Center of Mass and Collision

The total mass is 14 kg , divided in the ratio 2 : 2 : 3.

Let the common multiplier be x .

$ 2 x+2 x+3 x=14 \Rightarrow 7 x=14 \Rightarrow x=2 $

The masses of the three fragments are :

• 1. $\mathrm{m}_1=2 \times 2=4 \mathrm{~kg}$

• 2. $\mathrm{m}_2=2 \times 2=4 \mathrm{~kg}$

• 3. $\mathrm{m}_3=3 \times 2=6 \mathrm{~kg}$ (Heavier fragment)

The two fragments of 4 kg each fly off perpendicular to each other.

$\overrightarrow{\mathrm{p}}_1=\mathrm{m}_1 \overrightarrow{\mathrm{v}}_1=4 \times 18 \hat{\mathrm{i}}=72 \hat{\mathrm{i}} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$

$\Rightarrow $ $\overrightarrow{\mathrm{p}}_2=\mathrm{m}_2 \overrightarrow{\mathrm{v}}_2=4 \times 18 \hat{\jmath}=72 \hat{\jmath} \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}$

The resultant momentum of these two fragments is :

$ \mathrm{p}_{\mathrm{res}}=\sqrt{\mathrm{p}_1^2++\mathrm{p}_2^2}=\sqrt{72^2+72^2}=72 \sqrt{2} \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s} $

Initially, the body was at rest, so the total initial momentum is zero. Therefore, the final total momentum must also be zero:

$ \overrightarrow{\mathrm{p}}_1+\overrightarrow{\mathrm{p}}_2+\overrightarrow{\mathrm{p}}_3=0 \Rightarrow \overrightarrow{\mathrm{p}}_3=-\left(\overrightarrow{\mathrm{p}}_1+\overrightarrow{\mathrm{p}}_2\right) $

This means the momentum of the third (heavier) fragment must be equal in magnitude and opposite in direction to the resultant momentum of the first two.

$ \left|\overrightarrow{\mathrm{p}}_3\right|=\mathrm{p}_{\mathrm{res}}=72 \sqrt{2} \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s} $

We know $\mathrm{p}_3=\mathrm{m}_3 \times \mathrm{v}_3$ :

$6 \times v_3=72 \sqrt{2}$

$\Rightarrow $ $v_3=\frac{72 \sqrt{2}}{6}$

$\Rightarrow $ $ \mathrm{v}_3=12 \sqrt{2} \mathrm{~m} / \mathrm{s} $

Therefore, the velocity of the heavier fragment is $12 \sqrt{2} \mathrm{~m} / \mathrm{s}$.

Before Collision :

The masses of the spheres are, $\mathrm{m}_{\mathrm{A}}=15 \mathrm{~kg}, \mathrm{~m}_2=25 \mathrm{~kg}$

The initial velocities of the spheres are, $\mathrm{u}_{\mathrm{A}}=10 \mathrm{~m} / \mathrm{s}, \mathrm{u}_{\mathrm{B}}=-30 \mathrm{~m} / \mathrm{s}$

Hence, the kinetic energy of system before collision is,

$ \mathrm{K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}^2+\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}^2 $

$\Rightarrow $ $ \mathrm{K}_{\mathrm{i}}=\frac{1}{2}(15)(10)^2+\frac{1}{2}(25)(30)^2=750+11250=12000 \mathrm{~J} $

Since the collision is perfectly inelastic, the spheres stick together and move with a common final velocity $\mathrm{v}_{\mathrm{f}}$.

Using conservation of linear momentum,

$ m_A u_A+m_B u_B=\left(m_A+m_B\right) v_f $

$\Rightarrow $ $(15)(10)+(25)(-30)=(15+25) v_f$

$\Rightarrow $ $150-750=40 v_f$

$ -600=40 v_f \Rightarrow v_f=-15 \mathrm{~m} / \mathrm{s} $

(The negative sign indicates the combined mass moves in the direction of the 25 kg sphere.

So, the final kinetic energy of the system is

$ \mathrm{K}_{\mathrm{f}}=\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v}_{\mathrm{f}}^2 $

$\Rightarrow $ $\mathrm{K}_{\mathrm{f}}=\frac{1}{2}(40)(15)^2$

$\Rightarrow $ $ \mathrm{K}_{\mathrm{f}}=4500 \mathrm{~J} $

The loss in kinetic energy appears as heat (Q).

$ Q=K_i-K_f $

$\Rightarrow $ $Q=12000 \mathrm{~J}-4500 \mathrm{~J}$

$ Q=7500 \mathrm{~J} $

Using the heat capacity formula $\mathrm{Q}=\operatorname{mc} \Delta \mathrm{T}$.

Where, total mass (m) of the spheres is 40 kg , the specific heat (c) of the material of spheres is $31 \mathrm{cal} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$.

As $1 \mathrm{cal}=4.2 \mathrm{~J}$, so $\mathrm{c}=31 \times 4.2 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}=130.2 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$

Substituting the values in heat capacity formula,

$ 7500=40 \times 130.2 \times \Delta \mathrm{T} $

$\Rightarrow $ $7500=5208 \times \Delta \mathrm{T}$

$\Rightarrow $ $ \Delta \mathrm{T}=\frac{7500}{5208} \approx 1.44^{\circ} \mathrm{C} $

Therefore, the rise in temperature is $1.44^{\circ} \mathrm{C}$.

Hence, the correct option is (B).

Bob A is released from rest at an angle $\theta=60^{\circ}$. It swings down to the lowest point (vertical position) where it hits Bob B.

Length of rod is $\mathrm{l}=1 \mathrm{~m}$.

Height lost by $\mathrm{A}(\Delta \mathrm{h})$ :

$\Delta \mathrm{h}=\mathrm{l}-\mathrm{l} \cos \theta=\mathrm{l}\left(1-\cos 60^{\circ}\right)$

$\Rightarrow $ $ \Delta \mathrm{h}=1(1-0.5)=0.5 \mathrm{~m} $

Using conservation of energy for Bob A :

$ \mathrm{mg} \Delta \mathrm{~h}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{A}}^2 $

$\Rightarrow $ $\mathrm{v}_{\mathrm{A}}=\sqrt{2 \mathrm{~g} \Delta \mathrm{~h}}$

$\Rightarrow $ $ \mathrm{v}_{\mathrm{A}}=\sqrt{2 \mathrm{~g} \times 0.5}=\sqrt{\mathrm{g}} $

It is given that the collision is elastic. Both bobs are identical (mass m ).

Initial velocities: $\operatorname{Bob} A$ has $v_A=\sqrt{g}$, $\operatorname{Bob} B$ is at rest $\left(v_B=0\right)$.

In a perfectly elastic collision between two identical masses where one is initially at rest, their velocities are exchanged.

So, after collision, bob A comes to rest and bob B acquires the velocity of Bob A .

