 + 1600} = \frac{M}{4} \times 50$$

Therefore, the momentum of the third piece must satisfy

$$|\vec{p}_3| = |\vec{p}_1 + \vec{p}_2| = \frac{50M}{4}$$

Substituting into $$\frac{M}{2} \times v_3 = \frac{50M}{4}$$ and solving for $$v_3$$ yields

$$v_3 = \frac{50}{4} \times \frac{2}{1} = 25 \text{ m s}^{-1}$$

The correct answer is Option C: 25 m s$$^{-1}$$.

We have two bodies of masses $$m_1 = 1$$ kg and $$m_2 = 3$$ kg with position vectors $$\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$$ and $$\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$$ respectively.

The position vector of the centre of mass is $$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$$.

Computing the numerator: $$(1 - 9)\hat{i} + (2 - 6)\hat{j} + (1 + 3)\hat{k} = -8\hat{i} - 4\hat{j} + 4\hat{k}$$.

Dividing by 4: $$\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$$.

The magnitude is $$|\vec{r}_{cm}| = \sqrt{(-2)^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$.

Now we check the magnitudes of each option. For Option A: $$\hat{i} - 2\hat{j} + \hat{k}$$, the magnitude is $$\sqrt{1 + 4 + 1} = \sqrt{6}$$. This matches our answer.

Hence, the correct answer is Option A.

Consider a uniform semicircular wire of radius $$R$$ lying in the $$x$$-$$y$$ plane with its centre at the origin and its diameter along the $$x$$-axis. A small element of the wire at angle $$\theta$$ (measured from the positive $$x$$-axis) has coordinates $$(R\cos\theta,\, R\sin\theta)$$ and arc length $$dl = R\,d\theta$$, where $$\theta$$ ranges from $$0$$ to $$\pi$$ (upper semicircle).

By symmetry, the $$x$$-coordinate of the centre of mass is zero. For the $$y$$-coordinate:

$$y_{\text{cm}} = \frac{\displaystyle\int_0^{\pi} (R\sin\theta)\, R\,d\theta}{\displaystyle\int_0^{\pi} R\,d\theta} = \frac{R^2 \displaystyle\int_0^{\pi} \sin\theta\, d\theta}{R\pi}$$

$$\int_0^{\pi} \sin\theta\, d\theta = [-\cos\theta]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2$$

$$y_{\text{cm}} = \frac{R^2 \times 2}{R\pi} = \frac{2R}{\pi}$$

The position of the centre of mass is $$\left(0,\, \dfrac{2R}{\pi}\right)$$. Comparing with the given form $$\left(0,\, \dfrac{xR}{\pi}\right)$$, we get $$x = 2$$, so $$|x| = 2$$.

We need to find the value of the angle $$\theta$$ to the nearest integer after a collision between two balls.

Let the first ball have mass $$m_1 = 10\text{ kg}$$ and initial velocity along the $$X$$-axis $$\vec{v}_1 = 10\sqrt{3}\hat{i}\text{ m s}^{-1}$$.

The second ball has mass $$m_2 = 20\text{ kg}$$ and is initially at rest, so $$\vec{v}_2 = 0$$.

The total initial momentum of the system along the $$X$$-axis is: $$P_{initial, x} = m_1 v_1 = 10 \times 10\sqrt{3} = 100\sqrt{3}\text{ kg m s}^{-1}$$.

The total initial momentum of the system along the $$Y$$-axis is: $$P_{initial, y} = 0$$.

After the collision, the first ball comes to rest ($$\vec{v}_1' = 0$$). The second ball disintegrates into two equal pieces of mass $$m = 10\text{ kg}$$ each.

The first piece moves along the positive $$Y$$-axis at a speed of $$10\text{ m s}^{-1}$$. Its momentum vector is: $$\vec{p}_3 = 10 \times 10\hat{j} = 100\hat{j}\text{ kg m s}^{-1}$$.

The second piece moves at a speed of $$20\text{ m s}^{-1}$$ at an angle $$\theta$$ below the $$X$$-axis. Its momentum components are: $$p_{4, x} = m v_4 \cos\theta = 10 \times 20 \cos\theta = 200\cos\theta$$ and $$p_{4, y} = -m v_4 \sin\theta = -10 \times 20 \sin\theta = -200\sin\theta$$.

According to the law of conservation of linear momentum along the $$X$$-axis: $$P_{initial, x} = P_{final, x}$$.

Substituting the values gives: $$100\sqrt{3} = 200\cos\theta$$, which simplifies to: $$\cos\theta = \frac{100\sqrt{3}}{200} = \frac{\sqrt{3}}{2}$$.

According to the law of conservation of linear momentum along the $$Y$$-axis: $$P_{initial, y} = P_{final, y}$$.

Substituting the values gives: $$0 = 100 - 200\sin\theta$$, which simplifies to: $$\sin\theta = \frac{100}{200} = \frac{1}{2}$$.

From both conditions, we find that: $$\theta = 30^\circ$$.

Therefore, the value of $$\theta$$ to the nearest integer is 30.

