Question

A pendulum bob of ideal string mass `m` connected to the end of of length `l` is released from rest from horizontal position as shown in Fig. At the lowest point, the bob makes an elastic collision with a stationary block of mass `5m`, Which is kept on a frictionless surface. Mark out the comet statement(s) for the instant just after impact.

A. Tension in the string is `(17//9)mg`
B. tensiion in the string is `3mg`
C. the velocity of the block is `sqrt(2gl)//3`
D. The maximum height attained by the pendulum bob after impact is (measured from the lowest position) `4l//9`.

Answer 1

Correct Answer - A::D
The velocity of bob just before the impace is `v=sqrt(2gl)` along the horizontal direction.

From momentum conservation
`mv=-mv_(1)+5mv_(2)`
From coefficient of restitution equation
`1=(v_(1)+v_(2))/vimpliesv_(1)+v_(2)=v`
Solving above equations, we get `v_(1)=(2v)/3`
`v_(2)=v/3`
For tension in string `T-mg=(mv_(1)^(2))/l`
`implies T=17/9mg`
Let the maximum height attained by the bob be `h`, then `(mv_(1)^(2))/2=mgh`
`implies h=(4l)/9`