Question

A 0.02kg ball moving to the right at 0.25 m/s makes an elastic head-on collision with a 0.04kg ball moving to the left at 0.15m/s. After the collision, the smaller ball moves to the left at 0.16 m/s. What is the velocity of the 0.04 kg ball after collision?

Answer 1

Answer 2

Let's use the conservation of momentum and kinetic energy to solve this problem.

Conservation of momentum tells us that the total momentum before the collision is equal to the total momentum after the collision:

m1v1 + m2v2 = m1v1' + m2v2'

where m1 and m2 are the masses of the balls, v1 and v2 are their velocities before the collision, and v1' and v2' are their velocities after the collision.

Using the values given in the problem, we can substitute:

(0.02 kg)(0.25 m/s) + (0.04 kg)(-0.15 m/s) = (0.02 kg)(0 m/s) + (0.04 kg)(v2')

Simplifying and solving for v2', we get:

v2' = [(0.02 kg)(0.25 m/s) + (0.04 kg)(-0.15 m/s)] / (0.04 kg) = -0.175 m/s

Therefore, the velocity of the 0.04 kg ball after the collision is -0.175 m/s, which means it is still moving to the left but at a slower speed than before the collision.