Question

Two equal masses ${m_1}$ and ${m_2}$ moving along the same straight line with velocities \[ + 3{\text{ }}m/s\] and \[{\text{ - 5 }}m/s\] respectively collide elastically. Their velocities after the collision will be respectively:
A. \[ + 4{\text{ }}m/s\] for both
B. \[{\text{ - 3 }}m/s\] and \[ + 5{\text{ }}m/s\]
C. \[{\text{ - 4 }}m/s\] and \[ + 4{\text{ }}m/s\]
D. \[{\text{ - 5 }}m/s\] and \[ + 3{\text{ }}m/s\]

Answer 1

Hint: In this question, we are given the masses of two bodies and their masses are equal. Also, the initial velocity of both the particles are equal. We have to find the velocities after the elastic collision of both the bodies. To calculate velocity, we’ll apply the velocity formula of elastic collision i.e., ${v_1} = \dfrac{{\left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2}}}{{{m_1} + {m_2}}}$ and ${v_2} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2} + 2{m_1}{u_1}}}{{{m_1} + {m_2}}}$.

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,
Velocities of both the bodies are \[ + 3{\text{ }}m/s\] and \[{\text{ - 5 }}m/s\] respectively.
Initial velocity of the first body, ${u_1} = + 3{\text{ }}m/s$
Initial velocity of the second body, ${u_2} = {\text{ - 5 }}m/s$
Masses of the given bodies 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 ${u_1} = + 3{\text{ }}m/s$, ${u_2} = {\text{ - 5 }}m/s$ and ${m_1} = {m_2} = m$.
Velocity of P,
${v_1} = \dfrac{{\left( {m - m} \right)3 + 2m\left( { - 5} \right)}}{{m + m}} \\ $
$\Rightarrow {v_1} = \dfrac{{ - 10m}}{{2m}}$
$\Rightarrow {v_1} = - 5{\text{ }}m/s \\ $
And Velocity of Q,
${v_2} = \dfrac{{\left( {m - m} \right)\left( { - 5} \right) + 2m\left( 3 \right)}}{{m + m}} \\ $
$\Rightarrow {v_2} = \dfrac{{6m}}{{2m}} \\ $
$\therefore {v_2} = + 3{\text{ }}m/s$

Hence, option D is the correct answer i.e., \[{\text{ - 5 }}m/s\] and \[ + 3{\text{ }}m/s\].

Note: To solve such problems, one should always remember the concept of collision of elastic bodies that if the mass is the same and the bodies are perfectly elastic then after collision their velocities get interchanged. Consider the two particles or bodies of equal mass moving in opposite directions at the same velocity. In this case, both bodies come to rest after the collision, implying that no velocity exchange occurs.