$ \mathrm{v}_{\mathrm{B}}=\mathrm{v}_{\mathrm{A}}=\sqrt{\mathrm{g}} $

Bob B moves along the smooth horizontal surface and enters the vertical circular track of radius R.

The bob B just manages to complete the circular path up to a point Q .

For a particle to just complete a full vertical circle (reach the top), the minimum velocity at the bottom $(\sqrt{5 \mathrm{gR}})$ is required.

$ \mathrm{v}_{\text {bottom }} \geq \sqrt{5 \mathrm{gR}} $

$\Rightarrow \sqrt{g}=\sqrt{5 g R}$

$\Rightarrow $ $\Rightarrow \mathrm{g}=5 \mathrm{gR}$

$ \Rightarrow \mathrm{R}=\frac{1}{5} \mathrm{~m} $

Therefore, the required radius $R$ is $\frac{1}{5} \mathrm{~m}$. Hence, the correct option is (A).

For a mechanical system consisting of $n$ discrete particles, the total kinetic energy ( $K_{\text {total }}$ ) of the system is simply the algebraic sum of the individual kinetic energies of all the constituent particles.

If particle $i$ has mass $m_i$ and velocity $v_i$ relative to a given frame of reference, the total kinetic energy is:

$ \mathrm{K}_{\text {total }}=\sum \frac{1}{2} \mathrm{~m}_{\mathrm{i}} \mathrm{v}_{\mathrm{i}}^2 $

Therefore, Statement I is true.

The total kinetic energy of a system of particles is equal to the sum of translational kinetic energy of the centre of mass and internal kinetic energy.

Translational kinetic energy of the centre of mass is the kinetic energy the system would have if all its mass ( $\mathrm{M}_{\text {total }}$ ) were concentrated at the center of mass and moved with the velocity of the center of mass ( $\mathrm{V}_{\mathrm{cm}}$ ) relative to the origin. $\left(\mathrm{K}_{\mathrm{cm}}=\frac{1}{2} \mathrm{M}_{\text {total }} \mathrm{V}_{\mathrm{cm}}^2\right)$

Internal kinetic energy is the sum of the kinetic energies of all individual particles measured with respect to the centre of mass frame

$ \mathrm{K}_{\text {internal }}=\sum \frac{1}{2} \mathrm{~m}_{\mathrm{i}} \mathrm{v}_{\mathrm{iC}}^2 $

So, the total kinetic energy is

$ \mathrm{K}_{\text {total }}=\frac{1}{2} \mathrm{M}_{\text {total }} \mathrm{V}_{\mathrm{cm}}^2+\sum_\limits{\mathrm{i}=1} \frac{1}{2} \mathrm{~m}_{\mathrm{i}} \mathrm{v}_{\mathrm{iC}}^2 $

Therefore, Statement II is also true. Hence, the correct option is (D).

For two bodies of equal mass $\left(\mathrm{m}_1=\mathrm{m}_2=\mathrm{m}\right)$, the velocity of the COM is:

$ \overrightarrow{\mathrm{v}}_{\mathrm{cm}}=\frac{\mathrm{m} \overrightarrow{\mathrm{v}}_1+\mathrm{m} \overrightarrow{\mathrm{v}}_2}{\mathrm{~m}+\mathrm{m}}=\frac{\overrightarrow{\mathrm{v}}_1+\overrightarrow{\mathrm{v}}_2}{2} $

The velocity of body $A$ is $\vec{v}_1=4 \hat{i}$ and that of body $B$ it is $\vec{v}_2=4 \hat{j}$ :

So,

$ \overrightarrow{\mathrm{v}}_{\mathrm{cm}}=\frac{4 \hat{\imath}+4 \hat{\jmath}}{2}=(2 \hat{\imath}+2 \hat{\jmath}) \mathrm{m} / \mathrm{s} $

Similarly, for equal masses the acceleration of center of mass is,

$ \overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\overrightarrow{\mathrm{a}}_1+\overrightarrow{\mathrm{a}}_2}{2} $

The acceleration of body A is $\overrightarrow{\mathrm{a}}_1=6 \hat{i}+6 \hat{\mathrm{j}}$ and that of body B is $\overrightarrow{\mathrm{a}}_2=0$ :

$ \overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{(6 \hat{\imath}+6 \hat{\jmath})+0}{2}=(3 \hat{\imath}+3 \hat{\jmath}) \mathrm{m} / \mathrm{s}^2 $

A particle moves in a straight line if its acceleration vector is parallel (or anti-parallel) to its velocity vector. If they are not parallel, the path is generally a parabola (under constant acceleration).

Direction of velocity of centre of mass is,

$ \tan \theta_{\mathrm{v}}=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{2}{2}=1 \Rightarrow \theta_{\mathrm{v}}=\tan ^{-1}(1)=\frac{\pi}{4} \mathrm{rad} $

Direction of acceleration of centre of mass is,

$ \tan \theta_{\mathrm{a}}=\frac{\mathrm{a}_{\mathrm{y}}}{\mathrm{a}_{\mathrm{x}}}=\frac{3}{3}=1 \Rightarrow \theta_{\mathrm{a}}=\tan ^{-1}(1)=\frac{\pi}{4} \mathrm{rad} $

Since both vectors have the same direction (they both make an angle of $45^{\circ}$ with the positive x -axis), the acceleration is acting exactly along the line of motion.

Because $\overrightarrow{\mathrm{a}}_{\mathrm{cm}}$ is parallel to $\overrightarrow{\mathrm{v}}_{\mathrm{cm}}$, the center of mass moves in a straight line. Hence, the correct option is (C).

If we have multiple masses $\mathrm{m}_1, \mathrm{~m}_2, \ldots$ at positions $\left(\mathrm{x}_1, \mathrm{y}_1\right),\left(\mathrm{x}_2, \mathrm{y}_2\right), \ldots$, the coordinates of the COM are

$ \begin{aligned} \mathrm{x}_{\mathrm{cm}} & =\frac{\mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2+\mathrm{m}_3 \mathrm{x}_3 \ldots}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3 \ldots} \\ \mathrm{Y}_{\mathrm{cm}} & =\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2+\mathrm{m}_3 \mathrm{y}_3 \ldots}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3 \ldots} \end{aligned} $

Let the midpoint p be the origin $(0,0)$. From the geometry,

In right angled triangle $\mathrm{ACD}, \cos 60^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}} \Rightarrow \mathrm{CD}=\mathrm{AC} \cos 60^{\circ}=10 \times \frac{1}{2} \mathrm{~m}=5 \mathrm{~m}$

Point P is the midpoint of CD , so $\mathrm{PC}=\mathrm{PD}=\frac{\mathrm{CD}}{2}=\frac{5 \mathrm{~m}}{2}=2.5 \mathrm{~m}$

As P is the origin, the coordinate of 15 kg is $(0,2.5) \mathrm{m}$ and the coordinates of D is $(0,-2.5) \mathrm{m}$.

