Center of Mass and Collision

The mass of the soccer ball is given as:

$ m = 250 \, \text{g} = 0.25 \, \text{kg} $

Initially, the ball is moving horizontally to the left with a speed of:

$ 22 \, \text{m/s} $

After being kicked, the ball is moving to the right with a velocity of:

$ 30 \, \text{m/s} $

at an angle of $ 53^\circ $ with the horizontal, in the upward direction. The collision lasts for:

$ t = 0.01 \, \text{s} $

To solve for the average force, we need to find the horizontal and vertical components of the final velocity:

$ v_{f_x} = 30 \times \frac{3}{5} = 18 \, \text{m/s} $

$ v_{f_y} = 30 \times \frac{4}{5} = 24 \, \text{m/s} $

The change in momentum ($ \Delta p $) is calculated as:

For the horizontal component:

$ \Delta p_x = m(v_{f_x} - v_i) = 0.25[18 - (-22)] $

$ \Delta p_x = 0.25 \times 40 = 10 \, \text{kg} \, \text{m/s} $

For the vertical component:

$ \Delta p_y = m(v_{f_y} - 0) = 0.25 \times 24 $

$ \Delta p_y = 6 \, \text{kg} \, \text{m/s} $

The total change in momentum is:

$ \Delta p = \sqrt{\Delta p_x^2 + \Delta p_y^2} = \sqrt{(10)^2 + (6)^2} $

$ \Delta p = \sqrt{136} \, \text{kg} \, \text{m/s} $

Finally, the average force acting on the ball is:

$ F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{\sqrt{136}}{0.01} $

$ F_{\text{avg}} = 1166 \, \text{N} $

Given:

Mass of first body, $ m_1 = 30 \, \text{kg} $

Initial velocity of the first body, $ u_1 = 20 \, \text{m/s} $

Mass of the second body, $ m_2 = 30 \, \text{kg} $

Initial velocity of the second body, $ u_2 = -30 \, \text{m/s} $ (since it is moving in the opposite direction)

Since the collision is elastic, and the masses are equal, the principle of conservation of momentum and kinetic energy indicates that the velocities after the collision will be exchanged. Therefore:

Final velocity of the first body, $ v_1 = u_2 = -30 \, \text{m/s} $

The negative sign indicates that the first body moves in the opposite direction compared to its initial direction.

Final velocity of the second body, $ v_2 = u_1 = 20 \, \text{m/s} $

Similarly, this means the second body also moves in the opposite direction to its initial motion.

From the figure,

$ \begin{aligned} & m g \sin 60^{\circ}-T=m a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & T-m g \sin 30^{\circ}=m a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Adding both Eq. (i) and Eq. (ii), we get

$ m g\left(\sin 60^{\circ}-\sin 30^{\circ}\right)=2 m a $

$ \begin{aligned} &\begin{aligned} & a=\frac{g}{2}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right) \\ & a=\frac{g}{4}(\sqrt{3}-1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\begin{aligned} & \text { Given, } m_1=m_2=m \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & \mathbf{a}_1=-a \cos 60^{\circ} \hat{\mathbf{i}}-a \sin 60 \hat{\mathbf{j}}=\frac{a}{2}(-\hat{\mathbf{i}}-\sqrt{3 \mathbf{j}}) \\ & \mathbf{a}_2=-a \cos 30^{\circ} \hat{\mathbf{i}}+a \sin 30 \hat{\mathbf{j}} \\ & \\ & =\frac{a}{2}(-\sqrt{3 \hat{\mathbf{i}}}+1 \hat{\mathbf{j}}) \end{aligned} \end{aligned} $

