Center of Mass and Collision

Momentum imparted to the surface in one collision,

$\Delta p = ({p_i} - {p_f}) = mv - ( - mv) = 2mv$ ..... (i)

Force on the surface due to n collision per second, $F = {n \over t}(\Delta p) = n\Delta p$ ($\because$ $t = 1s$)

= 2 mnv [from Eq. (i)]

So, pressure on the surface,

$p = {F \over A} = {{2mnv} \over A}$

Here, m = 10^$-$26 kg, n = 10²² s^$-$1, v = 10⁴ ms^$-$1, A = 1 m²

$\therefore$ Pressure, $p = {{2 \times {{10}^{ - 26}} \times {{10}^{22}} \times {{10}^4}} \over 1} = 2$ N/m²

So, pressure exerted is of order of 10⁰.

We have following collision, where mass of $\alpha$ particle $=m$ and mass of nucleus $=M$


Let $\alpha$ particle rebounds with velocity $v_1$, then Given :

final energy of $\alpha=36 \%$ of initial energy

$ \begin{aligned} \Rightarrow \frac{1}{2} m v_1^2 =0.36 \times \frac{1}{2} m v^2 \\\\ \Rightarrow v_1 = 0.6 v .......(i) \end{aligned} $

As unknown nucleus gained $64 \%$ of energy of $\alpha$, we have

$ \begin{aligned} & \frac{1}{2} M v_2^2=0.64 \times \frac{1}{2} m v^2 \\ & \Rightarrow v_2=\sqrt{\frac{m}{M}} \times 0.8 v .........(ii) \end{aligned} $

From momentum conservation, we have

$ m v=M v_2-m v_1 $

Substituting values of $v_1$ and $v_2$ from Eqs. (i) and (ii), we have

$ \begin{array}{rlrl} m v =M \sqrt{\frac{m}{M}} \times 0.8 v-m \times 0.6 v \\\\ \Rightarrow 16 m v =\sqrt{m M} \times 0.8 v \\\\ \Rightarrow 2 m =\sqrt{m M} \\\\ \Rightarrow 4 m^2 =m M \Rightarrow M=4 m \end{array} $

v = $\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} $

v₁ = $\sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} $

By momentum conservation

m$\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)} $
$= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)} $

$ \Rightarrow $$m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}$

$ = $ MV_m

and  e = 1 = ${{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}$

$\sqrt {2g\ell } $ $\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right) $
$= $ V_m     . . .(I)

m$\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)$
$ = $ MV_M     . . .(II)

Dividing

${{\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)} \over {\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)}} = {M \over m}$

By componendo divided

${{m - M} \over {m + M}}$ = ${{\sqrt {1 - \cos {\theta _1}} } \over {\sqrt {1 - \cos {\theta _0}} }} = {{\sin \left( {{{{\theta _1}} \over 2}} \right)} \over {\sin \left( {{{{\theta _0}} \over 2}} \right)}}$

$ \Rightarrow $  ${M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}$

${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$

${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$

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

$\int_P^{3P} {dP} = \int_0^T {kt\,dt} $

$2p = {{K{T^2}} \over 2}$

$T = 2\sqrt {{P \over K}} $

velocity of 1 kg block just before it collides with 3kg block

= $\sqrt {2gh} = \sqrt {2000} $ m/s

Applying momentum conversation just before and just after collision.

1 $ \times $ $\sqrt {2000} $ = 4v $ \Rightarrow $ v = ${{\sqrt {2000} } \over 4}$ m/s



initial compression of spring

1.25 $ \times $ 10⁶ x₀ = 30 $ \Rightarrow $ x₀ $ \approx $ 0

applying work energy theorem,

W_g + W_sp = $\Delta $KE

$ \Rightarrow $   40 $ \times $ x + ${1 \over 2}$ $ \times $ 1.25 $ \times $ 10⁶ (0² $-$ x²)

= 0 $-$ ${1 \over 2}$ $ \times $ 4 $ \times $ v²

solving x $ \approx $ 4 cm

Time taken for the particles to collide,

t = ${f \over {{V_{rel}}}} = {{100} \over {100}} = 1$ sec

Speed of wood just before collision = gt = 10 m/s

& speed of bullet just before collision v-gt

= 100 $-$ 10 = 90 m/s

Now, conservation of linear momentum just before and after the collision -

$-$ (0.02) (1v) + (0.02) (9v) = (0.05)v

$ \Rightarrow $   150 = 5v

$ \Rightarrow $   v = 30 m/s

Max. height reached by body h = ${{{v^2}} \over {2g}}$



h = ${{30 \times 30} \over {2 \times 10}}$ = 45m

$ \therefore $  Height above tower = 40 m

As in elastic or in elastic clollision, momentum is conserved.