Let us consider Centroid 1 as

$$C_1=\left(x_1,\ y_1\right)$$

where $$C_1=\left(a,\ 0\right)$$

and Centroid 2 as

$$C_2=\left(x_2,\ y_2\right)$$

where 

$$C_2=\left(\frac{3a}{2},\ 0\right)$$

Area of bigger circle is $$A_1=\ \pi\ R^2$$

Area of the cut part is 

$$A_2=\ \ \frac{\pi R^2}{4}$$

Now formula is $$X\ =\ \frac{\left(A_1x_1+A_2x_2\right)}{A_1+A_2}$$

$$X\ =\ \frac{\left(\pi\ R^2\left(a\right)-\frac{\pi R^2}{4}\left(\frac{3a}{2}\right)\right)}{\pi\ R^2-\frac{\pi R^2}{4}}$$

$$X\ =\ \frac{\frac{5\pi R^2}{8}}{\frac{3\pi R^2}{4}}$$

$$X\ =\frac{5}{6}$$

Let the $$1.0\text{ kg}$$ mass be at the origin $$(0,0)$$.

Given particle positions from the geometry:

$$(x_1, y_1) = (0, 0) \text{ for } m_1 = 1.0\text{ kg}$$

$$(x_2, y_2) = (3, 0) \text{ for } m_2 = 1.5\text{ kg}$$

$$(x_3, y_3) = (0, 4) \text{ for } m_3 = 2.5\text{ kg}$$

Total mass of the system: $$M = m_1 + m_2 + m_3 = 1.0 + 1.5 + 2.5 = 5.0\text{ kg}$$

X-coordinate of the center of mass: $$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{1.0(0) + 1.5(3) + 2.5(0)}{5.0} = \frac{4.5}{5.0} = 0.9\text{ cm}$$

Y-coordinate of the center of mass: $$y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} = \frac{1.0(0) + 1.5(0) + 2.5(4)}{5.0} = \frac{10}{5.0} = 2.0\text{ cm}$$

Mass of the original complete disk: $$M_1 = \sigma \cdot \pi a^2$$

Mass of the square portion removed: $$M_2 = \sigma \cdot l^2 = \sigma \left(\frac{a}{2}\right)^2 = \sigma \frac{a^2}{4}$$

Positions of the centres of mass relative to the origin O:

$$x_1 = 0$$

$$x_2 = d = \frac{a}{2}$$

Centre of mass of the remaining portion: $$x_{\text{cm}} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$$

$$x_{\text{cm}} = \frac{(\sigma \pi a^2)(0) - \left(\sigma \frac{a^2}{4}\right)\left(\frac{a}{2}\right)}{\sigma \pi a^2 - \sigma \frac{a^2}{4}} = \frac{-\frac{a^3}{8}}{\pi a^2 - \frac{a^2}{4}} = \frac{-\frac{a}{8}}{\pi - \frac{1}{4}}$$

$$x_{\text{cm}} = \frac{-a}{8\pi - 2} = -\frac{a}{8(3.14) - 2} = -\frac{a}{25.12 - 2} = -\frac{a}{23.12} \approx -\frac{a}{23}$$

Divide the lamina into two rectangles: a vertical strip $$R_1$$ and a square flag piece $$R_2$$.

Vertical strip $$R_1$$ bounded by vertices $$(0,0)$$, $$(1,0)$$, $$(1,2)$$, $$(0,2)$$:

$$\text{Area } A_1 = 1 \times 2 = 2, \quad \text{Center of mass } (x_1, y_1) = (0.5, 1.0)$$

Square flag piece $$R_2$$ bounded by vertices $$(0,2)$$, $$(2,2)$$, $$(2,3)$$, $$(0,3)$$:

$$\text{Area } A_2 = 2 \times 1 = 2, \quad \text{Center of mass } (x_2, y_2) = (1.0, 2.5)$$

X-coordinate of the center of mass: $$x_{\text{cm}} = \frac{A_1 x_1 + A_2 x_2}{A_1 + A_2} = \frac{2(0.5) + 2(1.0)}{2 + 2} = \frac{1 + 2}{4} = 0.75\text{ m}$$

Y-coordinate of the center of mass: $$y_{\text{cm}} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{2(1.0) + 2(2.5)}{2 + 2} = \frac{2 + 5}{4} = 1.75\text{ m}$$

Relative acceleration and velocity before collision:

$$a_{\text{rel}} = (-g) - (-g) = 0$$

$$v_{\text{rel}} = u_2 - u_1 = \sqrt{2gh} - 0 = \sqrt{2gh}$$

Collision time $$t_1$$: $$t_1 = \frac{h}{v_{\text{rel}}} = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$$

Velocities and height at $$t_1$$: $$v_1 = u_1 - gt_1 = -g\sqrt{\frac{h}{2g}} = -\sqrt{\frac{gh}{2}}$$

$$v_2 = u_2 - gt_1 = \sqrt{2gh} - g\sqrt{\frac{h}{2g}} = \sqrt{\frac{gh}{2}}$$

$$y = u_2 t_1 - \frac{1}{2}gt_1^2 = \sqrt{2gh}\sqrt{\frac{h}{2g}} - \frac{1}{2}g\left(\frac{h}{2g}\right) = \frac{3h}{4}$$

Conservation of linear momentum: 

$$m v_2 + m v_1 = (2m) v_c \implies m\left(\sqrt{\frac{gh}{2}}\right) + m\left(-\sqrt{\frac{gh}{2}}\right) = 2m v_c \implies v_c = 0$$

Free fall of combined mass from height $$y$$:

$$y = \frac{1}{2}g t_2^2 \implies \frac{3h}{4} = \frac{1}{2}g t_2^2 \implies t_2 = \sqrt{\frac{3h}{2g}} = \sqrt{\frac{3}{2}}\sqrt{\frac{h}{g}}$$

To find the center of mass of the remaining portion after removing the shaded rectangle HBGO, we treat the removed section as a negative mass.

