Question

A ball possessing kinetic energy T collides head on with an initially stationary elastic dumb-bell (shown in figure) and rebounds in opposite direction with kinetic energy T. the masses of all the three balls are equal. The energy of the dumbbell oscillations after collision will be:
(A) $\dfrac{{T - {T'} - 2\sqrt {T{T'}} }}{2}$
(B) $\dfrac{{T - 3{T'} - 2\sqrt {T{T'}} }}{2}$
(C) $T + {T'} + 2\sqrt {T{T'}}$
(D) $T - 3{T'} - 2\sqrt {T{T'}}$

Answer 1

Since it is mentioned in the question that the dumbbell is elastic, we can use the fundamental law of law of conservation of energy. But the ball hitting the dumbbell is not mentioned as elastic. So, we will use the law of conservation of momentum, another fundamental law.
Sum of initial momentum of a dumbbell is equal to the sum of final momentum of the ball and dumbbell.

Complete step by step answer:
Since in every situation (elastic or inelastic problems) the momentum is conserved. So, using the law of conservation of momentum, we have,
$p_b^i + p_d^i = p_b^f + p_d^f$
Where,
$p_b^i$ is the initial momentum of the ball and $p_d^i$ is the initial momentum of the dumbbell. Similarly, is for the final momentums. So, the kinetic energy is described as,
$T = \dfrac{1}{2}m{v^2} \\\ \Rightarrow v = \sqrt {\dfrac{{2T}}{m}} \\\$
Now,
$p_b^i = m\sqrt {\dfrac{{2T}}{m}} \\\ p_b^f = - m\sqrt {\dfrac{{2T'}}{m}} \\\ p_d^i = 0 \\\ p_d^f = mv \\\$
The negative sign indicates that the ball has rebounded in the opposite direction and momentum being a vector quantity so will become negative (Right is taken as positive direction).
Substituting the given data in the momentum equation,
$m\sqrt {\dfrac{{2T}}{m}} + 0 = - m\sqrt {\dfrac{{2T'}}{m}} + mv \\\ v = \sqrt {\dfrac{{2T}}{m}} + \sqrt {\dfrac{{2T'}}{m}} \\\$
Now, using the kinetic energy formula we will get the oscillation energy of the dumbbell as,
${T_d} = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{\left[ {\sqrt {\dfrac{{2T}}{m}} + \sqrt {\dfrac{{2T'}}{m}} } \right]^2} \\\ \Rightarrow {T_d} = \dfrac{1}{2}m\left[ {\dfrac{{2T}}{m} + \dfrac{{2T'}}{m} + \dfrac{{2.2\sqrt {TT'} }}{m}} \right] \\\ \Rightarrow {T_d} = T + T' + 2\sqrt {TT'} \\\$
Where,
${T_d}$ is the final energy of the dumbbell used further for oscillations
Therefore, the correct answer is option (C).

Note:
Do Not get confused in the energy of the dumbbell oscillations and the kinetic energy. Here it is asked for kinetic energy. Since the dumbbell is at rest initially, whatever energy is given in the collision as kinetic energy is further used for oscillations. If it was asked how much the dumbbell gets compressed after collision then we would have equated the found energy with spring constant energy i.e. ($\dfrac{1}{2} kx^2$).