$ \therefore $   P_i = P_f

P_i = Initial momentum

P_f = Final Momentum

      mv = (2m + m) V_f

$ \Rightarrow $   V_f = ${{mv} \over {2m + M}}$

Here due to collision ${5 \over 6}$th of kinetic energy is lost.

$ \therefore $    Remaining kinetic energy,

      K_f = ${1 \over 6}$ Ki

$ \Rightarrow $    ${1 \over 2}$(2m + M) $ \times $ ${{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}}$ = ${1 \over 6} \times {1 \over 2}m{v^2}$

$ \Rightarrow $   ${{{m^2}{v^2}} \over {2m + M}}$ = ${1 \over 6}m{v^2}$

$ \Rightarrow $   ${m \over {2m + M}}$ = ${1 \over 6}$

$ \Rightarrow $   6m = 2m + M

$ \Rightarrow $   M = 4m

$ \Rightarrow $    ${M \over m}$ = 4

$ \therefore $   ${r_1} = \left( {{{{n \over 2}} \over {m + {m \over 2}}}} \right)l$

${r_1} = {l \over 3}$

When rod is rotated by $\theta $₀ and released, then it look this.


After roating $\theta $₀ and released, when the rod go through mean position it will have maximum speed.

V_max $=$ A$\omega $

$=$ r₁$\theta $₀$\omega $

$=$ $\left( {{l \over 3}{\theta _0}} \right)\omega $

We know,

$\omega $ $ = \sqrt {{K \over {\rm I}}} $

and    ${\rm I} = m{\left( {{l \over 3}} \right)^2} + {m \over 2}{\left( {{{2l} \over 3}} \right)^2}$

$ = {{m{l^2}} \over 9} + {{4m{l^2}} \over {18}}$

$ = {{2m{l^2} + 4m{l^2}} \over {18}}$

$ = {{m{l^2}} \over 3}$

$ \therefore $   $\omega = \sqrt {{{3K} \over {m{l^2}}}} $

$ \therefore $   Tension in the rod when it passes through the mean position is,

T = ${{mv_{\max }^2} \over {{r_1}}}$

$ = {{m{{\left( {{l \over 3}{\theta _0}\omega } \right)}^2}} \over {{l \over 3}}}$

$ = {{m{{\left( {{l \over 3}} \right)}^2}\theta _0^2{\omega ^2}} \over {{\ell \over 3}}}$

$ = m\left( {{l \over 3}} \right)\theta _0^2\left( {{{3K} \over {m{l^2}}}} \right)$

$ = {{K\theta _0^2} \over l}$

Considering one hydrogen molecule :



As collision is elastic so, e = 1

Initial momentum,

$\overrightarrow {{P_i}} $ = ${{mv} \over {\sqrt 2 }}$ $\widehat i$ $-$ ${{mv} \over {\sqrt 2 }}$ $\widehat j$

Final momentum,

$\overrightarrow {{P_f}} $ = ${{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$ $-$ ${{mv} \over {\sqrt 2 }}\widehat j$

$\therefore\,\,\,$ Change in momentum for single H molecule,

$\Delta $P = $\overrightarrow {{P_f}} $ $-$ $\overrightarrow {{P_i}} $

= ${{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$

$\therefore\,\,\,$ $\left| {\Delta P} \right|$ = ${{2mv} \over {\sqrt 2 }}$

Now for n hydrogen molecule total momentum changes per second,

= $\left( {{{2mv} \over {\sqrt 2 }}} \right)$ $ \times $ n

As we know,

Force (F) = ${{\Delta P} \over {\Delta t}}$

= ${{{2mv} \over {\sqrt 2 }}}$ $ \times $ n

$\therefore\,\,\,$ As direction of $\Delta $P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