Let the total mass of the original sheet ABCD be M.

• Original Area: $$A = a \times b$$
• Removed Area (HBGO): $$A_{rem} = \frac{a}{2} \times \frac{b}{2} = \frac{A}{4}$$
• Removed Mass: $$m = \frac{M}{4}$$

Center of Mass Coordinates

The center of mass (x, y) of a composite system is given by:

$$x_{cm} = \frac{M X - m x_1}{M - m}$$

1. Calculation for x-coordinate:

• Center of original sheet (X): $$a/2$$
• Center of removed portion ($$x_1$$): $$3a/4$$
• Remaining mass: $$M - M/4 = 3M/4$$

$$x = \frac{M \left(\frac{a}{2}\right) - \frac{M}{4} \left(\frac{3a}{4}\right)}{\frac{3M}{4}}$$

$$x = \frac{\frac{a}{2} - \frac{3a}{16}}{\frac{3}{4}} = \frac{\frac{5a}{16}}{\frac{3}{4}}$$

$$x = \frac{5a}{12}$$

2. Calculation for y-coordinate:

Due to symmetry between the x and y dimensions:

• Center of original sheet (Y): $$b/2$$
• Center of removed portion ($$y_1$$): $$3b/4$$

$$y = \frac{M \left(\frac{b}{2}\right) - \frac{M}{4} \left(\frac{3b}{4}\right)}{\frac{3M}{4}}$$

$$y = \frac{5b}{12}$$

Final Result:

The coordinates of the centre of mass of the remaining portion are:

$$\boxed{\left( \frac{5a}{12}, \frac{5b}{12} \right)}$$

$$(x_1, y_1) = (0, 0)$$

$$(x_2, y_2) = (1, 0)$$

$$(x_3, y_3) = (1\cos60^\circ, 1\sin60^\circ) = \left(0.5, \frac{\sqrt{3}}{2}\right)$$

$$X_{\text{cm}} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{M_{\text{total}}}$$

$$X_{\text{cm}} = \frac{50(0) + 100(1) + 150(0.5)}{300} = \frac{100 + 75}{300} = \frac{175}{300} = \frac{7}{12}\text{ m}$$

$$Y_{\text{cm}} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{M_{\text{total}}}$$

$$Y_{\text{cm}} = \frac{50(0) + 100(0) + 150\left(\frac{\sqrt{3}}{2}\right)}{300} = \frac{75\sqrt{3}}{300} = \frac{\sqrt{3}}{4}\text{ m}$$

$$X_{cm} = \frac{\int_{0}^{L} x \cdot dm}{\int_{0}^{L} dm}$$

$$M = \int_{0}^{L} \mu(x) dx = \int_{0}^{L} \left(a + \frac{bx}{L}\right) dx$$

$$M = \left[ ax + \frac{bx^2}{2L} \right]_{0}^{L} = aL + \frac{bL^2}{2L}$$

$$M = L \left( a + \frac{b}{2} \right)$$

$$\int_{0}^{L} x \cdot \mu(x) dx = \int_{0}^{L} x \left( a + \frac{bx}{L} \right) dx = \int_{0}^{L} \left( ax + \frac{bx^2}{L} \right) dx$$

$$ = \left[ \frac{ax^2}{2} + \frac{bx^3}{3L} \right]_{0}^{L} = \frac{aL^2}{2} + \frac{bL^3}{3L}$$ $$ = L^2 \left( \frac{a}{2} + \frac{b}{3} \right)$$

$$X_{cm} = \frac{L^2 \left( \frac{a}{2} + \frac{b}{3} \right)}{L \left( a + \frac{b}{2} \right)} = \frac{7}{12}L$$

$$\frac{\frac{a}{2} + \frac{b}{3}}{a + \frac{b}{2}} = \frac{7}{12}$$

$$\frac{3a + 2b}{3(2a + b)} = \frac{7}{12}$$ 

Solving the equation gives $$b = 2a$$

$$\lambda(x) = \lambda_0 + kx$$

$$M = \int_0^L \lambda(x) dx = \int_0^L (\lambda_0 + kx) dx = \left[ \lambda_0 x + \frac{kx^2}{2} \right]_0^L$$

$$M = \lambda_0 L + \frac{kL^2}{2} \implies kL^2 = 2(M - \lambda_0 L) \quad \dots (1)$$

$$x_{cm} = \frac{1}{M} \int_0^L x \lambda(x) dx$$

$$x_{cm} = \frac{1}{M} \int_0^L (x\lambda_0 + kx^2) dx = \frac{1}{M} \left[ \frac{\lambda_0 x^2}{2} + \frac{kx^3}{3} \right]_0^L$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{k L^3}{3} \right]$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{L}{3} (kL^2) \right]$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{L}{3} (2M - 2\lambda_0 L) \right]$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{2ML}{3} - \frac{2\lambda_0 L^2}{3} \right]$$

$$x_{cm} = \frac{2L}{3} + \frac{\lambda_0 L^2}{M} \left( \frac{1}{2} - \frac{2}{3} \right)$$

$$x_{cm} = \frac{2L}{3} - \frac{\lambda_0 L^2}{6M}$$

We have a boy of mass 20 kg standing on a cart of mass 80 kg. The cart is free to move and there is negligible friction between the cart and the ground. Initially, the boy is 25 m from a wall. The boy walks 10 m towards the wall on the cart. We need to find the final distance of the boy from the wall.