As 2 kg and 3 kg is on the same horizontal level as D, so y-coordinate of 2 kg and 3 kg is $\mathrm{y}_1=-2.5 \mathrm{~m}$

Also, $\sin 60^{\circ}=\frac{\mathrm{AD}}{10} \Rightarrow \mathrm{AD}=10 \sin 60^{\circ}=10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} \mathrm{~m}$

Hence, the $x$-coordinate of 2 kg is $-5 \sqrt{3} \mathrm{~m}$ and that of 3 kg it is $5 \sqrt{3} \mathrm{~m}$

Hence, the coordinates of 2 kg are $(-5 \sqrt{3},-2.5)$, the coordinate of 3 kg are $(5 \sqrt{3},-2.5)$, and that of 15 kg is (0,2.5).

The total mass of system is $\mathrm{M}=2+3+15=20 \mathrm{~kg}$.

So, the x -coordinate of centre of mass is:

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

And the y -coordinate of centre of mass is:

$ y_{c m}=\frac{(15 \times 2.5)+(3 \times(-2.5))+(2 \times(-2.5))}{20}=\frac{37.5-12.5}{20} $

$\Rightarrow $ $ \mathrm{y}_{\mathrm{cm}}=\frac{25}{20}=\frac{5}{4}=1.25 $

Therefore, the centre of mass of system of masses is $\left(\frac{\sqrt{3}}{4}, 1.25\right)$.

Hence, the correct option is (A).

$\begin{aligned} & \mathrm{x}_{\mathrm{com}}=\frac{2 \mathrm{~m}(10)+3 \mathrm{~m}(0)}{5 \mathrm{~m}}=4 \mathrm{~cm} \\ & \mathrm{y}_{\mathrm{com}}=\frac{2 \mathrm{~m}(0)+3 \mathrm{~m}(15)}{5 \mathrm{~m}}=9 \mathrm{~cm} \\ & \overrightarrow{\mathrm{r}}_{\mathrm{com}}=4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}} \end{aligned}$

To solve this problem, we start by using the conservation of linear momentum.

Initially, block A of mass $ m_1 = 10 \, \text{kg} $ is moving with velocity $ v_1 = 3 \, \text{m/s} $, while block B of mass $ m_2 = 5 \, \text{kg} $ is at rest. The blocks move together after the collision. The final velocity of the system is $ v_{\text{cm}} $. Applying the conservation of linear momentum, we have:

$ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{\text{cm}} $

Substituting the known values:

$ 10 \times 3 + 5 \times 0 = (10 + 5) v_{\text{cm}} $

$ 30 = 15 v_{\text{cm}} $

Solving for $ v_{\text{cm}} $:

$ v_{\text{cm}} = \frac{30}{15} = 2 \, \text{m/s} $

Next, we use the conservation of energy to find the compression in the spring. The initial kinetic energy of block A is given by:

$ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 10 \times 3^2 = 45 \, \text{J} $

The final kinetic energy of both blocks moving together at $ v_{\text{cm}} = 2 \, \text{m/s} $ is:

$ \frac{1}{2} (m_1 + m_2) v_{\text{cm}}^2 = \frac{1}{2} \times 15 \times 2^2 = 30 \, \text{J} $

The difference in kinetic energy is the energy stored in the compressed spring:

$ \frac{1}{2} k x^2 = 45 - 30 = 15 \, \text{J} $

Substituting the given spring constant $ k = 3000 \, \text{N/m} $ into the energy equation:

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

$ 1500 x^2 = 15 $

Solving for $ x^2 $:

$ x^2 = \frac{15}{1500} = \frac{1}{100} $

Finally, solving for $ x $:

$ x = \frac{1}{10} = 0.1 \, \text{m} $

Therefore, the compression in the spring is $ 0.1 \, \text{m} $.

Here, $r = \sqrt {{a^2} - {{{a^2}} \over 4}} $

$ \Rightarrow r = {{\sqrt 3 a} \over 2}$

Here, as ${\tau_{ext}} = 0$

we know, $\tau = {{dL} \over {dt}}$

$ \Rightarrow {{dL} \over {dt}} = 0$ $\Rightarrow$ L = Constant

So, we can apply angular momentum conservation. i.e. angular momentum of the system about any point (Let's take point C) is conserved.

$ \Rightarrow {L_f} = {L_i}$

$ \Rightarrow \overrightarrow {{L_f}} = o \times \overrightarrow {{P_A}} + \overrightarrow r \times m{V_0}\widehat BA + o \times \overrightarrow {{P_c}} $ (as ${P_B} = m{v_0}\widehat {BA}$ and $\overrightarrow L = \overrightarrow r \times \overrightarrow P $)

$ \Rightarrow \overrightarrow {{L_f}} = \left( {{{\sqrt 3 a} \over 2}} \right)m{v_0}\sin 90^\circ $ (As $\left| {\overrightarrow r } \right| = {{\sqrt 3 a} \over 2}$)

$ \Rightarrow \overrightarrow {{L_f}} = {{\sqrt 3 } \over 2}am{v_0}$ (outwards & $ \bot $ to plane)

$ \Rightarrow {L_f} = {{\sqrt 3 } \over 2}am{v_0}$

Hence, option D is correct.

We know that two bodies of the same mass exchange their velocities after an elastic head-on collision.

Given, initially,

Here, ${V_{AB}} = {v_A} - {v_B} = 5 - 2 = 3\,m/s$

${V_{CB}} = {v_C} - {v_B} = 4 - 2 = 2\,m/s$

first A will collide with B and the velocities will be exchanged.

i.e. now, V$_A$ = 2 m/s and V$_B$ = 5 m/s and V$_C$ = 4 m/s

As $V_B > V_C$

So, B will collide and exchange the velocities with C.

So, $V_B=4$ m/s. and $V_C=5$ m/s.

Hence, finally,

$\eqalign{ & {V_A} = 2\,m/s \cr & {V_B} = 4\,m/s \cr & {V_C} = 5\,m/s \cr} $

So, A is false but R is true.

Velocity of a just before hitting :

$\mathrm{u}=\sqrt{2 \mathrm{~g} \frac{\mathrm{R}}{2}}=\sqrt{\mathrm{gR}}$

Just after collision, let velocity of A and B are $\mathrm{v}_1$ and $v_2$ respectively

$\therefore \text { by COM: }$

$\begin{aligned} & \mathrm{mu}=\mathrm{mv}_1+\frac{\mathrm{m}}{2} \mathrm{v}_2 \\ & 2 \mathrm{v}_1+\mathrm{v}_2=2 \mathrm{u} \quad\text{...... (i)}\\ & \mathrm{e}=1=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}} \\ & \Rightarrow \mathrm{v}_2-\mathrm{v}_1=\mathrm{u} \quad\text{...... (ii)} \end{aligned}$

From (i) -(ii)

$\Rightarrow 3 \mathrm{v}_1=\mathrm{u} \Rightarrow \mathrm{v}_1=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}$

$\sigma = {{{\sigma _0}x} \over {ab}}$

Let's consider a strip of infinitesimal width dx at x distance away from O.

