Question

A body of mass $m$ having an initial velocity $v$, makes head-on collision with a stationary body of mass $M$. After the collision the body of mass m comes to rest and only the body having mass $M$ moves. This will happen only when:
A. $m > > M$
B. $m < < M$
C. $m = M$
D. $m = \dfrac{M}{2}$

Answer 1

Hint:In this question, we are given that there are two bodies with the masses $M$ and $m$. After collision their velocities get exchanged. We have to tell behind the exchange of their velocities what is the relation between their mass. If the mass of two bodies are the same then their velocities get exchanged after collision.

Formula used:
When two bodies collides (elastically) then the velocities are;
${v_1} = \dfrac{{\left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2}}}{{{m_1} + {m_2}}}$, ${v_2} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2} + 2{m_1}{u_1}}}{{{m_1} + {m_2}}}$
Here, ${m_1},{m_2}$ and ${u_1},{u_2}$ are the masses and initial velocities of both the bodies respectively.

Complete step by step solution:
Given that the mass of the bodies are $m$ and $M$ respectively. Before the collision the initial velocities of both the bodies were $v$ and $0$ respectively and after the collision their velocities were exchanged. Exchange of velocities during collision is possible if and only if $M = m$.

For Example:
Two perfectly elastic particles M and N of equal mass are travelling along the line joining them with velocities \[15{\text{ }}m/sec\] and \[10{\text{ }}m/sec\]. After collision, what will be their velocities?
Velocities of particles M and N are \[15{\text{ }}m/sec\] and \[10{\text{ }}m/sec\] respectively.
Initial velocity of M, ${u_1} = 15{\text{ }}m/sec$
Initial velocity of N, ${u_2} = 10{\text{ }}m/sec$
The Masses of the particles M and N are the same.
Therefore, let ${m_1} = {m_2} = m$

After collision their velocities will be,
Using the formula,
${v_1} = \dfrac{{\left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2}}}{{{m_1} + {m_2}}} \\ $
$\Rightarrow {v_2} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2} + 2{m_1}{u_1}}}{{{m_1} + {m_2}}} \\ $
Put the value of ${u_1} = 15{\text{ }}m/sec$, ${u_2} = 10{\text{ }}m/sec$ and ${m_1} = {m_2} = m$
It will be, velocity of M,
${v_1} = \dfrac{{\left( {m - m} \right)15 + 2m\left( {10} \right)}}{{m + m}} \\ $
$\Rightarrow {v_1} = \dfrac{{20m}}{{2m}} \\ $
$\Rightarrow {v_1} = 10{\text{ }}m/sec$
And Velocity of N,
${v_2} = \dfrac{{\left( {m - m} \right)10 + 2m\left( {15} \right)}}{{m + m}} \\ $
$\Rightarrow {v_2} = \dfrac{{30m}}{{2m}} \\ $
$\therefore {v_2} = 15{\text{ }}m/sec$

Hence, option C is the correct answer i.e., $m = M$.

Note: The system experiences no net loss of kinetic energy as a result of an elastic collision. Kinetic energy and momentum are both conserved in this case. When kinetic energy is lost, an inelastic collision happens. In an inelastic collision, momentum is preserved but kinetic energy is not. Because some kinetic energy was transferred to something else. Thermal energy, acoustic energy, and material deformation are the most likely causes.