Since there is no external force acting on the system (friction is negligible), the center of mass of the system (boy + cart) remains stationary. This is a key principle we will use.

Set up a coordinate system with the wall at position $$x = 0$$. Initially, the boy is at $$x_{\text{boy, initial}} = 25$$ m. Let the initial position of the center of mass of the cart be $$X$$. The center of mass of the entire system ($$x_{\text{cm}}$$) is given by:

$$ x_{\text{cm}} = \frac{m_{\text{boy}} \cdot x_{\text{boy, initial}} + m_{\text{cart}} \cdot X}{m_{\text{boy}} + m_{\text{cart}}} = \frac{20 \cdot 25 + 80 \cdot X}{100} = \frac{500 + 80X}{100} = 5 + 0.8X $$

Now, the boy walks 10 m towards the wall relative to the cart. Since the wall is to the left (negative x-direction), the boy moves $$-10$$ m relative to the cart. However, as the boy moves, the cart may also move. Let the displacement of the cart's center of mass be $$D$$ (positive if to the right). The displacement of the boy relative to the ground is then the displacement relative to the cart plus the displacement of the cart:

Displacement of boy relative to ground $$ = -10 + D $$

Thus, the final position of the boy is:

$$ x_{\text{boy, final}} = x_{\text{boy, initial}} + (-10 + D) = 25 - 10 + D = 15 + D $$

The final position of the cart's center of mass is:

$$ x_{\text{cart, final}} = X + D $$

The center of mass remains unchanged, so:

$$ x_{\text{cm}} = \frac{m_{\text{boy}} \cdot x_{\text{boy, final}} + m_{\text{cart}} \cdot x_{\text{cart, final}}}{m_{\text{boy}} + m_{\text{cart}}} = \frac{20 \cdot (15 + D) + 80 \cdot (X + D)}{100} $$

Set this equal to the initial center of mass:

$$ \frac{20(15 + D) + 80(X + D)}{100} = 5 + 0.8X $$

Multiply both sides by 100 to clear the denominator:

$$ 20(15 + D) + 80(X + D) = 100(5 + 0.8X) $$

Expand both sides:

Left side: $$20 \cdot 15 + 20 \cdot D + 80 \cdot X + 80 \cdot D = 300 + 20D + 80X + 80D = 300 + 100D + 80X$$

Right side: $$100 \cdot 5 + 100 \cdot 0.8X = 500 + 80X$$

So:

$$ 300 + 100D + 80X = 500 + 80X $$

Subtract $$80X$$ from both sides:

$$ 300 + 100D = 500 $$

Subtract 300 from both sides:

$$ 100D = 200 $$

Divide by 100:

$$ D = 2 $$

So the cart moves 2 m to the right. The final position of the boy is:

$$ x_{\text{boy, final}} = 15 + D = 15 + 2 = 17 \text{ m} $$

Therefore, the final distance of the boy from the wall is 17 m.

Hence, the correct answer is Option D.

We apply the condition of rotational equilibrium (balance of torques) about the pivot at the 40 cm mark.
Torque $$\tau$$ due to a force is given by $$\tau = F \times d$$, where $$F$$ is the weight and $$d$$ is the perpendicular distance from the pivot.

Identify the torques:

Rod’s own weight: mass = 250 g acts at the centre (50 cm).
Lever arm = $$(50 - 40) = 10\text{ cm}$$.
This produces a clockwise torque of magnitude $$250 \times 10\text{ (g·cm)}$$.

Suspended mass 400 g at 10 cm: lever arm = $$(40 - 10) = 30\text{ cm}$$.
This produces an anticlockwise torque of magnitude $$400 \times 30\text{ (g·cm)}$$.

Unknown mass $$m$$ at 90 cm: lever arm = $$(90 - 40) = 50\text{ cm}$$.
This produces a clockwise torque of magnitude $$m \times 50\text{ (g·cm)}$$.

For equilibrium, sum of anticlockwise torques = sum of clockwise torques:

$$400 \times 30 \;=\; 250 \times 10 \;+\; m \times 50 \quad\,-(1)$$

Compute the numerical values in $$(1)$$:

$$12000 \;=\; 2500 \;+\; 50m \quad\,-(2)$$

Rearrange $$(2)$$ to solve for $$m$$:

$$50m \;=\; 12000 - 2500 = 9500$$
$$m = \frac{9500}{50} = 190$$

Therefore, the required mass to be suspended at the 90 cm mark is 190 g. This corresponds to Option A.

Since the plate is uniform, we can find the COM by treating the object as a large rectangle with a smaller section removed from its top.

We consider the full $$3 \times 2$$ rectangle and subtract the $$1 \times 1$$ empty square region in the upper middle.

Total Rectangle ($$A_1$$): The area is $$3 \times 2 = 6$$, with its geometric center at $$(1.5, 1)$$.

Removed Square ($$A_2$$): The area is $$1 \times 1 = 1$$, with its geometric center at $$(1.5, 1.5)$$.

Coordinates of the Center of Mass

Because the figure is symmetric about the line $$x = 1.5$$, the $$x$$-coordinate is:

$$x_{cm} = \frac{A_1x_1 - A_2x_2}{A_1 - A_2} = \frac{6(1.5) - 1(1.5)}{5} = 1.5$$

The $$y$$-coordinate is determined by the distribution of the remaining area along the vertical axis:

$$y_{cm} = \frac{A_1y_1 - A_2y_2}{A_1 - A_2} = \frac{6(1) - 1(1.5)}{6 - 1} = \frac{4.5}{5} = 0.9$$

Taking the ratio of the $$x$$ and $$y$$ coordinates:

$$\text{Ratio} = \frac{x_{cm}}{y_{cm}} = \frac{1.5}{0.9}$$

$$\frac{15}{9} = \frac{n}{9}$$

The value of $$n$$ is 15.

