In the arrangement shown in Fig. the ball and the block have the same mass `m=1 kg` each, `theta=60^(@)` and length `l = 2.50m`. Coefficient of friction between the block and the floor is `0.5`. When the ball is released from the position shown in Fig. it collides with the block and the block stops after moving a distance `2.50 m`.

The velocity of block just after collision is

Options

A. a. `10m//s`
B. b. `5m//s`
C. c. `2.5m//s`
D. d. `3m//s`

Answer

B

Explanation

When ball is released, it moves along a vertical circle with centre at `A`. Kinetic energy of ball just before collision is equal to loss of its potential energy from point of release to the point of collision. Therefore, velocity `v_(1)` of ball, just before collision, is given by
`1/2mv_(1)^(2)mg(l-lcostheta)impliesv_(1)=5m//s`
After collision, the block starts to move towards right. But it is retarded by force of friction and ultimately it comes to rest. According to law of conservation of energy,
Kinetic energy of the block just after collision `=` Work done by it against friction.
Therefore, its velocity `v`, just after collision is given by
`1/2mv_(2)^(2)=mumgs`
where `m=0.54` and `s=2.50 m:.v_(2)=5m//s`
Coefficient of restitution `e=(v_(2)-v_(1))/(u_(2)-u_(1))`
where `u_(2)=0, u_(1)=5m/s` and `v_(1)=(5-5e)` …………i
Applying law of conservation momentum.
`"mu"_(1)+"mu"_(2)=mv_(1)+mv_(2)`
`5m+(mxx0)=mv_(1)+5m` or `v_(1)=0`
substituting in eqn i we get `e=1`