A 0.060-kg tennis ball, moving with a speed of 5.14 m/s , has a head-on collisio

Question

A 0.060-kg tennis ball, moving with a speed of 5.14 m/s , has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.62 m/s . Assume that the collision is perfectly elastic. 1. Determine the speed of the 0.060-kg ball after the collision.
2. Determine the speed of the 0.090-kg ball after the collision. A 0.060-kg tennis ball, moving with a speed of 5.14 m/s , has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.62 m/s . Assume that the collision is perfectly elastic. 1. Determine the speed of the 0.060-kg ball after the collision.
2. Determine the speed of the 0.090-kg ball after the collision. 1. Determine the speed of the 0.060-kg ball after the collision.
2. Determine the speed of the 0.090-kg ball after the collision.

Explanation / Answer

First, conserve momentum: initial p = final p
0.060kg * 5.14m/s + 0.090kg * 3.62m/s = 0.060kg * u + 0.090kg * v
where u, v are the post-collision velocities of the first, second ball
0.6342 m/s = 0.060u + 0.090v
Next, for an elastic, head-on collision, we know (from conservation of energy), that
the relative velocity of approach = relative velocity of separation, or
v - u = (5.14 - 3.62)m/s = 1.52 m/s
v = u + 1.52 m/s plug into momentum equation
0.6342 = 0.06u + 0.090(u + 1.52) = 0.15u + 0.1368
u = 0.4974 / 0.15 = 3.316 m/s first (smaller) ball
v = u + 1.52 m/s = 4.836 m/s second (larger) ball
Both balls are still moving in the original direction

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