Acceleration of the centre of mass

$ \mathbf{a}_{\mathrm{com}}=\frac{m_1 \mathbf{a}_1+m_2 \mathbf{a}_2}{m_1+m_2} $

From Eq. (iv), $\mathbf{a}_{\text {com }}=\frac{\mathbf{a}_1+\mathbf{a}_2}{2}$

$ \begin{gathered} \mathbf{a}_{\text {com }}=\frac{\frac{a}{2}(-\hat{\mathbf{i}}-3 \sqrt{3 \hat{\mathbf{j}}})+\frac{a}{2}(-\sqrt{3 \hat{\mathbf{i}}}+\hat{\mathbf{j}})}{2} \\ \left|\mathbf{a}_{\text {com }}\right|=\left\lvert\, \frac{a}{4}[(-1-\sqrt{3}) \hat{\mathbf{i}}+(1-\sqrt{3 \hat{\mathbf{j}})] \mid}\right. \\ \quad\left[\because|a \hat{\mathbf{i}}+b \hat{\mathbf{j}}|=\sqrt{a^2+b^2}\right] \end{gathered} $

$ \begin{aligned} &\begin{aligned} \left|\mathbf{a}_{\mathrm{com}}\right| & =\frac{a}{4}\left[\sqrt{(-1-\sqrt{3})^2+(1-\sqrt{3})^2}\right] \\ & =\frac{a}{4}[\sqrt{1+3+2 \sqrt{3}+1+3-2 \sqrt{3}}] \\ & =\frac{a}{4} 2 \sqrt{2} \end{aligned}\\ &\text { Putting value of a from Eq. (iii), }\\ &\begin{aligned} & =\frac{g}{4 \cdot 4} \cdot 2 \sqrt{2}(\sqrt{3}-1) \\ & =\frac{10 \times 2 \sqrt{2} \times(\sqrt{3}-1)}{4 \times \sqrt{2} \times 2 \sqrt{2}} \\ & =\frac{10}{4 \sqrt{2}}(\sqrt{3}-1) \\ \left|a_{\mathrm{com}}\right| & =\frac{5(\sqrt{3}-1)}{2 \sqrt{2}} \mathrm{~m} / \mathrm{s}^2 \end{aligned} \end{aligned} $

Given:

Initial height, $ h_i = 6.25 \, \text{m} $

Height after second bounce, $ h_2 = 81 \, \text{cm} = 0.81 \, \text{m} $

The formula for the coefficient of restitution $ e $ is:

$ e = \sqrt{\frac{h_{\text{after}}}{h_{\text{before}}}} $

Let $ h_1 $ be the height reached after the first bounce. According to the formula, we have:

$ h_1 = e^2 \cdot h_i $

After the second bounce, the height $ h_2 $ is given by:

$ h_2 = e^2 \cdot h_1 $

Substituting the expression for $ h_1 $, we get:

$ h_2 = e^2 \cdot (e^2 \cdot h_i) = e^4 \cdot h_i $

Rearranging the equation, we find:

$ e^4 = \frac{h_2}{h_i} $

Substituting the given values:

$ e^4 = \frac{0.81}{6.25} $

Simplifying further:

$ e^4 = \frac{81}{625} $

Taking the fourth root, we calculate $ e $:

$ e = \frac{3}{5} \Rightarrow e = 0.6 $

Given:

Mass of the first body, $ m_1 = 2 \, \text{kg} $

Mass of the second body, $ m_2 = 4 \, \text{kg} $

Initial relative velocity, $ u_{\text{rel}} = 10 \, \text{ms}^{-1} $

Final relative velocity, $ v_{\text{rel}} = 4 \, \text{ms}^{-1} $

First, calculate the coefficient of restitution ($ e $):

$ e = \frac{v_{\text{rel}}}{u_{\text{rel}}} = \frac{4}{10} = \frac{2}{5} $

The loss of kinetic energy ($ \Delta \mathrm{KE} $) in the system due to the collision is given by:

$ \Delta \mathrm{KE} = \frac{1}{2} \cdot \frac{m_1 m_2}{m_1 + m_2} \cdot (1 - e^2) \cdot \left(u_{\text{rel}}\right)^2 $

Substituting the given values:

$ \Delta \mathrm{KE} = \frac{1}{2} \cdot \frac{2 \times 4}{(2 + 4)} \cdot \left[1 - \left(\frac{2}{5}\right)^2\right] \cdot (10^2) $

$ = \frac{4}{6} \cdot \left(1 - \frac{4}{25}\right) \cdot 100 $

$ = \frac{4}{6} \cdot \frac{21}{25} \cdot 100 $

$ = \frac{336}{6} = 56 \, \text{J} $

Thus, the loss of kinetic energy of the system due to the collision is $ 56 \, \text{J} $.