We need to find how far $$m_2$$ must move to keep the centre of mass at its original position. The position of the centre of mass is $$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$, and for $$x_{\text{cm}}$$ to remain unchanged the net displacement must satisfy $$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$$.

Here $$m_1 = 3$$ kg moves towards the centre of mass by $$\Delta x_1 = 2$$ cm, so we have $$3 \times 2 = 2 \times d_2$$, giving $$d_2 = \frac{6}{2} = 3 \text{ cm}$$. Both particles move towards the CM, so both displacements reduce the distance to the CM.

The correct answer is 3 cm.

For the centre of mass to remain unchanged, the total displacement weighted by mass must be zero.

Let the 10 kg block move a distance of 6 cm towards the 30 kg block (i.e., in the positive x-direction), and let the 30 kg block move a distance $$d$$ in some direction.

For the centre of mass position to remain unchanged:

$$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$$

$$10 \times 6 + 30 \times \Delta x_2 = 0$$

$$60 + 30 \Delta x_2 = 0$$

$$\Delta x_2 = -2 \text{ cm}$$

The negative sign indicates that the 30 kg block must move in the opposite direction to the 10 kg block's movement. Since the 10 kg block moved towards the 30 kg block, the 30 kg block must move towards the 10 kg block.

Therefore, the 30 kg block must be moved 2 cm towards the 10 kg block.

The correct answer is Option C.

Let the length of the train be $$L$$ and the uniform acceleration be $$a$$. The front of the train (engine) passes the signal post with velocity $$u$$ and the last compartment passes with velocity $$v$$. Using the kinematic relation $$v^2 = u^2 + 2aL$$, we get $$a = \frac{v^2 - u^2}{2L}$$.

The middle point of the train is at a distance $$\frac{L}{2}$$ from the engine. When the middle point passes the signal post, the engine has travelled a distance $$\frac{L}{2}$$ beyond the post. Let the velocity of the middle point at the signal post be $$v_m$$. Using the kinematic equation for the engine travelling from the signal post to a distance $$\frac{L}{2}$$:

$$v_m^2 = u^2 + 2a \cdot \frac{L}{2} = u^2 + aL$$

Substituting $$aL = \frac{v^2 - u^2}{2}$$:

$$v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$$

Therefore $$v_m = \sqrt{\frac{u^2 + v^2}{2}}$$, which corresponds to Option (2).

For first segment (horizontal top bar of mass $$2m$$):

$$x_1 = L,\quad y_1 = L$$

For second segment (vertical bar of mass $$m$$):

$$x_2 = 2L,\quad y_2 = \frac{L}{2}$$

For third segment (horizontal bottom bar of mass $$m$$):

$$x_3 = \frac{5}{2}L,\quad y_3 = 0$$

Net center of mass coordinates:

$$x_{cm} = \frac{2m(L) + m(2L) + m\left(\frac{5}{2}L\right)}{2m + m + m} = \frac{13}{8}L$$

$$y_{cm} = \frac{2m(L) + m\left(\frac{L}{2}\right) + m(0)}{2m + m + m} = \frac{5}{8}L$$

Problem Analysis

• Masses of the particles: $$m_A = m$$, $$m_B = 2m$$, $$m_C = 3m$$, $$m_D = 4m$$
• Total mass of the system ($$M$$):
• Accelerations ($$\vec{a}$$): Each particle has an acceleration of magnitude $$a$$. Based on standard square corner configurations in these types of problems, their vector directions are:
  • o $$\vec{a}_A = -a\hat{i}$$
• o $$\vec{a}_B = a\hat{j}$$
• o $$\vec{a}_C = a\hat{i}$$
• o $$\vec{a}_D = -a\hat{j}$$
• Along $$\hat{i}$$ (x-axis): $$-ma + 3ma = 2ma$$
• Along $$\hat{j}$$ (y-axis): $$2ma - 4ma = -2ma$$
• Correct Option: (3) $$\frac{a}{5}(\hat{i} - \hat{j})$$

$$M = m_A + m_B + m_C + m_D = m + 2m + 3m + 4m = 10m$$

Step-by-Step Solution

Step 1: Formula for Acceleration of Centre of Mass ($$\vec{a}_c$$)

$$\vec{a}_c = \frac{m_A\vec{a}_A + m_B\vec{a}_B + m_C\vec{a}_C + m_D\vec{a}_D}{M}$$

Step 2: Substitute the Vector Values

Substitute the masses and acceleration vectors into the formula:

$$\vec{a}_c = \frac{m(-a\hat{i}) + 2m(a\hat{j}) + 3m(a\hat{i}) + 4m(-a\hat{j})}{10m}$$

$$\vec{a}_c = \frac{-ma\hat{i} + 2ma\hat{j} + 3ma\hat{i} - 4ma\hat{j}}{10m}$$

Step 3: Simplify Component-wise

Combine the terms:

$$\vec{a}_c = \frac{2ma\hat{i} - 2ma\hat{j}}{10m}$$

$$\vec{a}_c = \frac{2ma(\hat{i} - \hat{j})}{10m} = \frac{a}{5}(\hat{i} - \hat{j})$$

Conclusion

option (B)

Place point $$B$$ at the origin $$(0, 0)$$ with segment $$BC$$ along the x-axis.

