Question

A uniform smooth rod (mass $m$ and length $l$) placed on a smooth horizontal floor is hit by a particle (mass $m$) moving on the floor, at a distance $\frac{l}{4}$ from one end elastically $(e=1)$. The distance travelled by the centre of the rod after the collision when it has completed three revolutions will be:

• A. $2\pi l$
• B. cannot be determined
• C. $\pi l$
• D. none of these

Answer 1

Answer: (A)

$2\pi l$

Applying conservation of linear momentum,

$mv=m{v}^{\prime }+mV\Rightarrow v=v^{\prime }+V$ ....(i)

Applying conservation of angular momentum about the collision point.

$0=\left(\frac{m{l}^2}{12}\right)\omega -mV\left(\frac{l}{4}\right)\Rightarrow l\omega =3V$ .....(ii)

Applying restituting equation

$(u_1-u_2{)}_n=(v_2-v_1{)}_n\Rightarrow (v-0)=(V-v^{\prime })$ .....(iii)

Solving Eqs. (i), (ii), and (iii) we get $V=v$ and

$\omega =\frac{3v}{l}$

Time taken to complete three revolutions $(\theta =6\pi )$

$t=\frac{\theta }{\omega }=\frac{6\pi }{\omega }=\frac{6\pi l}{3v}=\frac{2\pi l}{v}$

Hence, distance travelled by the centre of the rod is

$s=Vt=v\left(\frac{2\pi l}{v}\right)=2\pi l$