Question 275: Centre of Mass and Rotational Motion

Question

A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $$\left( {{m \over M}} \right)$$ is $${1 \over x}$$. The value of 'x' will be ____________.

Accepted Answer

The given situation can be shown as
Before collision,
As the collision is perfectly elastic, therefore momentum is conserved, i.e.
p
initially
= p
finally
$$\Rightarrow$$ mu = Mv ..... (i)
Angular momentum will also be conserved about point O.
$$ \Rightarrow mv\,.\,{L \over 2} = {{M{L^2}} \over {12}}\omega $$
$$ \Rightarrow \omega  = {{6mv} \over {ML}}$$ .... (ii)
$$\because$$ Coefficient of restitution,
$$e = {{{\mathop{m Relative}
olimits} \,velocity\,after\,collision} \over {{\mathop{m Relative}
olimits} \,velocity\,before\,collision}}$$
$$ \Rightarrow 1 = {{v + {{\omega L} \over 2}} \over u}$$
$$ \Rightarrow v + {{\omega L} \over 2} = u$$ .... (iii)
From Eqs. (ii) and (iii), we get
$$v + {{3mu} \over M} = u$$
$$ \Rightarrow {{mu} \over M} + {{3mu} \over M} = u$$ [using Eq. (i)]
$$ \Rightarrow {{4mu} \over M} = u \Rightarrow {m \over M} = {1 \over 4}$$ .... (iv)
According to question,
ratio of masses $$\left( {{m \over M}} ight) = {1 \over x}$$.
Comparing it with Eq. (iv), we get x = 4.

Centre of Mass and Rotational Motion

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