Question

A 0.10 kg ball traveling at 9.6 m/s to the north collides and bounces off a second 0.12kg ball that is traveling 7.6 m/s in the opposite direction. If the 0.10 ball travels at 0.30 m/s to the south after the collision, what is the velocity of the 0.12kg ball after the collision?​

Answer 1

0.65 m/s NORTH

Step-by-step explanation:

North is +, south is -

let V = velocity of the 0.12 kg ball AFTER the collision

Law of conservation of momentum (P = mv) states that total momentum before the collision must equal total momentum after the collision.

(0.10 kg)(9.6 m/s) - (0.12 kg)(7.6 m/s) = -(0.10 kg)(0.30 m/s) + (0.12 kg)(V)

Before → After

0.96 kg·m/s - 0.912 kg·m/s = -0.03 kg·m/s + 0.12 kg(V)

0.048 kg·m/s + 0.03 kg·m/s = 0.12 kg(V)

0.078 kg·m/s = ·0.12 kg(V)

V = 0.078 kg·m/s / 0.12 kg = 0.65 m/s NORTH