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.

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

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

To find the number of bullets fired per second, we can use the concept of momentum. When the machine gun fires bullets, it experiences a backward force due to the conservation of momentum. The force applied on the machine gun is used to balance this backward force.

First, let's find the momentum of each bullet:

Momentum = Mass × Velocity

Bullet mass = $10 \mathrm{~g} = 0.01 \mathrm{~kg}$ Bullet velocity = $250 \mathrm{~m/s}$

Momentum per bullet = $0.01 \mathrm{~kg} \cdot 250 \mathrm{~m/s} = 2.5 \mathrm{~kg \cdot m/s}$

Now let's find the momentum per second that needs to be balanced by the applied force:

Force = $125 \mathrm{~N}$

Momentum per second = Force × Time

Since we are considering a time interval of 1 second, the momentum per second is equal to the applied force:

Momentum per second = $125 \mathrm{~N}$

Now, let's find the number of bullets fired per second:

Number of bullets = Momentum per second / Momentum per bullet

Number of bullets = $\frac{125 \mathrm{~N}}{2.5 \mathrm{~kg \cdot m/s}}$

Number of bullets = 50

Therefore, the machine gun fires 50 bullets per second.

When two particles collide, the total momentum before the collision is equal to the total momentum after the collision, according to the law of conservation of momentum. Therefore, we can write:

$mv = (m+2m) v_f$

where $v_f$ is the final velocity of the combined particles.

Simplifying the above equation, we get:

$v_f = \frac{mv}{3m} = \frac{v}{3}$

So, the correct option is $\frac{v}{3}$.

Momentum of bullets per unit time

$ = {{180 \times {{20} \over {1000}} \times 100} \over {60}}$ kg m/s$^2$

= 6 N

$\Rightarrow$ Force on gun = 6 N

We cannot calculate recoil velocity with the given data.

If we consider recoil velocity at $t = 1$ s, then

${V_{\mathrm{recoil}}} = u + at$

$ = 0 + {6 \over {10}} \times 1$ = 0.6 m/s

Mass is same and elastic collision, so speed gets exchanged, $v=2$ m/s

$F = {{dp} \over {dt}}$

$ \Rightarrow |F| = \left| {{{dp} \over {dt}}} \right| = |\mathrm{slope\,of\,p - t\,curve}|$

As we can see from graph,

$|{F_c}|$ is maximum and $|{F_b}|$ is minimum.

$\because$ Time of flight of each ball and bullet

$ = \sqrt {{{2H} \over g}} = \sqrt {{{2 \times 20} \over {10}}} = 2$ s

$\Rightarrow$ By applying linear momentum conservation

$100u + 200(0) = 200\left( {{{30} \over 2}} \right) + 10\left( {{{120} \over 2}} \right)$

$u = 360$ m/s

Let, initial momentum of body $({p_i}) = p$

$\therefore$ Final momentum $({p_f}) = {p_i} + 20\% $ of ${p_i}$

$ = p + 0.2~p$

$ = 1.2~p$

We know,

Kinetic energy $(E) = {{{p^2}} \over {2m}}$

$\therefore$ ${E_i} = {{{p^2}} \over {2m}}$

and ${E_f} = {{{{(1.2p)}^2}} \over {2m}} = {{1.44\,{p^2}} \over {2m}}$

$\therefore$ % Change in kinetic energy

$ = {{{E_f} - {E_i}} \over {{E_i}}} \times 100$

$ = {{{{1.44\,{p^2}} \over {2m}} - {{{p^2}} \over {2m}}} \over {{{{p^2}} \over {2m}}}} \times 100$

$ = {{{{{p^2}} \over {2m}}(1.44 - 1)} \over {{{{p^2}} \over {2m}}}} \times 100 = 0.44 \times 100 = 44$

${\overline r _{com}} = {{{m_1}{{\overline r }_1} + {m_2}{{\overline r }_2}} \over {{m_1} + {m_2}}}$

$ = {{(1 - 9)\widehat i + (2 - 6)\widehat j + (1 + 3)\widehat k} \over 4}$

$ = {{ - 8\widehat i - 4\widehat j + 4\widehat k} \over 4}$

${\overline r _{com}} = - 2\widehat i - \widehat j + \widehat k$

$\left| {\overline r } \right| = \sqrt {4 + 1 + 1} = \sqrt 6 $

$\left| {\widehat i + 2\widehat j + \widehat k} \right| = \sqrt 6 $

$\Delta$P = impulse = same since acceleration is different force acting will be different.

m = 10 kg

$\theta$ = 45$^\circ$

$y = x\tan \theta \left( {1 - {x \over R}} \right)$

$ \Rightarrow 10 = 20\left( {1 - {{20} \over R}} \right)$

$ \Rightarrow R = 40$

$40 = {{{u^2}} \over {10}} \Rightarrow u = 20$

$ \Rightarrow T = {{20 \times 20 \times {1 \over {\sqrt 2 }}} \over {10}} = {4 \over {\sqrt 2 }}s \Rightarrow t = 2\,s$

at $t = 2,\,\overrightarrow v = \left( {10\sqrt 2 \widehat i} \right) + \left( {10\sqrt 2 - 2 \times 10} \right)\widehat j$

$ \Rightarrow \overrightarrow p = 10\left[ {10\sqrt 2 \widehat i + \left( {10\sqrt 2 - 20} \right)\widehat j} \right]$

$ = 100\sqrt 2 \widehat i + \left( {100\sqrt 2 - 200} \right)\widehat j$

F = 100 N

$\Delta$P = 2 $\times$ 0.15 $\times$ 12

= 3.6

$\Rightarrow$ t = ${{3.6} \over {100}}$ = 0.036 s

$P = \sqrt {2m\,KE} $

$ \Rightarrow {{{P_1}} \over {{P_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} $

$ = \sqrt {{8 \over 2}} = {2 \over 1}$

Change in momentum of one ball

= 2 $\times$ (0.05)(10) kg m/s

= 1 kg m/s

$ \Rightarrow {F_{avg}} = {1 \over {\Delta t}} = {1 \over {0.005}}$ N

= 200 N

Given,

Mass of body A = 5 kg

Mass of body B = 8 kg

Momentum of body B is twice that of body A,

$\therefore$ ${P_B} = 2{P_A}$

We know,

Kinetic Energy $(K) = {{{P^2}} \over {2m}}$

$\therefore$ ${{{K_A}} \over {{K_B}}} = {\left( {{{{P_A}} \over {{P_B}}}} \right)^2} \times {{{m_B}} \over {{m_A}}}$

$ = {\left( {{1 \over 2}} \right)^2} \times {8 \over 5}$

$ = {1 \over 4} \times {8 \over 5}$

$ = {2 \over 5}$

For COM to remain unchanged,

m₁x₁ = m₂x₂

$\Rightarrow$ 10 $\times$ 6 = 30 $\times$ x₂

$\Rightarrow$ x₂ = 2 cm towards 10 kg block.

For a head on elastic collision

${v_2} = {{m{u_1}} \over {m + 5m}} + {{m{u_1}} \over {m + 5m}}$

$ = {{2{u_1}} \over 6}$ or ${{{u_1}} \over 3}$

Initial kinetic energy of first mass $ = {1 \over 2}mu_1^2$

Final kinetic energy of second mass

$ = {1 \over 2} \times 5m{\left( {{{{u_1}} \over 3}} \right)^2}$

$ = {5 \over 9}\left( {{1 \over 2}mu_1^2} \right)$

$\Rightarrow$ kinetic energy transferred = 55% of initial kinetic energy of first colliding mass

At maximum height it's velocity is zero. So momentum (mv) will be zero.