Question

A body of mass $2\text{kg}$ moving with a speed of $4\text{m/s}$ makes an elastic collision with another body at rest. It continues to move in the original direction but with one fourth of its initial speed. The speed of the center of mass of the two bodies is $\frac{x}{10}\text{m/s}$. Then the value of $x$ is

Answer 1

According to conservation of momentum principle,
$P_i=P_f$
$2\times 4=2\times 1m_2\times v_2$
$m_2{v}_2=\mathrm{6...}\left(i\right)$
and the coefficient of restitution is,
$e=\frac{v_2-v_1}{u_1-u_2}$
Here, $u_1=4,u_2=0,v_1=1$
$\therefore 1=\frac{v_2-1}{4}\Rightarrow v_2=5\text{m/s}$
from $\left(i\right)$,
$m_2\times 5=6$
$m_2=1.2\text{kg}$
Now, $V_{cm}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

$=\frac{2\times 1+1.2\times 5}{2+1.2}=\frac{8}{3.2}=\frac{x}{10}$
$\therefore x=25$