Question

High-speed stroboscopic photographs show the head of a 220g golf club is traveling at 49.9 m/s just before it struck a a 45.8g gold ball at rest on a tee. After the collision, the club head travels (in the same direction) at 33.9 m/s. Find the speed of the golf ball immediately after impact.

Please show your work and use this formula: m1v1+m2v2=m1v1f+m2v2f

Answer 1

76.9 m/s

Step-by-step explanation:

You want the speed of a 45.8 g golf ball after it has been hit by a 220 g club traveling 49.9 m/s, if the club is traveling 33.9 m/s in the same direction after the collision.

Momentum

The conservation of momentum requires ...

m1·v1 +m2·v2 = m1·v1f +m2·v2f

Rearranging this equation, we get ...

m1·(v1 -v1f) = m2·(v2f -v2)

When v2 = 0, we can solve for v2f:

m1(v1 -v1f)/m2 = v2f . . . . . . . divide by m2

Application

In the given scenario, we have ...

• m1 = 220 g
• m2 = 45.8 g
• v1 = 49.9 m/s
• v1f = 33.9 m/s

The speed of the driven golf ball will be ...

v2f = (220)(49.9 -33.9)/45.8 ≈ 76.9 . . . m/s

The speed of the golf ball after impact will be 76.9 m/s.