Question

A rigid ball with mass 1.0kg initially travels with a velocity 12 m s ˆı. It strikes and bounces off of a block of mass 2.0kg which was initially at rest and hanging from a light rope of length 3.0m. After the collision, the ball travels at −6.0 m s ˆı.

Immediately after the ball strikes, what is the speed of the block? (a) 3m s (b) 6m s (c) 9m s (d) 12m s (e) 18m s

Immediately after the ball strikes, what is the tension in the rope? (a) 20N (b) 34N (c) 54N (d) 74N (e) 96N

Answer 1

During collision between ball and Block, momentum of the system( ball + Block) is conserved.
Taking ball mass as M1 and it' velocity before and after collision U1 and V1 and Block mass as M2 and it' velocity before and after collision as U2 and V2 , we have
M1 U1 + M2 U2 = M1 V1 + M2 V2

1x12 + 0 = 1x(-6) + 2xV2
V2 = 9 m/s
data given is not correct ​as final relative velocity of 15 is gerater than initila relative velocity of 12.

As block is hanging with rope, it starts moving in the circle. It's acceleration towards center, that is upward is
V²/R.. Forces on block are Tension T(upward) and it's weight mg (downward)
Applying Newton's second law we get
T - mg = mV²/R
T = 2x 9.8 + 2x(9)²/3 = 73.6 N
option d is correct