$\therefore\,\,\,$ Force on the wall = ${{{2mv} \over {\sqrt 2 }}}$ $ \times $ n

$\therefore\,\,\,$ Pressure on the wall, P = ${F \over A}$

= ${{2mv\,n} \over {\sqrt 2 A}}$

= ${{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}$

= 2.35 $ \times $ 10³ N m^$-$2

From conservation of linear momentum,

mv₀ = mv₁ + mv₂

or v₀ = v₁ + v₂ ........(1)

According to the question,

K_f = ${3 \over 2}$K_i

$ \Rightarrow $ ${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2$

$ \Rightarrow $ $v_1^2 + v_2^2 = {3 \over 2}v_0^2$

Using eq (1) ${\left( {{v_1} + {v_2}} \right)^2} = v_0^2$

$ \Rightarrow $ $v_1^2 + v_2^2 + 2{v_1}{v_2}$ = $v_0^2$

$ \Rightarrow $ $2{v_1}{v_2}$ = $v_0^2 - {3 \over 2}v_0^2$ = $ - {1 \over 2}v_0^2$

Now, ${\left( {{v_1} - {v_2}} \right)^2}$ = ${\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}$

= $v_0^2 - \left( { - v_0^2} \right)$ = $2v_0^2$

$\therefore$ ${{v_1} - {v_2}}$ = $\sqrt 2 {v_0}$

In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v₁ is initial velocity.

(i) Before collision :

Now, in figure (ii), v₁ is final velocity of unknown particle and v₂ is final velocity of proton.

(ii) After collision :

By conservation of momentum, we have

Momentum before collision = Momentum after collision

Consider x-component, we have

$m{v_i} + m'\,.\,0 = m'{v_1}\cos 45^\circ + m{v_2}\cos 45^\circ $

$m{v_i} = {1 \over {\sqrt 2 }}(m'{v_1} + m{v_2})$ ..... (1)

Consider y-component, we have

$0 = m'{v_1}\sin 45^\circ - m{v_2}\sin 45^\circ $

${1 \over {\sqrt 2 }}(m'{v_i} - m{v_2}) = 0 \Rightarrow m'{v_1} = m{v_2}$ ...... (2)

Substitute Eq. (2) in Eq. (1), we get

$m{v_i} = {1 \over {\sqrt 2 }}(m{v_2} + m{v_2}) - \sqrt 2 m{v_2}$

$ \Rightarrow {v_i} = \sqrt 2 {v_2}$ ..... (3)

Using Eq. (2) and (3) in Eq. (1), we get

$m\sqrt 2 {v_2} = {1 \over {\sqrt 2 }}(m'{v_1} + m'{v_1})$

$2m{v_2} = 2m'{v_1}$ ($\because$ ${v_1} = {v_2}$)

$ \Rightarrow m = m'$

Using conservation of linear moments,

Along X-axis,

2MV'cos$\theta $ = MVsin30^o $-$ MVsin45^o . . . . . (1)

Along Y - axis,

2Mv' sin$\theta $ = Mv cos30^o + Mv cos45^o . . . . . (2)

Dividing (2) by (1), we get,

${{\sin \theta } \over {\cos \theta }}$ = ${{\cos {{30}^ \circ } + \cos {{45}^ \circ }} \over {\sin {{30}^ \circ } - \sin {{45}^ \circ }}}$

= ${{{{\sqrt 3 } \over 2} + {1 \over {\sqrt 2 }}} \over {{1 \over 2} - {1 \over {\sqrt 2 }}}}$

$\therefore\,\,\,$ tan$\theta $ = ${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$

Let, velocity offer collision = v₁

$ \therefore $   From conservation of momentum,

mv = (m + m) v₁

$ \Rightarrow $    v₁ = ${v \over 2}$

$ \therefore $   Loss in kinetic energy

= ${1 \over 2}$ mv² $-$ ${1 \over 2}$ (2m) $ \times $ ${\left( {{v \over 2}} \right)^2}$

= ${1 \over 4}\,$mv²

lost kinetic energy is used by the electron to jump from first orbit to second orbit.

$ \therefore $   ${1 \over 4}$mv² = (13.6 $-$ 3.4) eV = 10.2 eV

$ \Rightarrow $  ${1 \over 2}$mv² = 20.4 eV

Here   AB = x

and   BC = y

and   $\lambda $ = linear mass density.

