First slide

The steps in collision

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 [AIIMS 2018]

Moderate

A

2 v
B

3 v
C

v
D

$\frac{v}{2}$

Solution

From law of conservation of momentum,
$Mv + m \times 0 = {Mv}_{1} + {mv}_{2}$
$\Rightarrow M (v - v_{1}) = {mv}_{2}$ …(i)
Again, from the conservation of kinetic energy (as collision is of elastic nature),
   $\frac{1}{2} M v^{2} + \frac{1}{2} m \times 0 = \frac{1}{2} M v_{1}^{2} + \frac{1}{2} m v_{2}^{2}$
$\Rightarrow M (v^{2} - v_{1}^{2}) = m v_{2}^{2}$ …(ii)
On solving Eqs. (i) and (ii), we get
$\frac{M (v - v_{1})}{M (v + v_{1}) (v - v_{1})} = \frac{m v_{2}}{m v_{2}^{2}}$
   $v_{2} = v + v_{1}$
Now, solving Eqs. (i) and (iiil, we get
   $v_{1} = \frac{(M - m) v}{(M + m)}$
and $v_{2} = \frac{2 M v}{(M + m)}$
As, M >>m
So, $v_{1} = v and v_{2} = 2 v$
Hence, velocity of lighter body (m) is 2v.

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