As $\sigma$ depends only on x i.e. it's constant for a specific value of x so centre of mass of the strip will be $\left( {x,{b \over 2}} \right)$.

Now, we know,

$COM = \left( {{{\int {xdm} } \over {\int {dm} }},{{\int {ydm} } \over {\int {dm} }}} \right)$

$ = \left( {{{\int {x\sigma dA} } \over {\int {\sigma dA} }},{{\int {{b \over 2}\sigma dA} } \over {\int {\sigma dA} }}} \right)$

$ = \left( {{{\int {xbdx{{{\sigma _0}x} \over {ab}}} } \over {\int {{{{\sigma _0}x} \over {ab}}bdx} }},{{{b \over 2}\int {\sigma dA} } \over {\int {\sigma dA} }}} \right)$

For whole plate, x varies from o to a.

So, $ = \overleftarrow C OM = \left( {{{{{{\sigma _0}} \over a}\int_0^a {{x^2}dx} } \over {{{{\sigma _0}} \over a}\int_0^a {xdx} }},{b \over 2}} \right)$

$ = \left( {{{{1 \over 3}\left[ {{x^3}} \right]_0^a} \over {{1 \over 2}\left[ {{x^2}} \right]_0^a}},{b \over 2}} \right)$

$ = \left( {{2 \over 3}{{{a^3}} \over {{a^2}}},{b \over 2}} \right)$

$ \Rightarrow COM = \left( {{2 \over 3}a,{b \over 2}} \right)$

Let's solve this step by step.

The original disc has a radius of 20 cm and is uniformly dense. Its mass (or weight) is proportional to its area:

$M = \pi \times 20^2 = 400\pi.$

A hole of radius 5 cm is cut out. Similarly, its mass (if it were a complete disc) would be:

$m = \pi \times 5^2 = 25\pi.$

Since the hole cuts through the disc, we treat it as having a negative mass. The center of the original disc is at the origin (0,0), and we need to locate the center of the hole.

The problem states that the edge of the hole touches the edge of the disc. This means that the distance between the centre of the hole and the centre of the large disc is:

$20 \text{ cm (disc radius)} - 5 \text{ cm (hole radius)} = 15 \text{ cm}.$

For convenience, we can assume that the hole is positioned along the positive x-axis. Thus, the centre of the hole is at:

$\vec{r}_\text{hole} = (15, 0).$

Now, the centre of mass (COM) of the remaining shape (the disc with the hole) can be calculated using the idea of superposition:

$\vec{R}_{\text{cm}} = \frac{M \cdot \vec{r}_\text{disc} - m \cdot \vec{r}_\text{hole}}{M - m}.$

Here, $\vec{r}_\text{disc} = (0,0)$ (since the large disc is centered at the origin).

Plug in the values:

$\vec{R}_{\text{cm}} = \frac{400\pi \cdot (0,0) - 25\pi \cdot (15,0)}{400\pi - 25\pi} = \frac{-(25\pi)(15,0)}{375\pi} = \frac{-375\pi \cdot (1,0)}{375\pi} = (-1,0).$

This tells us that the centre of mass of the remaining piece is 1 cm from the origin, along the negative x-axis.

Thus, the distance of the centre of mass from the origin is:

$\boxed{1.0\text{ cm}}.$

So, the correct answer is Option D.

The ball is dropped from a height of 42 m. The coefficient of restitution, $e$, is 0.4. This number tells us how much energy the ball keeps after bouncing.

At first, the ball is dropped. So, its starting speed ($u$) is 0.

Height after first bounce:

The ball comes back up to a height equal to $e^2$ times the height it fell from. So:

$ h_1 = (0.4)^2 \times 42 $

$ h_1 = 0.16 \times 42 = 6.72~\mathrm{m} $

Height after later bounces:

Each time the ball bounces, it goes up to $e^2$ times the previous bounce's height.

$ h_2 = (0.4)^2 \times 6.72 = 0.16 \times 6.72 $

Total Distance:

When the ball is dropped, it first travels 42 m down.

After that, on every bounce, it goes up and then comes down the same height. So, for all the bounces, we call the sum of all the heights $h_1, h_2, h_3, \ldots$.

Total distance = distance of first fall + 2 × (sum of all bounce heights)

$ = 42 + 2 \times [h_1 + h_2 + h_3 + \ldots] $

Summing the Series:

All the bounce heights make a series called a geometric progression (GP). The first term $a = 42$ and the common ratio $r = 0.16$.

The sum of an infinite GP is given by:

Sum $= \frac{a}{1 - r} = \frac{42}{1 - 0.16}$

$ \text{Sum} = \frac{42}{0.84} = 50 $

Final Calculation:

Now, total distance is:

$ \text{Total distance} = 42 + 2 \times (0.4)^2 \times 50 $

$ = 42 + 2 \times 0.16 \times 50 $

$ = 42 + 16 = 58~\mathrm{m} $

According to conservation of linear momentum

$ \begin{aligned} & \Rightarrow \quad m v+2 m \times 0=m v_1+2 m v_2 \\ & \Rightarrow \quad v=v_1+2 v_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Coefficient of restitution

$ \begin{aligned} e & =\frac{v_2-v_1}{u_1-u_2} \\ \Rightarrow \quad e \quad e & =\frac{v_2-v_1}{v-0} \\ \Rightarrow \quad e v & =v_2-v_1 \\ \Rightarrow \quad v_1 & =v_2-e v \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} & v=v_2-e v+2 v_2 \\ \quad & v=3 v_2-e v \\ \Rightarrow \quad & v(1+e)=3 v_2 \\ \Rightarrow \quad & v_2=\frac{v(1+e)}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Now, from Eq. (ii) and (iii), we get,

$ \begin{aligned} v_1 & =\frac{v(1+e)}{3}-e v=\frac{v(1-2 e)}{3} \\ \Rightarrow \quad v_1 & =\frac{v(1-2 e)}{3} \end{aligned} $

Thus, $\frac{v_1}{v_2}=\frac{1-2 e}{1+e}$

Let the centre of the sphere be at origin ( $x=0$ ).

Then, the centre of the cylinder lies at a distance of $x=0.3 \mathrm{~m}$ from the sphere centre.