As centre of mass is below point A, so horizontal distance of the centre of mass from B is = xcos60^o = ${x \over 2}$

$ \therefore $   X_CM = ${x \over 2}$ = ${{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}$

$ \Rightarrow $   ${x \over 2}$ = ${{\left( {\lambda x} \right)\left( {{x \over 2}} \right)\cos {{60}^o} + \left( {\lambda y} \right)\left( {{y \over 2}} \right)} \over {\lambda \left( {x + y} \right)}}$

$ \Rightarrow $   ${x \over 2}$ = ${{{{{x^2}} \over 4} + {{{y^2}} \over 2}} \over {x + y}}$

$ \Rightarrow $   x² + xy = ${{{x^2}} \over 2} + {y^2}$

$ \Rightarrow $   x² + 2xy $-$ 2y² = 0

$ \therefore $   x = ${{ - 2y \pm \sqrt {{{\left( {2y} \right)}^2} - 4.1\left( { - 2{y^2}} \right)} } \over {2.1}}$

=    ${{ - 2y \pm \sqrt {12{y^2}} } \over 2}$

=   $-$ y $ \pm $ $\sqrt 3 $y

${x \over y} \ne - \sqrt 3 - 1$  as  ${x \over y}$ = positive.

$ \therefore $   ${x \over y}$ = $\sqrt 3 - 1$

$ \Rightarrow $   ${y \over x}$ = ${1 \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$

=   ${{\sqrt 3 + 1} \over 2}$

=   ${{2.732} \over 2}$

=   1.366 $ \simeq $ 1.37

Let the density of solid cone $\rho $.

$dm = \rho \pi {r^2}dy$

${y_{cm}} = {{\int {ydm} } \over {\int {dm} }}$

$ = {{\int\limits_0^h {\pi {r^2}} dy\rho \times y} \over {{1 \over 3}\pi {R^2}h\rho }}$

$ = {{3h} \over 4}$

Applying conservation of linear momentum in x direction

$2mv = \left( {2m + m} \right){V_x}$

$ \Rightarrow {V_x} = {2 \over 3}v$

Applying conservation of linear momentum in y direction

$2mv = \left( {2m + m} \right){V_y}$

$ \Rightarrow {V_y} = {2 \over 3}v$

Final speed of 3m mass, ${V_f}$ = $\sqrt {V_x^2 + V_y^2} $

= $\sqrt {{{4{v^2}} \over 9} + {{4{v^2}} \over 9}} $

= $\sqrt {{{8{v^2}} \over 9}} $

Initial kinetic energy

${E_i} = {1 \over 2}m{\left( {2v} \right)^2} + {1 \over 2}\left( {2m} \right){\left( v \right)^2}$

$ = 2m{v^2} + m{v^2}$

= $3m{v^2}$

Final kinetic energy,

${E_f} = {1 \over 2}\left( {3m} \right)$${V_f^2}$

$ = {{3m} \over 2}\left[ {{{8{v^2}} \over 9}} \right]$

$ = {{4m{v^2}} \over 3}$

Energy loss = ${E_i} - {E_f}$

= $3m{v^2} - {{4m{v^2}} \over 3}$

= ${{5m{v^2}} \over 3}$

Percentage loss in the energy during the collision

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

= ${{{{5m{v^2}} \over 3}} \over {3m{v^2}}}$$ \times 100$

= ${5 \over 9} \times 100$

$ \simeq 56\% $

Initial energy = ${{{P^2}} \over {2m}}$, where $P$ is the momentum and m is the mass of the moving particle.

Loss of energy is maximum when collision is inelastic means when the particles get stuck together as a result of the collision.

So after collision energy = ${{{P^2}} \over {2\left( {m + M} \right)}}$

$\therefore$ Maximum energy loss $ = {{{P^2}} \over {2m}} - {{{P^2}} \over {2\left( {m + M} \right)}}$.

$\left[ \right.$ As $\left. {K.E. = {{{P^2}} \over {2m}} = {1 \over 2}m{v^2}\,\,} \right]$

$ = {{{P^2}} \over {2m}}\left[ {{M \over {\left( {m + M} \right)}}} \right] = {1 \over 2}m{v^2}\left\{ {{M \over {m + M}}} \right\}$

$\therefore$ $f = \left( {{M \over {m + M}}} \right)$

So statement $I$ is wrong.

