Question

A body of mass $2.0 \,kg$ makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed $ \upsilon $ . What is the mass of the other body and the speed of the two body center of mass?

• A. 1.0 kg and $ \frac{2}{3}v $
• B. 1.2 kg and $ \frac{5}{8}v $
• C. 1.4 and $ \frac{10}{17}v $
• D. 1.5 kg and $ \frac{4}{7}v $

Answer 1

Answer: (B)

Given that $m_{1}=2\, kg \,m_{2}=?$
$ u_{1}=v \,v_{2}=? $
$v_{1}=v / 4\, v_{c m}=?$
From conservation of momentum
$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} $
$2 v+0=2 \times \frac{v}{4}+m_{2} v_{2}$
$\frac{3}{2} v=m_{2} v_{2} \ldots$ (i)
Now, from Newtons law of restitution
$v_{1}-v_{2}=e\left(u_{2}-u_{1}\right)$
$(\because e=1$ for elestic collision)
$\frac{v}{4}-v_{2}=-v v_{2}=\frac{5}{4} v$
From Eq. (i), we get
$m_{2}=\frac{6}{5}=1.2\, kg$
$v_{c m}=\frac{m_{1} v_{1}+m_{2} v_{2}}{m_{1}+m_{2}}$
$=\frac{2 \times \frac{v}{4}+1.2+\frac{5}{4} v}{2+12}$
$=\frac{(0.5+15) v}{32}=\frac{5}{8} v$