Segment $$AB$$ (Length $$L_1$$): Center of mass is at its midpoint. $$x_1 = \frac{L_1 \cos 60^\circ}{2} = \frac{L_1}{4}$$

Segment $$BC$$ (Length $$L_2$$): Center of mass is at its midpoint. $$x_2 = \frac{L_2}{2}$$

Point $$A$$: The x-coordinate is $$L_1 \cos 60^\circ = \frac{L_1}{2}$$.

The problem states the system COM lies vertically below $$A$$, so $$X_{com} = x_A$$:

$$\frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{L_1}{2}$$

$$\frac{L_1(\frac{L_1}{4}) + L_2(\frac{L_2}{2})}{L_1 + L_2} = \frac{L_1}{2}$$

$$L_1^2 + 2L_2^2 = 2L_1(L_1 + L_2)$$

$$2L_2^2 - 2L_1L_2 - L_1^2 = 0$$

$$r = \frac{L_2}{L_1}$$: $$2r^2 - 2r - 1 = 0$$

$$r = \frac{2 + \sqrt{4 - 4(2)(-1)}}{2(2)} = \frac{2 + \sqrt{12}}{4} = \frac{1 + \sqrt{3}}{2}$$

$$r \approx \frac{1 + 1.732}{2} = \mathbf{1.366}$$

Since the masses interact only via internal gravitational forces, there is no external torque, and the total angular momentum ($$\vec{L}$$) remains constant from $$t=0$$ until the point of collision.

By symmetry, the three masses will move in identical spiral paths and collide at the centroid ($$G$$) of the equilateral triangle. It is most convenient to calculate the angular momentum about this point.

For an equilateral triangle of side $$a$$, the distance from any vertex to the centroid ($$r$$) is $$r = \frac{a}{\sqrt{3}}$$

The velocity vector $$\vec{V}_A$$ is directed along the side $$AC$$. The angle between the position vector $$\vec{r}_A$$ (from $$G$$ to $$A$$) and the side $$AC$$ is $$30^\circ$$. Therefore, the angle $$\phi$$ between the position vector and the velocity vector is $$150^\circ$$

$$L_1 = mvr \sin \phi$$

$$L_1 = m V_0 \left( \frac{a}{\sqrt{3}} \right) \sin(150^\circ)$$

$$L_1 = \frac{maV_0}{2\sqrt{3}}$$

$$L_{net} = 3 \times L_1 = 3 \times \frac{maV_0}{2\sqrt{3}}$$

$$L_{net} = \frac{\sqrt{3}}{2} amV_0$$

In a binary star system, both stars revolve around their common centre of mass. Crucially, both stars complete one full orbit in the same time — they always remain on opposite sides of the centre of mass. This is a direct consequence of Newton's third law: the gravitational force each exerts on the other is equal and opposite, and since they orbit the same centre of mass with the same angular velocity, their periods must be identical.

Although the orbital radii $$r_A$$ and $$r_B$$ may differ (depending on the mass ratio $$m_A r_A = m_B r_B$$), and $$r_A \neq r_B$$ in general, the angular velocity $$\omega$$ is the same for both, so $$T_A = T_B$$.

Therefore $$T_A = T_B$$.

For a point mass outside a uniform solid sphere, the entire mass $$M$$ acts as if it is concentrated at the center $$O$$.

$$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$$

The cavity has a radius $$r = R/3$$. Since mass is proportional to volume ($$M \propto R^3$$),

Mass of cavity ($$m_c$$): $$M \cdot (\frac{R/3}{R})^3 = \frac{M}{27}$$

Distance to $$P$$ ($$d_c$$): The center of the cavity $$O'$$ is at a distance $$R - R/3 = 2R/3$$ from $$O$$. Thus, its distance from point $$P$$ is $$2R - \frac{2R}{3} = \frac{4R}{3}$$.

The force exerted by this removed mass is $$F_{cav} = \frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{27} \cdot \frac{9}{16R^2} = \frac{GMm}{48R^2}$$

The force from the remaining part ($$F_2$$) is $$F_2 = F_1 - F_{cav} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$$

$$F_2 = \frac{12GMm - 1GMm}{48R^2} = \frac{11GMm}{48R^2}$$

$$\frac{F_1}{F_2} = \frac{\frac{GMm}{4R^2}}{\frac{11GMm}{48R^2}} = \frac{1}{4} \times \frac{48}{11} = \frac{12}{11}$$

$$The\ mass\ per\ unit\ area\ is\ \sigma=\frac{\sigma_0x}{ab}.\ We\ integrate\ over\ the\ area\ ofthe\ plate\ (x\in[0,a],\ y\in[0,b]):$$

$$M = \int_{0}^{a} \int_{0}^{b} \sigma \, dy \, dx = \int_{0}^{a} \int_{0}^{b} \frac{\sigma_0 x}{ab} \, dy \, dx$$

$$M = \frac{\sigma_0}{ab} \left[ \frac{x^2}{2} \right]_0^a \left[ y \right]_0^b = \frac{\sigma_0}{ab} \left( \frac{a^2}{2} \right) (b) = \frac{\sigma_0 a}{2}$$

