# Physics

A golf ball, mass 0.048 kg, rests on a tee. It is struck by a golf club with an effective mass of 0.225 kg and a speed of 44 m/s. Assuming the collision is elastic, find the speed of the ball when it leaves the tee (remember: you can not assume the club is at rest after the collision).

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1. Because this is an elastic collision, you will need 2 equations. I'm sure you will be familiar with these:

m1v1 + m2v2 = m1v1' + m2v2'
v1 - v2 = v2' - v1'

Let's make it that...
m1 = mass of golf ball
v1 = initial velocity of golf ball
v1' = final velocity of golf ball
m2 = mass of golf club
v2 = initial velocity of golf club
v2' = final velocity of golf club

Okay, now that is settled, plug in what you know.

(0.048 kg)(0 m/s) + (0.225 kg)(44 m/s) = (0.048 kg)v1' + (0.225 kg)v2'
(0 m/s) - (44 m/s) = v2' - v1'

I'll leave the algebra to you. You should get that the velocity of the golf ball (v1') is 72.5 m/s.

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