Question

A body of $1\text{kg}$ makes an elastic collision with a second body at rest and continues to move in the original direction, but with half of its original speed. What is the mass of the second body?

• A. $\frac{1}{2}\text{kg}$
• B. $2\text{kg}$
• C. $\frac{1}{3}\text{kg}$
• D. $3\text{kg}$

Answer 1

The correct option is C $\frac{1}{3}\text{kg}$

Given that
$m_1=1\text{kg}$
$m_2=?$


Let $v_0$ be the velocity of mass $m_1$ before collision.
Now, we can apply momentum conservation because no net external force is acting on the system.
$m_1{v}_0+(m_2\times 0)=m_1\times \frac{v_0}{2}+m_{2}v$
$\Rightarrow m_1{v}_0=\frac{m_1{v}_0}{2}+m_{2}v$
$\Rightarrow v=\frac{m_1{v}_0}{2m_2}=\frac{v_0}{2m_2}$ ... (1) $[\because m_1=1\text{kg}]$

From energy conservation,
$\frac{1}{2}{m}_1{v}_0^2+0=\frac{1}{2}{m}_1{\left(\frac{v_0}{2}\right)}^2+\frac{1}{2}{m}_2{v}^2$
$\Rightarrow \frac{v_0^2}{2}=\frac{v_0^2}{8}+\frac{1}{2}{m}_2{v}^2$
$\Rightarrow m_2{v}^2=\frac{3}{4}{v}_0^2$

Putting the value of $v$ from eq. (1),
$m_2{\left(\frac{v_0}{2m_2}\right)}^2=\frac{3}{4}{v}_0^2$
$\Rightarrow m_2\times \frac{v_0^2}{4m_2^2}=\frac{3}{4}{v}_0^2$
$\Rightarrow \frac{v_0^2}{4}\times \frac{4}{3v_0^2}=m_2$
$\Rightarrow m_2=\frac{1}{3}\text{kg}$