Question

A rigid massless rod of length $L=1m$ joins two particles each of mass $m=1kg$. The rod lies on a frictionless table, and is struck by a particle of equal mass $m=1kg$ and velocity$v_0=7\sqrt{2}$ moving as shown in the figure. After the collision the partcle moves straight back. Calculate the angular velocity of mass after collision, assuming that collision is perfectly elastic.

Answer 1

Since the collision is perfectly elastic the momentum of object A will be equal to the momentum pf object B.

Momentum of A=mass$\times$velocity=1 $\times$$7\sqrt{2}$

$=7\sqrt{2}kgm/s$

The velocity of the object B=

$=7\sqrt{2}m/s$ (To satisfy law of conservation of momentum)

Now from figure we observe that velocity of object B is experienced at an angle of ${ 30 }^{ 0 }$.

Now as we know linear velocity acts tangential to the circular motion than linear velocity of the object will be

$=Vcos{ 45 }^{ 0 }=7\sqrt { 2 } \times \frac { 1 }{ \sqrt { 2 } } =\quad 7\quad m/s$

Also angular velocity $\omega =\frac { V }{ R } =7\quad rad/s\quad $

as lenght of the rod =1m