Question

A 170 g cue ball is hit directly to the right with an initial velocity of [v] m/s. It undergoes an elastic collision with a 155 g pool ball at rest.

Part A

Find the final velocity of the cue ball.

nothing

Part B

Find the final velocity of the pool ball.

Answer 1

m1 = 170 gm = 0.17 Kg

m2 = 155 gm = 0.155 kg

initial velocity, u1 = v

and u2 =0

let final velocities be v1 and v2

then

As per the law of conservation of momentum,

momentum before collision = momentum after collision

m1u1+m2u2 = m1v1+m2v2

0.17*v+0 = 0.17v1+0.155v2 (1)

0.155*v2 = 0.17 (v-v1)

v2 = 0.17(v-v1)/0.155 = 1.096(v-v1)

also kinetic energy before collision = KE after collision

0.5*m1*u12 +0.5*m2*u22 = 0.5*m1*v12 +0.5*m2*v22

0.17*v2 +0 = 0.17*v12+0.155*v22    (2)

v1 = ((m1-m2)u1+2m2*u2)/(m1+m2) = ((0.17-0.155)v+0)/(0.17+0.155) = 0.015v/0.325 = 0.046 v m/sec = final velocity of cue ball

v2 = [2m1*u1-(m1-m2)u2]/(m1+m2) = [2*0.17*v-0]/(0.325) = 1.046 v m/sec = final velcoity of pool ball