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,
$Mv+m\times 0=M{v}_1+m{v}_2$
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 $v_2=\frac{2Mv}{(M+m)}$
As $M>>m$
So, $v_1=v\Rightarrow v_2=v+v=2v$