Question

A sphere of mass 'm' moving with velocity 'v' collides head-on on another sphere of same mass which is at rest. The ratio of final velocity of second sphere to the initial velocity of the first sphere is (e is coefficient of restitution and collision is inelastic)

• A. $\frac{e-1}{2}$
• B. $\frac{e}{2}$
• C. $\frac{e + 1}{2}$
• D. e

Answer 1

Answer: (C)

$\frac{e + 1}{2}$

Answer 2

Given, mass of sphere $=m$
According to the question, we can draw the following diagram

Applying the conservation law of momentum, we get
$m v+m(0) =m v_{1}+m v_{2}$
$m v =m v_{1}+m v_{2}$
$m v =m\left(v_{1}+v_{2}\right)$
$v =v_{1}+v_{2}$
$=>\,\,\,\,\,\,v_{1} =v-v_{2} \,\,\,\,\,\,\,\,\, ....(i)$
We know that, coefficient of restitution
$=\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}$
or $\,\,\,\,\, e =\frac{ v _{2}- v _{1}}{ v -0} \,\,\,\,\, \text { or } e v = v _{2}- v _{1}$
From E (i), we get
$\Rightarrow \,\,\, e v=v_{2}-\left(v-v_{2}\right)$
$\Rightarrow \,\,\, e v=v_{2}-v+v_{2}$
$\Rightarrow \,\,\, e v=2 v_{2}-v \Rightarrow e v+v=2 v_{2}$
$\Rightarrow \,\,\, v(1+e)=2 v_{2}$
$\Rightarrow \,\,\, \frac{v_{2}}{v}=\frac{e+1}{2}$