Question

A very massive object travelling at $10m{s}^{-1}$ strikes a very light object, initially at rest and the light object moves off in the direction of travel of the heavy object. If the collision is elastic, the speed of the lighter object is

• A. $5m{s}^{-1}$
• B. $10m{s}^{-1}$
• C. $15m{s}^{-1}$
• D. $20m{s}^{-1}$

Answer 1

The correct option is B

$10m{s}^{-1}$

Given data and assumptions:

Let the mass of a massive object being $m_1$

Mass of lighter object be $m_2$ So, $m_1>m_2$

The velocity of a massive object, $u_1=10m{s}^{-1}$

The lighter object is at rest. So, the initial velocity $u_2=0$

Let the final velocities of massive and lighter objects are $v_1$ and $v_2$

Apply conservation principle

For elastic collision, both momentum and kinetic energy are conserved.

$$\begin{gathered} m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2 \\ {m}_1(u_1-v_1)=m_2(v_2-u_2)-----\left(1\right) \\ \frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_1{v}_2^2 \\ {m}_1({u^2}_1-{v^2}_1)=m_2({v^2}_2-{v^2}_1)-----\left(2\right) \end{gathered}$$

Divide 2 by 1

$$\begin{gathered} \frac{m_1({u^2}_1-{v^2}_1)}{m_1(u_1-v_1)}=\frac{m_2({v^2}_2-{u^2}_2)}{m_2(v_2-u_2)} \\ {u}_1+v_1=v_2+u_2 \end{gathered}$$

Substituting the given values, $10+v_1=v_2+0$

velocity of the lighter object, $v_2=10+v_1$

If the massive object is coming to rest after collision, $v_1=0$

Thus, the velocity of the lighter object, $v_2=10m{s}^{-1}$

The velocity of the lighter object after the collision, $v_2=10m{s}^{-1}$

Option(b) is correct.