CM of the system

$ \begin{aligned} X_{\mathrm{CM}} & =\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\ & =\frac{2 \times 0.3+0.5 \times 0}{2+0.5} \\ X_{\mathrm{CM}} & =\frac{0.6}{2.5}=0.24 \mathrm{~m} \end{aligned} $

or $X_{\mathrm{CM}}=24 \mathrm{~cm}$ (From the centre of sphere)

For block $m$ (moving down)

$ m g-T=m a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For block $n$ (moving up)

$ T-n g=n a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we get

$ \begin{aligned} m g-n g & =m a+n a \\ (m-n) g & =(m+n) a \\ a & =\frac{(m-n)}{(m+n)} g \end{aligned} $

$ \therefore \quad a_m=+a \text { and } a_n=-a $

then, $a_{\mathrm{CM}}=\frac{m a-n a}{m+n}=\frac{(m-n)}{(m+n)} \times a$

$ a_{\mathrm{CM}}=\left(\frac{m-n}{m+n}\right)^2 g $

Velocity of mass 2 kg is $20 \mathrm{~m} / \mathrm{s}$ North.

So, $\quad v_1=20 \hat{\mathbf{j}}$

$So,\,\,\,\,\, v_2=10 \hat{\mathbf{i}} $

Now, velocity of centre of mass

$ \begin{aligned} & v_{\mathrm{CM}}=\frac{m_1 v_1+m_2 v_2}{\left(m_1+m_2\right)} \\ & v_{\mathrm{CM}}=\frac{2 \times 20 \hat{\mathbf{j}}+3 \times 10 \hat{\mathbf{i}}}{(2+3)} \\ & v_{\mathrm{CM}}=8 \hat{\mathbf{j}}+6 \hat{\mathbf{i}} \end{aligned} $

So, magnitude of velocity

$ \left|v_{\mathrm{CM}}\right|=\sqrt{8^2+6^2}=10 \mathrm{~m} / \mathrm{s} $

For a single bounce, the height after impact $\left(h_n\right)$ is related to the heigil before impact $\left(h_{n-1}\right)$ by the equation

$ h_n=e^2 h_{n-1} $

Height after the first impact,

$ h_1=e^2 h_0 $

$ \begin{aligned} &\text { Height after the second impact }\\ &\begin{array}{rlrl} h_2 & =e^2 h_1=e^2\left(e^2 h_0\right) \\ h_2 & =e^4 h_0 \\ \therefore \quad & \frac{h_2}{h_0} =e^4=(0.8)^4 \\ & =0.4096=\frac{4096}{10000}=\frac{256}{625} \\ \Rightarrow \quad & h_2: h_0 =256: 625 \end{array} \end{aligned} $

The two bodies move due to their mutual attraction. Force are internal forces within the system. Since no external forces act on the system, the net external force is zero.

Initially, both bodies are at rest, thus, their Initial momentum is zero.

According to the principle of conservation of linear momentum, with out any external force total linear momentum of the system remains constant. So final momentum is also zero

Consequently, the velocity of centre of mass also remains constant.

Since, the initial velocity of the centre of mass $=0$

Thus, final velocity of the centre of mass $=0$

According to conservation of momentum

$ m v_i+2 m \times 0=(m+2 m) v_f $

$ \Rightarrow \quad v_f=\frac{v_i}{3} $

∴ Loss in kinetic energy

$ \begin{aligned} & \triangle \mathrm{KE}=\frac{1}{2} m v_i^2-\frac{1}{2}(3 m) v_f^2 \\ & =\frac{1}{2} m v_i^2-\frac{3}{2} m\left(\frac{v_i}{3}\right)^2 \\ & =\frac{1}{2} m v_i^2-\frac{1}{6} m v_i^2=\frac{1}{3} m v_i^2 \end{aligned} $

∴ Fraction of kinetic energy lost

$ =\frac{\Delta \mathrm{KE}}{K_i}=\frac{\frac{1}{3} m v_i^2}{\frac{1}{2} m v_i^2}=\frac{2}{3} $

As disc is floating in air without falling, force by bullets balances gravitational force.

$ \Rightarrow \quad\left(\frac{N \Delta p}{\Delta t}\right)=m_d g $

Here, $N=$ number of bullets fired $\frac{\Delta p}{\Delta t}=$ Rate of change of one bullet $m_d=$ mass of disc

Hence, bullets rebound with same speed, momentum change for each bullet $=m(v-(-v))=2 m v$

$ \therefore \quad \frac{N}{t} \cdot 2 m_b v=m_d g $

Substituting given values,

$ \begin{aligned} & 10 \times 2 \times 0.05 \times v=0.2 \times 10 \\ & \therefore \quad v=\frac{0.2 \times 10}{10 \times 2 \times 0.05}=2 \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \text { Before collision } $

Let initial velocities are $u$ and $\frac{u}{2}$ and final velocities are $v_A$ and $v_B$. Then, momentum is conserved,

$ m u+\frac{m u}{2}=m v_A+m v_B $

or

$ v_A+v_B=\frac{3 u}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Coefficient of restitution is 0.5 ,

$ \begin{aligned} & \Rightarrow \quad e=\frac{v_B-v_A}{u_A-u_B}=\frac{v_B-v_A}{u-\left(\frac{u}{2}\right)} \\ & \Rightarrow \quad v_B-v_A=\frac{u}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From (i) and (ii)

$ v_B=\frac{7}{8} \cdot u \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

And from (i) and (iii)

$ v_A=\frac{5}{8} u \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Hence, ratio $\frac{v_A}{v_B}=\frac{\frac{5}{8}}{\frac{7}{8}}=\frac{5}{7}$

If $\sigma$ be the mass per unit area, Then,

$ \begin{aligned} \sigma & =\frac{3}{2 \times 1+1 \times 1} \\ & =1 \mathrm{~kg} / \mathrm{m}^2 \end{aligned} $

$\therefore$ mass of rectangle-1,

$ m_1=\sigma A_1=1 \times 2 \times 1=2 \mathrm{~kg} $

mass of rectangle-2,

$ \begin{aligned} & m_2=\sigma A_2=1 \times 1 \times 1=1 \mathrm{~kg} \\ & \therefore X_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\ & \quad=\frac{2 \times \frac{1}{2}+1 \times \frac{3}{2}}{2+1}=\frac{5}{6} \mathrm{~m} \end{aligned} $

and $Y_{\mathrm{CM}}=\frac{m_1 y_1+m_2 y_2}{m_1+m_2}$

$ =\frac{2 \times 1+1 \times \frac{1}{2}}{2+1}=\frac{5}{6} \mathrm{~m} $

$\therefore$ Centre of mass

$ =\left(X_{\mathrm{CM}}, Y_{\mathrm{CM}}\right)=\left(\frac{5}{6} \mathrm{~m}, \frac{5}{6} \mathrm{~m}\right) $

First we calculate centre of mass of system let block $A$ is at origin ( $x=0$ )

$ \begin{aligned} & x_{C M}=\frac{10(0)+25(10)+15(20)}{10+25+15} \\ & x_{C M}=\frac{550}{50}=11 \mathrm{~m} \end{aligned} $

Now, new position of block $A$ with respect to origin is $x=2 \mathrm{~m}$ and new position of block $C, x=20-3=17 \mathrm{~m}$.

