EXAMPLE 6.1 Teeing Off GOAL Use the impulse-momentum theorem to estimate the average force exerted during an impact. Ted Kinsman/Photo Researchers, Inc. During impact, the club head momentarily flattens the side of the golf ball. PROBLEM A golf ball with mass 5.0 x 10"- kg is struck with a club as in the figure. The force on the ball varies from zero when contact is made up to some maximum value (when the ball is maximally deformed) and then back to zero when the ball leaves the club, as in a graph of force vs. time. Assume that the ball leaves the club face with a velocity of +44 m/s. (a) Find the magnitude of the impulse due to the collision. (b) Estimate the duration of the collision and the average force acting on the ball. STRATEGY In part (a), use the fact that the impulse is equal to the change in momentum. The mass and the initial and final speeds are known, so this change can be computed. In part (b), the average force is just the change in momentum computed in part (a) divided by an estimate of the duration of the collision. Estimate the distance the ball travels on the face of the club (about 2.0 cm, roughly the same as the radius of the ball). Divide this distance by the average velocity (half the final velocity) to get an estimate of the time of contact. SOLUTION (A) Find the impulse delivered to the ball. The problem is essentially one I = Ap = Pf - P; = (5.0 x 10-2 kg)(44 m/s) - 0 dimensional. Note that v, = 0, and = +2.2 kg . m/s calculate the change in momentum, which equals the impulse. (B) Estimate the duration of the collision and the average force acting on the ball. Estimate the time interval of the At = - Ax 2.0 x 10- m _ 1 = 9.1 x 10 4 s collision, At, using the approximate Vav 22 m/s displacement (radius of the ball) andEstimate the average force. Ap 2.2 kg . m/s = +2.4 x 10 N At 9.1 x 104 5 LEARN MORE REMARKS This estimate shows just how large such contact forces can be. A good golfer achieves maximum momentum transfer by shifting weight from the back foot to the front foot, transmitting the body's momentum through the shaft and head of the club. This timing, involving a short movement of the hips, is more effective than a shot powered exclusively by the arms and shoulders. Following through with the swing ensures that the motion isn't slowed at the critical instant of impact. QUESTION What average club speed would double the average force? m/s PRACTICE IT Use the worked example above to help you solve this problem. A golf ball with mass 4.60 x 10 2 kg is struck with a club as shown in the figure above. The force on the ball varies from zero when contact is made up to some maximum value (when the ball is maximally deformed) and then back to zero when the ball leaves the club, as in the graph of force vs. time in the figure below. Assume that the ball leaves the club face with a velocity of +49 m/s. Fav Area = Fay At if A (a) Find the magnitude of the impulse due to the collision. kg . m/s (b) Estimate the duration of the collision and the average force acting on the ball. (Assume the distance the ball travels on the face of the club is 2 cm, roughly the same as the radius of the ball.) duration average forceEXERCISE HINTS: GETTING STARTED | I'M STUCK! A 0.188 kg baseball, thrown with a speed of 38.8 m/s, is hit straight back at the pitcher with a speed of 49.8 m/s. (a ) What is the magnitude of the impulse delivered by the bat to the baseball? kg . m/s (b) Find the magnitude of the average force exerted by the bat on the ball if the two are in contact for 2.06 x 10 3 s. KN