X-coordinate of Center of Mass:

$$x_{cm} = \frac{1}{M} \int \int x \sigma \, dA$$

$$x_{cm} = \frac{1}{M} \int_{0}^{a} \int_{0}^{b} x \left( \frac{\sigma_0 x}{ab} \right) \, dy \, dx = \frac{1}{M} \frac{\sigma_0}{ab} \int_{0}^{a} x^2 \, dx \int_{0}^{b} dy$$

$$x_{cm} = \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0}{ab} \left( \frac{a^3}{3} \right) (b) = \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0 a^2}{3} = \frac{2}{3}a$$

Y-coordinate of Center of Mass:

$$Since\ the\ density\ \sigma\ depends\ only\ on\ x$$

the mass is distributed uniformly in the y-direction for any given x. Therefore: $$y_{cm}=\frac{b}{2}$$

The center of mass is at: $$(x_{cm},y_{cm})=\left(\frac{2}{3}a,\frac{b}{2}\right)$$

Let the incident (heavier) particle have mass $$m_1 = 2m$$ and initial speed $$u$$ along the +x-axis. The target particle has mass $$m_2 = m$$ and is initially at rest in the laboratory (lab) frame.

Step 1: Velocity of the centre of mass (COM)
$$V = \frac{m_1 u + m_2 \times 0}{m_1 + m_2} = \frac{2m\,u}{2m+m} = \frac{2u}{3}$$

Step 2: Velocities in the COM frame before collision
Velocity of $$m_1$$ in COM frame: $$u_{1c} = u - V = u - \frac{2u}{3} = \frac{u}{3}$$ (along +x)
Magnitude of this velocity: $$|u_{1c}| = \frac{u}{3}$$

Step 3: Velocities in the COM frame after an elastic collision
In a perfectly elastic, two-body collision the magnitudes of the velocities in the COM frame remain unchanged; only their directions change. Hence the heavier particle still has speed $$\dfrac{u}{3}$$ in the COM frame, but its direction makes some angle $$\phi$$ with the +x-axis.

Write the post-collision COM-frame velocity of $$m_1$$ as $$\vec{u}_{1c}' = \frac{u}{3}\bigl(\cos\phi\,\hat{i} + \sin\phi\,\hat{j}\bigr)$$

Step 4: Transform back to the lab frame
Lab-frame velocity of $$m_1$$ after collision: $$\vec{v}_{1}' = \vec{V} + \vec{u}_{1c}' = \frac{2u}{3}\,\hat{i} + \frac{u}{3}\bigl(\cos\phi\,\hat{i} + \sin\phi\,\hat{j}\bigr)$$
Components:
$$v_{x} = \frac{2u}{3} + \frac{u}{3}\cos\phi$$ $$v_{y} = \frac{u}{3}\sin\phi$$

Step 5: Angular deviation of the heavier particle
Let $$\theta$$ be the angle between $$\vec{v}_{1}'$$ and the initial direction (+x). Then $$\tan\theta = \frac{v_{y}}{v_{x}} = \frac{\dfrac{u}{3}\sin\phi}{\dfrac{2u}{3} + \dfrac{u}{3}\cos\phi} = \frac{\sin\phi}{2 + \cos\phi} \quad -(1)$$

Step 6: Maximising $$\theta$$
Because $$0 \lt \theta \lt \dfrac{\pi}{2}$$, maximising $$\theta$$ is equivalent to maximising $$\tan\theta$$. Define $$f(\phi) = \dfrac{\sin\phi}{2 + \cos\phi}$$.

Differentiating and setting $$\dfrac{df}{d\phi}=0$$ gives $$\frac{df}{d\phi} = \frac{\cos\phi(2+\cos\phi) + \sin^{2}\phi}{(2+\cos\phi)^{2}} = \frac{2\cos\phi + \cos^{2}\phi + \sin^{2}\phi}{(2+\cos\phi)^{2}} = \frac{2\cos\phi + 1}{(2+\cos\phi)^{2}}$$

Setting the numerator to zero:
$$2\cos\phi + 1 = 0 \;\;\Longrightarrow\;\; \cos\phi = -\frac{1}{2} \;\;\Longrightarrow\;\; \phi = \frac{2\pi}{3}$$ (Only the value within $$0 \le \phi \le \pi$$ is relevant.) For $$\phi \lt \dfrac{2\pi}{3}$$, the derivative is positive and for $$\phi \gt \dfrac{2\pi}{3}$$ it is negative, so $$\phi = \dfrac{2\pi}{3}$$ indeed gives the maximum.

Step 7: Maximum value of $$\theta$$
Insert $$\phi = \dfrac{2\pi}{3}$$ into equation $$(1)$$:
$$\tan\theta_{\max} = \frac{\sin(2\pi/3)}{2 + \cos(2\pi/3)} = \frac{\dfrac{\sqrt{3}}{2}}{2 - \dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{3}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$

Therefore $$\theta_{\max} = \tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$

Option D which is: $$\dfrac{\pi}{6}$$

A circular disc of radius 20 cm has a circular hole of radius 5 cm cut such that the edge of the hole touches the edge of the disc. We need the distance of the center of mass of the remaining disc from the origin.

Place the center of the original disc at the origin. Since the edge of the hole touches the edge of the disc, the center of the hole is at distance $$20 - 5 = 15$$ cm from the origin along the x-axis, so the hole center is at $$(15, 0)$$.

Let $$\sigma$$ be the surface mass density. For a uniform disc, the mass of the original disc is $$M = \sigma \pi (20)^2 = 400\pi\sigma$$, the mass of the removed portion is $$m = \sigma \pi (5)^2 = 25\pi\sigma$$, and the mass of the remaining disc is $$M - m = 375\pi\sigma$$.