Let new position of block $B$ with respect to origin is $Y$, so that centre of mass remains at same position

$ \begin{aligned}Then,\,\, x_{\mathrm{CM}} & =11=\frac{10(2)+25(y)+15(17)}{50} \\ 550 & =20+25 y+255 \\ 25 y & =550-275=275 \\ y & =\frac{275}{25}=11 \mathrm{~m} \end{aligned} $

So, block $B$ should move $11-10=1$ metre towards block $C$.

Impulse on ball $=$ Change in momentum

$ \begin{aligned} & =m(v-u) \\ & =0.25[16-(-16)] \\ & =8 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned} $

According to conservation of momentum.

$ \begin{aligned} & m u=m_1 v_1+m_2 v_2 \\ & 10 \times 5 \hat{\mathbf{i}}=4 \times 10 \hat{\mathbf{i}}+6 \times v_2 \\ & \Rightarrow \quad v_2=\frac{5}{3} \hat{\mathbf{i}}=1.67 \hat{\mathbf{i}} \mathrm{~ms}^{-1} \end{aligned} $

Given:

The final velocity of the first sphere, $v_1 = \frac{7}{9}u_1$

$u_2 = 0$ (the second sphere is at rest)

The mass of a sphere is proportional to the cube of its radius: $m \propto R^3$

Let $m_1$ and $m_2$ be the masses, and $R_1$ and $R_2$ be the radii of the first and second spheres.

Step 1: Use the formula for an elastic collision

After the collision, the velocity of the first sphere is:

$ v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2} $

Step 2: Substitute $u_2 = 0$

$ v_1 = \frac{(m_1 - m_2)u_1}{m_1 + m_2} $

We know $v_1 = \frac{7}{9}u_1$:

$ \frac{7}{9}u_1 = \frac{(m_1 - m_2)u_1}{m_1 + m_2} $

Divide both sides by $u_1$ (since $u_1 \neq 0$):

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

Step 3: Solve for $m_1$ in terms of $m_2$

Multiply both sides by $(m_1 + m_2)$:

$ 7(m_1 + m_2) = 9(m_1 - m_2) $

$7m_1 + 7m_2 = 9m_1 - 9m_2$

$7m_2 + 9m_2 = 9m_1 - 7m_1$

$16m_2 = 2m_1$

$m_1 = 8m_2$

Step 4: Relate masses to radii

Since mass of a sphere, $m \propto R^3$:

$ \frac{m_1}{m_2} = \left(\frac{R_1}{R_2}\right)^3 $

$8 = \left(\frac{R_1}{R_2}\right)^3$

Step 5: Solve for $R_2$

Take the cube root of both sides:

$\frac{R_1}{R_2} = 2$

$R_1 = 2R_2$

$R_2 = \frac{R_1}{2}$

Plug in $R_1 = 1.2$ cm:

$R_2 = \frac{1.2}{2} = 0.6 \ \mathrm{cm}$

$ \begin{aligned} \text { } \begin{aligned} v_{\mathrm{CM}} & =\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\ = & \frac{2 \times 20 \hat{\mathbf{i}}+3 \times 10 \hat{\mathbf{j}}}{2+3}=\frac{40 \hat{\mathbf{i}}+30 \hat{\mathbf{j}}}{5} \\ v_{\mathrm{CM}} & =8 \hat{\mathbf{i}}+6 \hat{\mathbf{j}} \\ \therefore\left|v_{\mathrm{CM}}\right| & =\sqrt{8^2+6^2}=\sqrt{100}=10 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned} $

Step 1: Find the final velocities after the collision

The first particle has mass $8\ \mu \mathrm{g}$ and moves with speed $u_1$. The second particle has mass $4\ \mu \mathrm{g}$ and is at rest, so $u_2 = 0$.

The formula for the final speed of the first particle after an elastic, one-dimensional collision is: $ v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2} $ Plug in the numbers:

$ v_1 = \frac{(8 - 4)u_1 + 2 \times 4 \times 0}{8 + 4} = \frac{4u_1}{12} = \frac{u_1}{3} $

The formula for the final speed of the second particle is: $ v_2 = \frac{2m_1 u_1 + (m_2 - m_1)u_2}{m_1 + m_2} $ Plug in the numbers:

$ v_2 = \frac{2 \times 8 u_1 + (4 - 8) \times 0}{8+4} = \frac{16u_1}{12} = \frac{4}{3}u_1 $

Step 2: Find the de-Broglie wavelength for each particle

The de-Broglie wavelength formula is: $ \lambda = \frac{h}{mv} $ where $h$ is Planck’s constant, $m$ is mass, and $v$ is velocity.

To compare their wavelengths, find the ratio: $ \frac{\lambda_2}{\lambda_1} = \frac{m_1 v_1}{m_2 v_2} $

Plug in the values found earlier: $ \frac{\lambda_2}{\lambda_1} = \frac{8 \times \frac{u_1}{3}}{4 \times \frac{4}{3}u_1} $

Simplify the expression: $ = \frac{8 \times \frac{u_1}{3}}{4 \times \frac{4u_1}{3}} = \frac{8u_1/3}{16u_1/3} = \frac{8u_1}{16u_1} = \frac{1}{2} $

So, $ \frac{\lambda_1}{\lambda_2} = 2 $ which means the ratio of their wavelengths after the collision is $2 : 1$.

A stationary particle breaks into two parts with masses $m_A$ and $m_B$, which then move with velocities $v_A$ and $v_B$, respectively. We need to determine the ratio of their kinetic energies $K_A$ and $K_B$.

Since the initial momentum of the particle is zero, the momentum of the two parts must be equal and opposite to conserve momentum:

$ m_A v_A = m_B v_B $

Here, the ratio of kinetic energies is given by:

$ \frac{K_A}{K_B} = \frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2} = \frac{m_A v_A^2}{m_B v_B^2} $

However, using the momentum relationship, we can substitute $m_A v_A = m_B v_B$ into the kinetic energy ratio, leading us to simplify:

$ \frac{K_A}{K_B} = \frac{v_A}{v_B} $

Therefore, the ratio of their kinetic energies $\left(K_B: K_A\right)$ is:

$ \frac{K_B}{K_A} = \frac{v_B}{v_A} $

$\begin{aligned} & \left|\overrightarrow{\mathrm{p}_1}\right|=\left|\overrightarrow{\mathrm{p}_2}\right| \\ & \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text { same } \\ & \mathrm{KE} \propto \frac{1}{\mathrm{~m}} \\ & \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1} \end{aligned}$

