Question

use the worked example above to help you solve this problem. a golf ball with mass ( 4.20\times10^{-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 ).

(a) find the magnitude of the impulse due to the collision.
( square kgcdot 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.)

exercise
hints: getting started | im stuck!
a ( 0.152 kg ) baseball, thrown with a speed of ( 39.8 m/s ), is hit straight back at the pitcher with a speed of ( 49.0 m/s ).
(a) what is the magnitude of the impulse delivered by the bat to the baseball?
( square kgcdot m/s )

(b) find the magnitude of the average force exerted by the bat on the ball if the two are in contact for ( 1.90\times10^{-3} s ).
( square kn )

Explanation:

Step1: Calculate impulse for part (a)

Impulse \(J = \Delta p=m(v - u)\). The initial velocity \(u = 0\) (ball at rest before collision), \(m = 4.20\times10^{-2}\text{ kg}\), \(v=49\text{ m/s}\).
\(J=(4.20\times 10^{-2})\times(49 - 0)\)

Step2: Calculate duration for part (b)

Use the kinematic equation \(x=\frac{u + v}{2}t\). Here \(x = 0.02\text{ m}\), \(u = 0\), \(v = 49\text{ m/s}\). So \(t=\frac{2x}{v}\)

Step3: Calculate average force for part (b)

From \(J = F_{av}\Delta t\), so \(F_{av}=\frac{J}{\Delta t}\)

Step4: Calculate impulse for exercise (a)

Impulse \(J=\Delta p=m(v - u)\). Let the initial velocity \(u = 39.8\text{ m/s}\), final velocity \(v=- 49.0\text{ m/s}\), \(m = 0.152\text{ kg}\). \(J=0.152\times(-49.0 - 39.8)\)

Step5: Calculate average force for exercise (b)

From \(J = F_{av}\Delta t\), \(\Delta t=1.90\times 10^{-3}\text{ s}\), so \(F_{av}=\frac{J}{\Delta t}\)

Answer:

• Problem (a): \(J=(4.20\times 10^{-2})\times49 = 2.06\text{ kg}\cdot\text{m/s}\)
• Problem (b):
• \(t=\frac{2\times0.02}{49}\approx8.16\times 10^{-4}\text{ s}\)
• \(F_{av}=\frac{2.06}{8.16\times 10^{-4}}\approx2520\text{ N}\)
• Exercise (a): \(J = 0.152\times(-49.0 - 39.8)=-13.5\text{ kg}\cdot\text{m/s}\), magnitude is \(13.5\text{ kg}\cdot\text{m/s}\)
• Exercise (b): \(F_{av}=\frac{13.5}{1.90\times 10^{-3}}\approx7.11\times 10^{3}\text{ N}=7.11\text{ kN}\)