Question

A body of mass $M$ moves with velocity $v$ and collides elastically with another body of mass $m$ $(M \gg>m)$ at rest, then the velocity of body of mass $m$ is :

• A. $ v $
• B. $ \text{2v} $
• C. $ \text{v/2} $
• D. zero

Answer 1

The Correct Option is B

In an elastic collision, linear momentum and kinetic energy are conserved.
Initial momentum $=$ Final momentum
i.e., $m_{1} \,u_{1}+m_{2}\, u_{2}=m_{1}\, v_{1}+m_{2} v_{2}$
$\therefore M \times v+m \times 0=M v_{1}+m v_{2}$
or $M v=M v_{1}+M v_{2}$
or $M\left(v-v_{1}\right)=m v_{2} \,\,\,...(i)$
Again kinetic energy is also conserved.
$\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} \,v_{1}^{2}+\frac{1}{2} m_{2}\, v_{2}^{2}$
$\therefore M v_{2}+m \times 0=M v_{1}^{2}+m v_{2}^{2}$
or $M v^{2}=M v_{1}^{2}+m v_{2}^{2}$
or $M\left(v^{2}-v_{1}^{2}\right)=m v_{2}^{2} ?(i i)$
Dividing E(ii) by E (i), we get
$\frac{M\left(v^{2}-v_{1}^{2}\right)}{M\left(v-v_{1}\right)}=\frac{m v_{2}^{2}}{m v_{2}}$
or $v+v_{1}=v_{2}$ As $M>>m$
so $v_{1} \approx v$
$\therefore v_{2}=v+v=2 v$
Note: The conservation of momentum and the conservation of total energy holds for all the three types of collisions, but KE conservation law holds only for perfectly elastic collision.