$ m_q=20 \mathrm{~kg} $

In the above problem, centre of mass of the system (plank + girl) remains at rest as no external force acts on the system. We have,

$ \Delta X_{\mathrm{cm}}=\frac{m_g \Delta X_1+m_b \Delta X_2}{m_g+m_b} $

Since, $\Delta X_{\mathrm{cm}}=0$ (COM remains at rest, i.e., $v_{\mathrm{cm}}=0$ )

Let the forward direction of movement of girl is negative and the direction of opposite to it is positive.

Suppose the plank moves $n \mathrm{~cm}$, when girl walks from one end to the other end of the plank, then,

$ \begin{aligned} & m_g[-(3.3-x)]+m_b x=0 \\ & m_b x=m_g(3.3-x) \\ & 90 x=20 \times 3.3-20 x \\ & 110 x=20 \times 3.3 \\ & x=\frac{20 \times 3.3}{110}=0.6 \mathrm{~m} \\ & x=60 \mathrm{~cm} \end{aligned} $

To find the total distance a ball travels before coming to rest after repeatedly bouncing, we use the following approach:

Initial setup: The ball falls freely from rest, with a velocity just before the first bounce of $7 \, \mathrm{ms}^{-1}$, and the coefficient of restitution $e = 0.75$. The acceleration due to gravity is $10 \, \mathrm{ms}^{-2}$.

Distance traveled: The ball falls a height $h$ before the first bounce. After each bounce, it reaches successively lower heights $h_1, h_2, \ldots$, where:

$h_1 = e^2 h$

$h_2 = e^4 h$

And so on.

Total distance formula: The total distance $d$ covered by the ball is:

$ d = h + 2h_1 + 2h_2 + \ldots $

This can be rewritten using geometric series:

$ d = h + 2e^2h (1 + e^2 + e^4 + \ldots) $

Using the sum of an infinite geometric series $S = \frac{1}{1 - e^2}$, we have:

$ d = h + 2e^2h\left(\frac{1}{1 - e^2}\right) $

Simplifying, we get:

$ d = h\left(\frac{1 + e^2}{1 - e^2}\right) $

Plugging in the values: Given $e = \frac{3}{4}$, the distance $d$ becomes:

$ d = h \left(\frac{1 + \left(\frac{3}{4}\right)^2}{1 - \left(\frac{3}{4}\right)^2}\right) = \frac{25}{7} \times h $

Calculating $h$:

Using the kinematic equation:

$ v^2 = u^2 + 2gh \quad \Rightarrow \quad (7)^2 = 0 + 2 \times 10 \times h $

Solving for $h$, we find:

$ \frac{49}{20} = h $

Finding total distance $d$:

Using the relation obtained:

$ d = \frac{25}{7} \times \frac{49}{20} = 8.75 \, \mathrm{m} $

Therefore, the total distance traveled by the ball before it comes to rest is $8.75 \, \mathrm{m}$.

For mass $2 m$,

$ 2 m a=2 m g-T...(i) $

For mass $m$,

$ m a=T-m g...(ii) $

On solving Eqs. (i) and (ii), we get

$ \begin{array}{ll} \therefore & a=\frac{g}{3} \\ \because & v=u+a t \\ \Rightarrow & v=0+\frac{g}{3} \times 5.4 (\because u=0)\\ \Rightarrow & v=\frac{10}{3} \times 5.4=18 \mathrm{~m} / \mathrm{s} \end{array} $

In an inelastic collision, although the momentum is conserved, kinetic energy is not. Some of the initial kinetic energy is converted into other forms of energy, such as heat, sound, or energy used in deforming the colliding bodies. This means that the kinetic energy after the collision is less than before the collision.

