Answer: Option D. -> e4h
:
D
The velocity attained after a fall through a height h is given by $v^2=2gh$
Thus $h\propto v^2$ The velocity after first rebound is ev. Therefore, the height attained after first rebound = $e^{2}h$. Velocity after second rebound is $e^{2}v$. Hence the height attained after second rebounds is $e^{4}h$. Thus the correct choice is (d).

Answer: Option B. -> v22g
:
B
In an elastic collision between two bodies of the same mass with one of them initially at rest, the movingbody is brought to rest and the other moves in the same direction with the same speed. Thus the ballwill come to rest and the bob of the pendulum acquires a speed v. At this speed, it will rise to height h givenby $h=\frac{v^2}{2g}$. Hence the correct choice is (b).

Answer: Option D. -> - 2 mv
:
D
Change in momentum = $m{\overrightarrow{v}}_2-m{\overrightarrow{v}}_1=-mv-mv=-2mv$

Answer: Option B. -> Two
:
B
The first collision will be between balls 1 and 2. Since both have the same mass, after the collision ball 1will come to rest and ball 2 will move with speed $v_1$. This ball will collide with the stationary ball 3.After this second collision, let $v_2$and $v_3$ be the speeds of balls 2 and 3 respectively. Since the collisionsare elastic,$v_2$ and $v_3$ are given by (see Sec. 9)
$v_2=\left(\frac{m-M}{m+M}\right)v_1$ (i)
and $v_3=\left(\frac{2m}{m+M}\right)v_1$ (ii)
If M < m,it follows from (i) and (ii) that v₂ < v₃ and both have the same direction.
Therefore, ball 2 cannot collide with ball 3 again. Hence there are only two collisions.
Thus the correct choice is (b).

Answer: Option C. -> v√2
:
C

Initial momentum of the system
${\overrightarrow{P}}_i=mv\hat{i}+mv\hat{j}$
$|{\overrightarrow{P}}_i|$ = $\sqrt{2}mv$
Final momentum of the system = 2mV
By the law of conservation of momentum
$\sqrt{2}mv=2mV\Rightarrow V=\frac{v}{\sqrt{2}}$

Answer: Option A. -> (A−1A+1)2
:
A
Mass of neutron $\left(m_1\right)=1unit$. Mass of nucleus $\left(m_2\right)=Aunits$.
The velocity of the neutron after the collision is
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u=\left(\frac{1-A}{1+A}\right)u$
KE of neutron after collision = $\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}\times 1\times {\left(\frac{1-A}{1+A}\right)}^2{u}^2$
KE of neutron before collision = $\frac{1}{2}m{u}^2=\frac{1}{2}\times 1\times u^2=\frac{1}{2}{u}^2$.
Their ratio is ${\left(\frac{1-A}{1+A}\right)}^2$, which is choice (a)

Answer: Option A. -> Bob A comes to rest at B and bob B move to the left attaining a maximum height h.
:
A
Suppose the bob A acquires a velocity v on reaching the bob B. In a head-on elastic collision between two bodies of the same mass, the velocities are exchanged after the collision. Hence the bob A will come to rest at the lowermost position (occupied by B before collision) and the bob B will move to the left attaining a maximum height h. hence the correct choice is (a).

Answer: Option D. -> v28g
:
D
In an inelastic collision, two bodies stick together. After the collision, the speed of the ball and the bob(sticking together) is $v^{{}^{\prime }}=\frac{v}{2}$.
The height to which they will rise is given by $v^{{}^{\prime }}=\sqrt{2g{h}^{{}^{\prime }}}or{h}^{{}^{\prime }}=\frac{v^{{}^{\prime }2}}{2g}=\frac{v^2}{8g}$
Hence the correct choice is (d).

Answer: Option D. -> 9.89 m/s
:
D

$P_x=m\times v_x=1\times 21=21kgm/s$
$P_y=m\times v_y=1\times 21=21kgm/s$
$\therefore$ Resultant = $\sqrt{P_x^2+P_y^2}=21\sqrt{2}kgm/s$
The momentum of heavier fragment should be numerically equal to resultant of ${\overrightarrow{P}}_x$ and ${\overrightarrow{P}}_y$.
$3\times v=\sqrt{P_x^2+P_y^2}=21\sqrt{2}\therefore v=7\sqrt{2}=9.89m/s$

Answer: Option B. -> 0
:
B
Coefficient of restitution is the ratio between Relative velocity of separation to Relative velocity of approach of colliding bodies.
In perfectly inelastic collision, the two bodies stay togethere after collision and hence the relative velocity of separation is zero.
Therefore, coefficient of restitution is zero.