Question

A particle of mass $1\text{kg}$ suffers one - dimensional elastic collision with an unknown mass at rest. It rebounds directly backward by losing $64\%$ of its initial kinetic energy. The unknown mass $\left(\text{in kg}\right)$ is

Answer 1

Given,
Mass of particle $\left(m\right)=1\text{kg}$
Let us suppose initial kinetic energy of the particle is $k_0$, then we can say that,
$\frac{1}{2}m{u}_1^2=k_0...........\left(1\right)$
where $u_1$ is the initial speed of the particle.
After collision, kinetic energy lost by $1\text{kg}$ mass is $64\%$ of its initial value.
$\therefore$ Final kinetic energy,
$\frac{1}{2}m{v}_1^2=k_0-0.64k_0$
$\Rightarrow \frac{1}{2}m{v}_1^2=0.36\times k_0............\left(2\right)$
Using $\left(1\right)$ in $\left(2\right)$ we get
$\frac{1}{2}m{v}_1^2=0.36\times \frac{1}{2}m{u}_1^2$
$\Rightarrow u_1=\frac{5v_1}{3}---\left(2\right)$
Let the unknown mass be $m_2$.
On applying Law of conservation of linear momentum,
$\overrightarrow{p_i}=\overrightarrow{p_f}$
For two particles, we can write this as,
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$\Rightarrow m{u}_1+m_2\times 0=-m{v}_1+m_2{v}_2$
$\Rightarrow 1\times \frac{5v_1}{3}=-1\times v_1+m_2{v}_2\left[\text{from}\right(1\left)\right]$
$\Rightarrow m_2{v}_2=\frac{8v_1}{3}$
$\Rightarrow m_2=\frac{8v_1}{3v_2}.............\left(3\right)$

For a perfectly elastic collision,
Co-efficient of restitution $=1$
$\Rightarrow \frac{\text{Velocity of separation}}{\text{Velocity of approach}}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\overrightarrow{u_1}-\overrightarrow{u_2}}=1$
$\Rightarrow \frac{v_2+v_1}{u_1}=1$
$\Rightarrow v_2+v_1=u_1=\frac{5v_1}{3}\left[\text{using equation}\right(2\left)\right]$
$\Rightarrow v_2=\frac{2v_1}{3}$
$\Rightarrow \frac{v_1}{v_2}=\frac{3}{2}............\left(4\right)$
Substituting (4) in (3),
$\Rightarrow m_2=\frac{8}{3}\times \frac{3}{2}$
$\Rightarrow m_2=4\text{kg}$
Hence, the unknown mass $\left(m_2\right)=4\text{kg}$
$\therefore$ Correct answer is 4.