So, the correct answer is:

$\text{Option B: is less than before collision}$

To summarize:

In an inelastic collision, momentum is conserved.

Kinetic energy decreases because some energy is lost to other forms (like heat or sound).

Therefore, the kinetic energy after collision is less than the kinetic energy before the collision.

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

Given,

Initial mass, $m=16 \mathrm{~kg}$

Mass of $1^{\text {st }}$ piece, $m_1=4 \mathrm{~kg}$

Mass of $2^{\text {nd }}$ piece, $m_2=12 \mathrm{~kg}$

Velocity of $2^{\text {nd }}$ piece, $v_2=4 \mathrm{~m} / \mathrm{s}$

Using law of conservation of linear momentum.

Initial momentum = Final momentum

$ \begin{aligned} & m \times v=m_1 v_1+m_2 v_2 \\ & \quad(v=0 \text { Initially at rest }) \\ & m_1 v_1=-m_2 v_2 \\ & 4 \times v_1=12 \times 4 \\ & v_1=-12 \mathrm{~m} / \mathrm{s} \end{aligned} $

Kinetic energy is given by for $2^{\text {nd }}$ particle

$ \begin{aligned} (\mathrm{KE}) & =\frac{1}{2} m_2 v_2^2 \\ & =\frac{1}{2} \times 4 \times(12)^2=\frac{1}{2} \times 4 \times(12)^2 \\ & =288 \mathrm{~J} \end{aligned} $

Hence, the correct option is (d).

Given,

Height $=h$

Coefficient of restitution $=e$

The ball reaches the $u=\sqrt{2 g h}$, when dropped from height $h$.

If collision on the floor is elastic, the ball will rebound with same velocity. But here coefficient of restitution is $e$; the velocity $v$ (less than $u$ ) is given by

$ v=e u \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

As $v$ is velocity of separation and $u$ is the velocity of approach.

In above case ball will rebound to a height $h_1$.

$ \begin{aligned} h_1 & =v^2 / 2 g \\ v & =e u \Rightarrow h_1=\frac{e^2 u^2}{2 g} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Initial height

$ h=\frac{u^2}{2 g} $

Put in $2^{\text {nd }}$ equation

$ h_1=e^2 h $

So, total distance travel by ball before $2^{\text {nd }}$ drop is $h+e^2 h+e^2 h=h\left(1+2 e^2\right)$

Acceleration of the given system will be

$ a=\frac{m_2-m_1}{m_2+m_1} \mathrm{~g} $

Putting values of $m_1$ and $m_2$, we get

$ a=\frac{g}{3} \mathrm{~m} / \mathrm{s}^2 $

Acceleration of centre of mass will be

$ a_{\mathrm{CM}}=\left(\frac{m_2-m_1}{m_2+m_1}\right) a=\frac{g}{9} $

Putting values of $a_{\mathrm{CM}}$ into second equation, we get

$ \begin{aligned} S_{\mathrm{CM}} & =\frac{1}{2} a_{\mathrm{CM}} t^2 \\ & =\frac{1}{2}\left(\frac{g}{9}\right)(2)^2 \\ & =2.22 \mathrm{~m} \end{aligned} $

Total mechanical energy is conserved in this collision

$ \begin{aligned} & \text { } \begin{aligned}Therefore, (\mathrm{KE}+\mathrm{PE})_i= & (\mathrm{KE}+\mathrm{PE})_f \\ & =\Delta K+\Delta U=0 \\ \Rightarrow \quad \Delta K & =-\Delta U \end{aligned} \end{aligned} $

Let initial velocity be $v$, then

$ \mathrm{KE}_1=\frac{1}{2} m v^2 $

Final potential energy due to spring,

$ (\mathrm{PE})_f=\frac{1}{2} k x^2 $

From conservation of energy,

$ \begin{array}{rlrl} & & \frac{1}{2} m v^2 & =\frac{1}{2} K x^2 \\ \Rightarrow & v & =\sqrt{\frac{K}{m}} x \end{array} $

