A golf club exerts an average horizontal force of 1000 N on a 0.045-kg golf ball that is initially at rest on the tee.The club is in contact with the ball for 1.8 ms.What is the speed of the golf ball just as it leaves the tee?
A)30 m/s
B)35 m/s
C)40 m/s
D)45 m/s
E)50 m/s

Question

Quizplus · Accepted Answer

The impulse imparted to the ball is equal to the change in momentum. Impulse = Force x time = 1000 N x 0.0018 s = 1.8 Ns. The change in momentum is equal to the mass x change in velocity. Therefore, 1.8 Ns = 0.045 kg x v. Solving for v gives v = 40 m/s.

A Golf Club Exerts an Average Horizontal Force of 1000