Statement ${\rm I}{\rm I}$ says "Maximum energy loss occurs when the particles get stuck together as a result of the collision." This is a case of perfectly inelastic collision.

Hence statement ${\rm I}$${\rm I}$ is correct.

In completely inelastic collision,

${m_1}{v_1} + {m_2}{v_2} = {m_1}v + {m_2}v$

after collision both particle have common velocity $v$, so all energy is not lost

$\therefore$ Statement - $1$ is true

The principle of conservation of momentum applicable for all kinds of collisions.

So statement - $2$ is also true.

Statement - $2$ explains statement - $1$ correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body.

Motion is uniform as graph is straight line. In x -t graph, velocity is ${x \over t}$ and the slope of the graph tells the direction of the velocity. Here because of impulse the direction of velocity or slope of the graph changes.

During each collision

Initial velocity ${v_1} = {2 \over 2} = 1\,m{s^{ - 1}}$

Final velocity ${v_2} = - {2 \over 2} = - 1\,m{s^{ - 1}}$

Impulse $=$ Change in momentum

$ = m\left| {{v_2} - {v_1}} \right| = 0.4 \times 2 = 0.8\,\,Ns$

Let $m$ = 0.50 kg and $M$ = 1.0 kg

Initial kinetic energy of the system when 1 kg mass is at rest,

$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$

$ = {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$

For collision, applying conservation of linear momentum

$\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v$

$\therefore$ $0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s$

Final kinetic energy of the system is

$K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}$

$ = {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J$

$\therefore$ Energy loss during collision

$ = \left( {1 - {1 \over 3}} \right)J = 0.67J$

Given The linear mass density $\lambda = k{\left( {{x \over L}} \right)^n}$

${x_{CM}} = {{\int\limits_0^L {x{\mkern 1mu} dm} } \over {\int\limits_0^L {dm} }}$

$ = {{\int\limits_0^L {x\left( {\lambda {\mkern 1mu} dx} \right)} } \over {\int\limits_0^L {\lambda {\mkern 1mu} dx} }}$

$ = {{\int\limits_0^L {k{{\left( {{x \over L}} \right)}^n}} .xdx} \over {\int\limits_0^L {k{{\left( {{x \over L}} \right)}^n}{\mkern 1mu} dx} }}$

= ${{k\left[ {{{{x^{n + 2}}} \over {\left( {n + 2} \right){L^n}}}} \right]_0^L} \over {\left[ {{{k{\mkern 1mu} {x^{n + 1}}} \over {\left( {n + 1} \right){L^n}}}} \right]_0^L}}$

$ = {{L\left( {n + 1} \right)} \over {n + 2}}$

For $n=0,$ ${x_{CM}} = {L \over 2};n = 1,$

${x_{CM}} = {{2L} \over 3};n = 2,\,{x_{CM}} = {{3L} \over 4};\,...$

From here you can see only option (A) can be the right answer.

Let the mass per unit area be $\sigma .$

Then the mass of the complete disc
$ = \sigma \left[ {\pi {{\left( {2R} \right)}^2}} \right] = 4\pi \sigma {R^2}$

The mass of the removed disc $ = \sigma \left( {\pi {R^2}} \right) = \pi \sigma {R^2}$

So mass of the remaining disc = $4\pi \sigma {R^2}$ - $\pi \sigma {R^2}$ = $3\pi \sigma {R^2}$

Let center of mass of $3\pi \sigma {R^2}$ mass is at x distance from origin O.

$\therefore$ ${{3\pi {R^2}\sigma .x + \pi {R^2}\sigma .R} \over {4\pi {R^2}\sigma }} = 0$

As center of mass of full disc is at Origin.

$\therefore$ $x = - {R \over 3}$

According to the question, $x$ = $\alpha R$

$\therefore$ $\alpha = - {1 \over 3}$

$ \Rightarrow $ $\left| \alpha \right| = {1 \over 3}$

We know, Force$ \times $ time = Impulse = Change in momentum

$\therefore$ $F \times t = m\left( {v - u} \right)$

$ \Rightarrow $ $F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$

Initially,

$0 = {{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}} \over {{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}$

Finally,

mass m₁ moved towards the center a distance d so the distance of mass m₁ from the origin is x₁ - d, and now let mass m₂ need to move d' to keep the center at the origin.