Where, $x=$ Compression $=L$

Therefore, $v=L \sqrt{\frac{K}{m}}$

Hence, maximum momentum,

$ p=m v=L \sqrt{M K} \quad[\because m=M] $

If a ball is dropped from height $h$, then it will strike the floor with velocity $(\eta)=\sqrt{2 g h}$

Now, after striking the floor, it will bounce back with velocity $(u)=e v$ Now, it will travel total $2 h$ distance before again striking the ground where

$ h_1=\frac{e^2 v^2}{2 g} $

Again the ball will bounce back with $e^2 v$ velocity and now it will travel $2 h_2$ distance where, $h_2=\frac{e^4 v^2}{2 g}$

This process continuous for long till ball comes to rest.

So, total distance travelled, $d$

$ \begin{aligned} &=h+1 h_1+2 h_2+\ldots \\ & d=\frac{v^2}{2 g}+\frac{2 e^2 v^2}{2 g}+\frac{2 e^4 v^2}{2 g}+\frac{2 e^6 v^2}{2 g} \ldots \\ &=\frac{v^2}{2 g}\left(1+e^2+e^4+e^6+\ldots\right) \\ &=\frac{v^2}{2 g}+\frac{2 e^2 v^2}{g}\left(\frac{1}{1-e^2}\right) \\ &=\left(\frac{1-e^2+2 e^2}{1-e^2}\right) \frac{v^2}{2 g}=\left(\frac{1+e^2}{1-e^2}\right) \frac{v^2}{2 g} \\ &=h\left(\frac{1+e^2}{1-e^2}\right) \end{aligned} $

$ \text { According to FBD given below, } $

$ \begin{aligned} &T-\frac{m g}{2}=m a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ &\text { (for left side block) } \end{aligned} $

$ \begin{aligned} &\frac{\sqrt{3}}{2} m g-T=m a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { (for right side block) } \end{aligned} $

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

$ a=\frac{1}{2 m}\left(\frac{\sqrt{3} m g}{2}-\frac{m g}{2}\right)=\left(\frac{\sqrt{3}-1}{4}\right) g $

Acceleration COM was $k_e$ as follows.

So net acceleration on COM is

$ \mathbf{a}_{\mathrm{CM}}=\frac{\mathbf{a}_1+\mathbf{a}_2}{2}=\frac{\sqrt{2} a}{2}=\frac{a}{\sqrt{2}} $

$\Rightarrow \quad \mathbf{a}_{\mathrm{CM}}=\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) g$ in $x$ direction

Given,

Two masses $m_1$ and $m_2$ moves toward CM by $d$ distance.

We know position of CM

$ X_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} $

Shift in position of centre of mass,

$ \Delta x_{\mathrm{CM}}=\frac{m_1 \Delta x_1+m_2 \Delta x_2}{m_1+m_2} $

But $\Delta x_{C M}=0$ (Given)

$ \begin{aligned} m_1 \Delta x_1+m_2 \Delta x_2 & =0 \\ \left(\frac{m_1}{m_2}\right) \Delta x_1 & =-\Delta x_2 \\ \Delta x_2 & =-\frac{m_1}{m_2} d \end{aligned} $

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.

Since, collision is head on, then particle $A$ will continue moving along the same line as before the collision, but there will be a change in the magnitude of its velocity vector. Let particle $A$ starts moving with velocity $v_1$ and particle $B$ with velocity $v_2$ after collision, then from conservation of linear momentum,

$ \begin{aligned} m u & =m v_1+m v_2 \\or\,\,\,\,\,\,\,\, u & =v_1+v_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

From the given condition,

$ \begin{aligned} & \text { i.e } \quad \eta=\frac{\frac{1}{2} m u^2-\left(\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2\right)}{\frac{1}{2} m u^2} \\ & \Rightarrow \quad \eta=1-\frac{v_1^2+v_2^2}{u^2} \\ & \Rightarrow \quad v_1^2+v_2^2=(1-\eta) u^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Now solving Eqs. (i) and (ii), we get $v_1=5 \mathrm{~m} / \mathrm{s}$

In an elastic collision of two billiard balls, both kinetic energy and linear momentum remain conserved.

