Question

Body of mass $M$ is much heavier than the other body of mass $m$. The heavier body with speed $v$ collides with the lighter body, which was at rest initially, elastically. The speed of lighter body after collision is

• A. $2v$
• B. $3v$
• C. $v$
• D. $v / 2$

Answer 1

Answer: (A)

From conservation of momentum,
$M v+m \times 0=M v_{1}+m v_{2}$
where, $v_{1}$ and $v_{2}$ be the velocities of $M$ and $m$ after collision.
$M\left(v-v_{1}\right)=m v_{2}$ ...(i)
Again, from the conservation of kinetic energy (as collision is of elastic nature),
$\frac{1}{2} M v^{2}+\frac{1}{2} m \times 0=\frac{1}{2} M v_{1}^{2}+\frac{1}{2} m v_{2}^{2}$
$\Rightarrow M\left(v^{2}-v_{1}^{2}\right)=m v_{2}^{2}$ ...(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{M\left(v-v_{1}\right)}{M\left(v+v_{1}\right)\left(v-v_{1}\right)}=\frac{m v_{2}}{m v_{2}^{2}}$
$v_{2}=v+v_{1}$ ...(iii)
Now, solving Eqs. (i) and (iii), we get
$M\left(v-v_{1}\right)=m\left(v+v_{1}\right)$
$v_{1}=\frac{(M-m) v}{(M+m)}$
and $v_{2}=\frac{2 M v}{(M+m)}$
As $M>>m$
So, $v_{1}=v \Rightarrow v_{2}=v+v=2 v$