$\therefore$ $0 = {{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)} \over {{m_1} + {m_2}}}$

$ \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'$

$ \Rightarrow d' = {{{m_1}} \over {{m_2}}}d$

Here linear momentum is conserved as no external force is acting on the bomb.

Let the velocity and mass of $4$ $kg$ piece be ${v_1}$ and ${m_1}$ and that of $12$ $kg$ piece be ${v_2}$ and ${m_2}$.

Applying conservation of linear momentum

$0 = {m_2}{v_2} - {m_1}{v_1}$

$\Rightarrow {v_1} = {{12 \times 14} \over 4} = 12\,m{s^{ - 1}}$

$\therefore$ $K.E{_1} = {1 \over 2}{m_1}v_1^2 = {1 \over 2} \times 4 \times 144 = 288\,J$

This is a case of translation motion without rotation, the force $\overrightarrow F $ has to be applied at center of mass.

i.e. the point $'P'$ has to be at the centre of mass

$y = {{{m_1}{y_1} + {m_2}{y_2}} \over {{m_1} + {m_2}}}$

$ = {{m \times 2\ell + 2m \times \ell } \over {3m}}$

$ = {{4\ell } \over 3}$

Elastic energy stored in the spring = ${1 \over 2}k{L^2}$

And kinetic energy of the block = ${1 \over 2}M{v^2}$

$\therefore$ ${1 \over 2}M{v^2} = {1 \over 2}k{L^2}$

$ \Rightarrow v = \sqrt {{k \over M}} .L$

$\therefore$ Momentum $ = M \times v = M \times \sqrt {{k \over M}} .L = \sqrt {kM} .L$

The center of mass does not shift as no external force is applied horizontally. So the center of mass of the system continues its original path. It is only the internal forces which comes into play while breaking.

Assume speed of second mass = ${v_1}$

As momentum is conserved,

In $x$-direction, $mv = m{v_1}\cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(1)$

In $y$-direction, ${{mv} \over {\sqrt 3 }} = m{v_1}\,\sin \theta \,\,\,\,\,\,\,\,\,\,...(2)$

Squaring and adding eqns.$(1)$ and $(2)$

${\left( {m{v_1}\cos \theta } \right)^2} + {\left( {m{v_1}\sin \theta } \right)^2}$$ = {\left( {mv} \right)^2} + {\left( {{{mv} \over {\sqrt 3 }}} \right)^2}$

$ \Rightarrow $ $v_1^2 = {v^2} + {{{v^2}} \over {\sqrt 3 }} $

$\Rightarrow {v_1} = {2 \over {\sqrt 3 }}v$

Assume the man can fire $n$ bullets in one second.

$\therefore$ change in momentum per second $ = n \times mv = F$

[ $m=$ mass of bullet, $v=$ velocity, $F$ = force) ]

$\therefore$ $n = {F \over {mv}} = {{144 \times 1000} \over {40 \times 1200}} = 3$

The correct answer is Option A : $A$ does not imply $B$ and $B$ does not imply $A$.

Here's why :

Statement $A:$ The linear momentum of a system of particles being zero does not mean that the kinetic energy is also zero. For example, consider two equal mass particles moving with the same speed but in opposite directions. The linear momentum of the system will be zero because momentum is a vector quantity and the two momenta will cancel out. However, kinetic energy is a scalar quantity and does not cancel out in this way. Each particle has kinetic energy due to its motion, so the total kinetic energy of the system is not zero.

Statement $B:$ The kinetic energy of a system of particles being zero also does not imply that the linear momentum is zero. If the kinetic energy is zero, it means that all the particles are at rest (since kinetic energy is associated with motion). However, the linear momentum will also be zero in this case because momentum depends on both mass and velocity, and the velocity of each particle is zero.

But remember that zero linear momentum doesn't exclusively mean all particles are at rest. As explained above, it could be the scenario where particles have equal and opposite momenta, thereby cancelling each other. Hence, $B$ doesn't imply $A$.

So, neither statement implies the other. Hence, Option A is correct.

The velocity of center of mass of two particle system is

${v_c} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$

$ = {{m\left( {2v} \right) + m\left( { - v} \right)} \over {m + m}}$

$= {v \over 2}$