In elastic collision, momentum is conserved but total energy of any one of ball is not conserved i.e. exchange of energy takes place.

Hence, A is true but R is false.

Let the mass of the moving particle $=m$

∴ The mass of the stationary particle

$ =\frac{1}{n} \times m=\frac{m}{n} $

Also, let the speed of moving particle $=v$

Now, by the law of conservation of linear momentum

$ \begin{aligned} & \therefore \quad m v+\left(\frac{m}{n} \times 0\right)=m v^{\prime}+\frac{m}{n} v^{\prime} \\ & \Rightarrow \quad m v=m\left(v^{\prime}+\frac{v^{\prime}}{n}\right) \\ & \Rightarrow \quad v=v^{\prime}+\frac{v^{\prime}}{n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, coefficient of restitution,

$ e=\frac{-\left(v^{\prime \prime}-v^{\prime}\right)}{(0-v)} $

Since, it is elastic collision,

$ \begin{array}{ll} \therefore & e=1 \\ \Rightarrow & 1=\frac{-\left(v^{\prime \prime}-v^{\prime}\right)}{-v} \Rightarrow v^{\prime \prime}=v+v^{\prime} \\ \Rightarrow & v^{\prime}=v^{\prime \prime}-v \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

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

$ \begin{gathered} v=v^*+\frac{v^*}{n} \\ v^*=\left[\frac{2 v}{1+\frac{1}{n}}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered} $

Now, kinetic energy transferred to the stationary particle

$=(\text { kinetic energy })_{\text {final }} -(\text { kinetic energy })_{\text {initial }}$

$ \Rightarrow \quad K_f^{\prime}-K_i^{\prime}=\frac{1}{2} \times \frac{m}{n} \times v^{\prime 2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Also, initial kinetic energy of mass $m$

$ =\frac{1}{2} m v^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

∴ The fraction of KE that gets transferred to the stationary particle from the moving particle

$ \begin{aligned} & =\frac{\frac{1}{2} \times \frac{m}{n} \times v^{n^2}}{\frac{1}{2} m v^2}=\frac{v^{v^2}}{n v^2}=\frac{1}{n}\left(\frac{v^0}{v}\right)^2 \\ & =\frac{1}{n}\left(\frac{2 v}{\left(1+\frac{1}{n}\right) v}\right)^2 \quad \text { [From Eq. (iii)] } \\ & =\frac{1}{n} \frac{4 v^2}{\left(1+\frac{1}{n}\right)^2 v^2}=\frac{4}{n\left(\frac{n+1}{n}\right)^2}=\frac{4 n}{(1+n)^2} \end{aligned} $

$ \text { The given situation is shown as } $

We know,

$ \begin{aligned} & X_{C M}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+m_4 x_4}{m_1+m_2+m_3+m_4} \\ & =\frac{(3 M \times-1)+(2 M \times 0)+(M \times 1)+(4 M \times 0)}{3 M+2 M+M+4 M} \\ & =\frac{-3 M+0+M+0}{10 M}=\frac{-2 M}{10 M}=\frac{-1}{5} \\ & \text { Also, } Y_{C M}=\frac{m_1 y_1+m_2 y_2+m_3 y_3+m_4 y_4}{m_1+m_2+m_3+m_4} \\ & =\frac{(3 M \times 0)+(2 M \times 1)+(M \times 0)+(4 M \times-1)}{10 M} \\ & =\frac{0+2 M+0-4 M}{10 M}=\frac{-2 M}{10 M}=\frac{-1}{5} \end{aligned} $

∴ Position of centre of mass from the centre of the circle

$ r_{C M}=X_{C M} \hat{\mathbf{i}}+Y_{C M} \hat{\mathbf{j}}=\left(\frac{-1}{5} \hat{\mathbf{i}}-\frac{1}{5} \hat{\mathbf{j}}\right) $

