Question

2 Small balls A and B each of mass m are joined rigidly at the ends of a light rod of length L,they are placed on a frictionless horizontal surface .Another ball of mass 2m moving with a speed of u towards one of the balland perpendicular to the length of the rod,on the horizontal frictionless surface as shown in the figure.If the coefficient of restitution is 1/2, then the angular speed of the rod after collision will be

Answer 1

Answer:

Angular speed of the rod will be $\omega =\frac{u}{L}$

Explanation:

Here mass 2m collide with ball A, during collision between object momentum of object is conserved.

here initial velocity of mass 2m is u

initial velocity of ball A and ball B is zero

Let final velocity of mass 2m is $v_{2}$ and final velocity of ball A is $v_{1}$  

final velocity of ball B just after collision will be zero

so form conservation of momentum between mass 2m and ball A

initial momentum before collision =  final momentum after collision

$2mu = mv_{1} + 2m v_{2}$       ...(a)

from newton law of collision

coefficient of restitution  e =$\frac{relative velocity of separation}{relative velocity of approach}$

$\frac{1}{2}= \frac{v_{1} - v_{2}}{u}$      ...(b)

on solving the equation a and equation b

$v_{1}$  = u

from here we got the velocity of ball A is while velocity of ball B is zero just after collision

so angular speed of rod $\omega= \frac{tangential velocity}{perpendicular distance}$

$\omega= \frac{u}{L}$