Using the principle of superposition, we write $$M \cdot 0 = (M-m) \cdot x_{cm} + m \cdot 15$$, which gives $$0 = 375\pi\sigma \cdot x_{cm} + 25\pi\sigma \cdot 15$$ and hence $$x_{cm} = -\frac{25 \times 15}{375} = -\frac{375}{375} = -1$$ cm.

The distance from the origin is $$|x_{cm}| = 1$$ cm, so the correct answer is Option 3: 1.0 cm.

$$\text{Total Mass } (M) = 2 + 3 + 15 = 20\text{ kg}$$

$$X_{\text{cm}} = \frac{2 \times (-5\sqrt{3}) + 3 \times (5\sqrt{3}) + 15 \times 0}{20} = \frac{\sqrt{3}}{4}\text{ cm}$$

$$Y_{\text{cm}} = \frac{2 \times \left(-\frac{5}{2}\right) + 3 \times \left(-\frac{5}{2}\right) + 15 \times \left(\frac{5}{2}\right)}{20} = \frac{25}{20} = 1.25\text{ cm}$$

$$\vec{v}_{\text{cm}} = \frac{m\vec{v}_1 + m\vec{v}_2}{m + m} = \frac{\vec{v}_1 + \vec{v}_2}{2}$$

$$\vec{v}_{\text{cm}} = \frac{4\hat{i} + 4\hat{j}}{2} = 2\hat{i} + 2\hat{j}\text{ m/s}$$

$$\vec{a}_{\text{cm}} = \frac{m\vec{a}_1 + m\vec{a}_2}{m + m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$$

$$\vec{a}_{\text{cm}} = \frac{6\hat{i} + 6\hat{j} + 0}{2} = 3\hat{i} + 3\hat{j}\text{ m/s}^2$$

Direction of $$\vec{v}_{\text{cm}} = 2(\hat{i} + \hat{j})$$

Direction of $$\vec{a}_{\text{cm}} = 3(\hat{i} + \hat{j})$$

Since $$\vec{a}_{\text{cm}}$$ is a scalar multiple of $$\vec{v}_{\text{cm}}$$ ($$\vec{a}_{\text{cm}} = \frac{3}{2}\vec{v}_{\text{cm}}$$), the acceleration vector is perfectly parallel to the initial velocity vector.

When the acceleration of a particle or a system's center of mass is collinear with its initial velocity, the direction of motion never turns, meaning it moves along a straight line.

Let the heavier block have mass $$m_1 = 2\; \text{kg}$$ and the lighter block have mass $$m_2 = 1\; \text{kg}$$. The system is an ideal Atwood machine (light, inextensible string and light, frictionless pulley).

Step 1 - Acceleration of the system
For an Atwood machine, the magnitude of the common acceleration is
$$a = \frac{(m_1 - m_2) g}{m_1 + m_2}$$
$$\Rightarrow a = \frac{(2 - 1)\,g}{2 + 1} = \frac{g}{3} = \frac{10}{3}\;\text{m s}^{-2}$$

The 2 kg block accelerates downward, the 1 kg block upward, both with the same magnitude $$a$$.

Step 2 - Individual displacements in 2 s
Starting from rest, the distance travelled in time $$t$$ under constant acceleration $$a$$ is
$$s = \frac{1}{2}\,a\,t^{2}$$
For $$t = 2\;\text{s}$$,
$$s = \frac{1}{2}\,\left(\frac{10}{3}\right)\,(2)^{2} = \frac{1}{2}\,\left(\frac{10}{3}\right)\,4 = \frac{20}{3}\;\text{m} \approx 6.67\;\text{m}$$

Thus, after 2 s:
• the 2 kg block has moved $$+s$$ (downward)
• the 1 kg block has moved $$-s$$ (upward)

Step 3 - Position of the centre of mass
Choose the initial horizontal level as origin and take downward as positive. The centre-of-mass coordinate after 2 s is
$$y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$$
$$\Rightarrow y_{\text{cm}} = \frac{\,2(+s) + 1(-s)}{2 + 1} = \frac{(2 - 1)s}{3} = \frac{s}{3}$$
Substituting $$s = \frac{20}{3}\;\text{m}$$:
$$y_{\text{cm}} = \frac{1}{3}\left(\frac{20}{3}\right) = \frac{20}{9}\;\text{m} \approx 2.22\;\text{m}$$

The centre of mass has shifted downward by $$\mathbf{2.22\;m}$$ in 2 s.

Option C which is: $$2.22$$

The bent rod can be treated as two uniform rods of lengths $$3L$$ and $$2L$$ joined at the corner.

For the horizontal rod, its centre of mass is at  $$x_1=L,\qquad y_1=0$$

For the vertical rod, its centre of mass is at  $$x_2=0,\qquad y_2=\frac{3L}{2}$$

Since the rod is uniform, mass is proportional to length.

Thus,  $$x_{cm}=\frac{(2L)(L)+(3L)(0)}{5L}$$

$$x_{cm}=\frac{2L}{5}$$

Similarly,  $$y_{cm}=\frac{(2L)(0)+(3L)\left(\frac{3L}{2}\right)}{5L}$$

$$y_{cm}=\frac{9L}{10}$$

Given,  $$L=10\ \mathrm{cm}$$

Therefore,  $$x_{cm}=4\ \mathrm{cm}$$

$$y_{cm}=9\ \mathrm{cm}$$

Hence, the position vector of the centre of mass is  $${4\hat{i}+9\hat{j}\ \mathrm{cm}}$$

Hence, the correct option is D.