For ball $A$,

$\begin{aligned} & m_A=1 \mathrm{~kg}, v_A=4 \mathrm{~ms}^{-1} \\ & \text { for ball } B, m_B=3 \mathrm{~kg}, v_B=0 \end{aligned}$

$\therefore$ Total momentum before collision,

$\begin{aligned} p_i & =m_A v_A+m_B v_B \\ & =1 \times 4+3 \times 0=4 \mathrm{~kg}-\mathrm{ms}^{-1} \end{aligned}$

Since, after collision, both body stick together, therefore, it is the case of perfectly enelastic collision. Let $v$ be the common velocity, then total momentum after collision,

$\begin{aligned} p_f & =\left(m_A+m_B\right) v \\ & =(1+3) v=4 v \end{aligned}$

According to law of conservation of linear momentum,

$\begin{array}{rlrl} & \qquad p_i =p_f \\ & \Rightarrow 4 =4 v \Rightarrow v=1 \mathrm{~ms}^{-1} \end{array}$

Time of impact, $\Delta t=0.1 \mathrm{~s}$

Force exerted on the body $B$,

$\begin{aligned} & F_B=\frac{\Delta p_B}{\Delta t}=\frac{m_B\left(v_B-u_B\right)}{\Delta t}=\frac{3(v-0)}{0.1} \\ \Rightarrow \quad & F_B=\frac{3(1-0)}{0.1}=\frac{3}{0.1}=30 \mathrm{~N} \end{aligned}$

According to first situation

Total momentum before collision,

$\begin{aligned} p_i & =m v_A+2 m v_B \\ & =m \times 150+2 m \times v=150 m+2 m v \end{aligned}$

Since, coefficient of restitution, $e=1$

Hence, after collision, body $A$ transfer its complete velocity to body $B$. hence, after collision velocity of body $B, v^{\prime}=150-v$.

Total momentum after collision,

$\begin{aligned} p_f & =2 m v^{\prime}+m \times 0 \\ & =2 m v^{\prime} \\ & =2 m(150-v) \end{aligned}$

According to law of conservation of momentum,

$\begin{aligned} p_i & =p_f \\ 150 m+2 m v & =2 m(150-v) \\ 150+2 v & =300-2 v \Rightarrow 4 v=150 \\ \Rightarrow \quad v & =37.5 \mathrm{~m} / \mathrm{s} \end{aligned}$

Given, volume of block $A=0.25 \mathrm{~m}^3$

Let the density of iron be $\rho \mathrm{kg} / \mathrm{m}^3$.

Thus, mass of iron block, $A=0.25 \rho\mathrm{~kg}$

When block $A$ is attached to the spring elongation produced $(x)=0.2 \mathrm{~m}$.

Thus, force on spring of spring constant $k$ is

$F=-k x$

($-$ve sign denotes force is in opposite direction of elongation)

$\begin{aligned} m g & =-k(0.2) \\ 0.25 \rho g & =-0.2 k \Rightarrow k=1.25 \rho g \mathrm{~N} / \mathrm{m} \end{aligned}$

Now, the same spring is attached to another block $B$.

Volume of block $B=0.75 \mathrm{~m}^3$

Mass of block $B=0.75 \rho \mathrm{~kg}$

FBD of the spring system on inclined surface is given as

$\begin{aligned} & \text { Force on block } B, F=-(k x) \\ & M g \sin \theta=-k x \\ & 0.75 \rho g \sin 30^{\circ}=-(-1.25 \rho g) x \\ & \frac{0.75}{1.25} \sin 30^{\circ}=x \Rightarrow \frac{0.75}{1.25} \times \frac{1}{2}=x \\ & \Rightarrow \quad x=0.3 \mathrm{~m} \\ \end{aligned}$

Thus, the distance of block from the top = original length of spring + elongation = 1 m + 0.3 m = 1.3 m