Question

PRACTICE IT Use the worked example above to help you solve this problem. A golf ball with mass 5.70 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 +48 m/s. Area _r_s (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 force EXERCISE HINTS: GETTING STARTED I M STUCK A 0.158 kg baseball, thrown with a speed of 39.7 m/s, is hit straight back at the pitcher with a speed of 48.1 m/s. (a) What is the magnitude of the impulse delivered by the bat to the baseball? the change in momentum, Since all other forces are negligible compared to the

Answer 1

1)

a)

I = Δp = pf - pi

I = (5.70 x 10^-2 kg)(48 m/s) - 0

I = +2.736 kg.m/s

b)

ΔT = Δx/ΔVaverage

ΔT = 0.02m/24 m/s

ΔT = 8.33 x 10^-4 s

Faverage = Δp/Δt

Faverage = (2.736 kg.m/s^2) / (8.33 x 10^-4)

Faverage= +3.2845 x 10^3 N

2)

I = Δp = pf - pi

I = (0.158 kg)[48.1 m/s - 39.7 m/s]

I = +1.3272 kg.m/s