$\begin{aligned} \vec{I} & =\Delta \vec{P}=\vec{P}_f-\vec{P}_i \\ \mathrm{M} & =0.1 \mathrm{~kg} \\ I & =\Delta P=0.1(\sqrt{2 \times 9.8 \times 5}-(-\sqrt{2 \times 9.8 \times 10})) \\ & =0.1(14+7 \sqrt{2}) \approx 2.39 \mathrm{~kg} \mathrm{~ms}^{-1} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{P}_1^2}{2 \mathrm{~m}_1}=\frac{\mathrm{P}_2^2}{2 \mathrm{~m}_2} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\sqrt{\frac{\mathrm{m}_1}{\mathrm{~m}_2}}=\frac{2}{5} \end{aligned}$

Momentum will remain conserve

$\begin{aligned} & 1000 \times 6=1200 \times v \\ & v=5 \mathrm{~m} / \mathrm{s} \end{aligned}$

Given,

Mass of ball, $ m_{A}=1.2 \mathrm{~kg} $

Mass of ball, $ m_{B}=3.6 \mathrm{~kg} $

Initial velocity of ball, $ v_{A_{i}}=8.4 \mathrm{~ms}^{-1} $

Initial velocity of ball, $ v_{B_{i}}=0 \mathrm{~ms}^{-1} $

In a one-dimensional elastic collision, the final velocities ( $ v_{A_{f}} $ and $ v_{B_{f}} $ ) of the two masses can be calculated using the following formula

$ v_{A_{f}}=\frac{\left(m_{A}-m_{B}\right) \cdot v_{A_{i}}+2 m_{B} v_{B_{i}}}{m_{A}+m_{B}} $

$ \begin{array}{l} \text { After putting values, we get } \\\\ \qquad \begin{aligned} v_{A_{f}} & =\frac{(1.2-3.6) 8.4}{1.2+3.6}=\frac{-2.4 \times 8.4}{4.8} \\\\ v_{A_{f}} & =-4.2 \mathrm{~m} / \mathrm{s} \\\\ \text { Similarly, } v_{B_{f}} & =\frac{\left(m_{B}-m_{A}\right) v_{B_{i}}+2 m_{A} v_{A_{i}}}{m_{A}+m_{B}} \\\\ v_{B_{f}} & =\frac{0+2 \times 1.2 \times 8.4}{1.2+3.6} \\\\ & =\frac{2.4 \times 8.4}{4.8} \\\\ V_{B_{F}} & =4.2 \mathrm{~m} / \mathrm{s} \end{aligned} \end{array} $

Initial kinetic energy of ball $ A $ is,

$ \begin{aligned} \mathrm{KE}_{A_{i}} & =\frac{1}{2} m_{A}\left(v_{A_{i}}\right)^{2}=\frac{1}{2} \times 1.2 \times 8.4 \times 8.4 \\\\ & =42.33 \mathrm{~J} \end{aligned} $

Final kinetic energy of ball $ B $ is,

$ \begin{aligned} \mathrm{KE}_{B_{f}} & =\frac{1}{2} m_{B}\left(v_{B_{f}}\right)^{2} \\\\ & =\frac{1}{2} \times 3.6 \times 4.2 \times 4.2=31.75 \mathrm{~J} \end{aligned} $

The percentage of kinetic energy

$ \begin{array}{l} \text { transferred from ball } A \text { to ball } B \text { is, } \\\\ \qquad \begin{aligned} \mathrm{KE}_{\text {transferred }} & =\frac{\mathrm{KE}_{A_{i}}}{\mathrm{KE}^{\prime}} \times 100 \% \\\\ & =\frac{31.75}{42.33} \times 100 \%=75 \% \end{aligned} \end{array} $

Let mass of meter scale be $ m $.

Mass of each coin $ =9 \mathrm{~g} $

Total mass of coin $ =18 \mathrm{~g} $

Length of meter scale $ =100 \mathrm{~cm} $

The new balance point is at 35 cm

The torque due to the coins placed at the 10 cm mark

Distance from the new balance point to the coins $ =(35-10) \mathrm{cm} $

$ =25 \mathrm{~cm} $

Torque due to coins

$ =F \cdot r=m g r=18 \times g \times 25 $

The centre of mass of the meter scale is at its mid-point ( 50 cm mark).

Distance from the new balance point to the $ =(50-35) \mathrm{cm} $

Centre of the meter scale $ =15 \mathrm{~cm} $

Torque due to meter scale $ =m \times g \times 15 $

For the scale to be balanced, the net torque will conserver. So, from Eq. (i) and Eq. (ii), we get

$ \begin{array}{l} 18 \times g \times 25=m \times g \times 15 \\\\ m=\frac{18 \times 25}{15} \\\\ m=30 \mathrm{~g} \end{array} $

$ u_1=10 \mathrm{~m} / \mathrm{s} \quad u_2=0 \mathrm{~m} / \mathrm{s} $

Coefficient of restitution,

$ e=0.4 $

Velocity of $m_1$ (ball $P$ ) after collision is given by

$ \begin{aligned} & v_1=\left(\frac{m_1-e m_2}{m_1+m_2}\right) u_1 \\\\ & v_1=\frac{(0.5-0.4 \times 1)}{1.5} \times 10 \\\\ & v_1=\frac{2}{3} \mathrm{~m} / \mathrm{s} \end{aligned} $

Velocity of ball $ Q $ (initially rest) after collision

$ v_{2}=\left[\frac{m_{2}-e m_{1}}{m_{1}+m_{2}}\right] u_{2}+\frac{(1+e) m_{1} u_{1}}{m_{1}+m_{2}} $

$ v_{2}=\frac{(1+e) m_{1} u_{1}}{m_{1}+m_{2}} $

(as $ u_{2}=0 $ )

$ v_{2}=\frac{(1+0.4)(0.5)(10)}{1.5} $

$ v_{2}=\frac{14}{3} \mathrm{~m} / \mathrm{s} $

Ratio of $ v_{1} $ and $ v_{2} $

$ \begin{aligned} \frac{v_{1}}{v_{2}} & =\frac{2}{3} \times \frac{3}{14}=\frac{1}{7} \\\\ v_{1}: v_{2} & =1: 7 \end{aligned} $

When the section is cut out we use area formula of COM.

$ X_{\mathrm{cm}}=\frac{A_{1} x_{1}-A_{2} x_{2}}{A_{1}-A_{2}} $

where $ A_{1} $ area of plate $ P, A_{2} $ area of cut out portion

$ \begin{array}{l} x_{1}=0, x_{2}=2 r, A_{1}=\pi(4 r)^{2}, A_{2}=\pi(r)^{2} \\\\ X_{\mathrm{cm}}=\frac{0-\pi r^{2} \cdot 2 r}{\pi(4 r)^{2}-\pi r^{2}} \\\\ X_{\mathrm{cm}}=\frac{-\pi r \cdot 2}{\pi \mathrm{l} 5} \end{array} $

$ \begin{aligned} & X_{\mathrm{cm}}=-\frac{2 r}{15} \\\\ & \text { Distance }=\left|X_{\mathrm{cm}}\right|=\frac{2 r}{15} \end{aligned} $

Given,

Mass $ m_{1}=1.2 \mathrm{~kg}, u_{1}=12 \mathrm{~m} / \mathrm{s} $. $ m_{2}=1.2 \mathrm{~kg}, u_{2}=0 $

Coefficient of restitution $ =1 / \sqrt{2} $

Now apply conservation of linear momentum equation for the collision.

$ m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} $

( $ v_{1}, v_{2}= $ velocity after collision)

$ m\left[u_{1}+u_{2}\right]=m\left[v_{1}+v_{2}\right] \quad \text { [same masses] } $

$ \text { as } u_{2}=0, u_{1}=12 \mathrm{~m} / \mathrm{s} $

$ 12=v_{2}+v_{1} $

From coefficient of restitution

$ \begin{aligned} e & =\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \Rightarrow \frac{1}{\sqrt{2}}=\frac{v_{2}-v_{1}}{12} \\\\ \Rightarrow 6 \sqrt{2} & =v_{2}-v_{1} \end{aligned} $

Solving Eqs. (i) and (ii)

$ v_{1}=(6-3 \sqrt{2}) \mathrm{m} / \mathrm{s} $

and

$ v_{2}=(3 \sqrt{2}+6) \mathrm{m} / \mathrm{s} $

Kinetic energy before collision

$ =\frac{1}{2} m_{1} u_{1}^{2}=\frac{1}{2}(m)(144) $

Kinetic energy after collision

$ \begin{array}{l} =\frac{1}{2} m\left[v_{1}^{2}+v_{2}^{2}\right] \\\\ =\frac{1}{2} m[108] \end{array} $

Ratio of final (after collision) kinetic energy to initial

$ =\frac{\frac{1}{2} m[108]}{\frac{1}{2} m[144]}=\frac{3}{4}=3: 4 $

Initial condition,

Centre of mass of two rod, ( $ Y $-axis)

$Y_{\mathrm{cm}}=\frac{m_1 y_1+m_2 y_2}{m_1+m_2}$

As rod are identical $ m_{1}=m_{2}=m $

$ \begin{array}{l} y_{1}=\frac{L}{2}, y_{2}=L+\frac{a}{2} \\\\ Y_{\mathrm{cm}}=\frac{m\left[\frac{L}{2}+L+\frac{a}{2}\right]}{2 m}=\frac{3 L+a}{4} \end{array} $

Final case,

Centre of mass of two rod,

$ \begin{aligned} Y_{\mathrm{cm}}^{\prime} & =\frac{m_{1} y_{1}^{\prime}+m_{2} y_{2}^{\prime}}{m_{1}+m_{2}} \\\\ y_{1}^{\prime} & =\frac{a}{2}, y_{2}^{\prime}=a+\frac{L}{2} \end{aligned} $

$ Y_{\mathrm{cm}}^{\prime}=\frac{1}{2}\left[\frac{3 a+L}{2}\right]=\frac{3 a+L}{4} $

Difference in shift

$ Y_{\mathrm{cm}}-Y_{\mathrm{cm}}^{\prime}=\frac{1}{2}[L-a] $

From the figure, velocity of centre of mass,

$ \begin{aligned} v_{\mathrm{COM}} & =\frac{m \times(6 \hat{\mathbf{j}})+2 m(-12 \hat{\mathbf{j}})+3 m(8 \hat{\mathbf{i}})}{m+2 m+3 m} \\\\ & =\frac{6 m \hat{\mathbf{j}}-24 m \hat{\mathbf{j}}+24 m \hat{\mathbf{i}}}{6 m} \\\\ & =-3 \hat{\mathbf{j}}+4 \hat{\mathbf{i}} \end{aligned} $

Magnitude of velocity of centre of mass

$ \begin{aligned} & =\sqrt{3^2+4^2} \\\\ & =5 \mathrm{~m} / \mathrm{s} \end{aligned} $

Here,

$ \begin{aligned} m \cdot v_{A_1} & =m \cdot v_{B_2}+2 m \cdot v_{B_2} \\\\ v_{B_2}-v_{A_2} & =0.4 \times v_{A_1} \\\\ m \cdot v_{A_1} & =m \cdot v_{A_2}+2 m \times\left(0.4 \times v_{A_1}+v_{A_2}\right) \end{aligned} $

For $A$

$ \begin{aligned} & v_{A_1}=3 \cdot v_{A_2}+0.8 \times v_{A_1} \\\\ & v_{A_2}=\frac{0.2}{3} \times v_{A_1} \end{aligned} $

For $B$

$ \begin{aligned} v_{B_2} & =\frac{0.4 \times 3+0.2}{3} \times v_{A_1} \\\\ & =\frac{1.4}{3} v_{A_1} \end{aligned} $

The ratio of both masses

$ \begin{aligned} & \frac{v_{A_2}}{v_{B_2}}=\frac{\left(\frac{0.2}{3}\right) \times v_{A_1}}{\left(\frac{1.4}{3}\right) \times v_{A_1}} \\\\ & \frac{v_{A_2}}{v_{B_2}}=\frac{1}{7} \end{aligned} $

Original system (with 4 particles)

Let coordinates of the particle as

$ \begin{aligned} & A=(0, a) \\\\ & B=(a, a) \\\\ & C=(a, 0) \\\\ & D=(0,0) \end{aligned} $

The centre of mass of system

$ \begin{gathered} X_{\text {com }}=\frac{m x_A+m x_B+m x_C+m x_D}{4 m} \\\\ Y_{\text {com }}=\frac{m y_A+m y_B+m y_C+m y_D}{4 m} \end{gathered} $

Solving,

$ \begin{aligned} & X_{\mathrm{com}}=\frac{0+a+a+0}{4}=\frac{a}{2} \\\\ & Y_{\mathrm{com}}=\frac{a+a+0+0}{4}=\frac{a}{2} \end{aligned} $

The centre of mass of original system at $\left(\frac{a}{2}, \frac{a}{2}\right)$

New centre of mass, when one particle (B) is removed,

$ \begin{gathered} X_{\text {com }}^{\prime}=\frac{m \cdot x_A+m \cdot x_C+m \cdot x_D}{3 m} \\\\ Y_{\text {com }}^{\prime}=\frac{m \cdot y_A+m \cdot y_C+m \cdot y_D}{3 m} \\\\ X_{\text {con }}^{\prime}=\frac{0+a+a}{3}=\frac{2 a}{3} \\\\ Y_{\text {com }}^{\prime}=\frac{a+0+0}{3}=\frac{a}{3} \end{gathered} $

Now, shift in position

$ \sqrt{\left(\frac{2 a}{3}-\frac{a}{2}\right)^2+\left(\frac{a}{3}-\frac{a}{2}\right)^2}=\frac{a}{3 \sqrt{2}} $

Thus, centre of mass of 3 system shift to $\frac{a}{3 \sqrt{2}}$