Question

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 dimensional. note that ( v_{i}=0 ), and calculate the change in momentum, which equals the impulse. ( i=delta p=p_{f}-p_{i}=(5.0 \times 10^{-2} mathrm{~kg})(44 mathrm{~m} / mathrm{s})-0=+2.2 mathrm{~kg} cdot mathrm{m} / mathrm{s} ) (b) estimate the duration of the collision and the average force acting on the ball. estimate the time interval of the collision, ( delta t ), using the approximate displacement (radius of the ball) and its average speed (half the maximum speed). ( delta t=\frac{delta x}{v_{mathrm{av}}}=\frac{2.0 \times 10^{-2} mathrm{~m}}{22 mathrm{~m} / mathrm{s}}=9.1 \times 10^{-4} mathrm{~s} ) estimate the average force. ( f_{mathrm{av}}=\frac{delta p}{delta t}=\frac{2.2 mathrm{~kg} cdot mathrm{m} / mathrm{s}}{9.1 \times 10^{-4} mathrm{~s}}=+2.4 \times 10^{3} mathrm{~n} ) 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 bodys 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 isnt slowed at the critical instant of impact. question what average club speed would double the average force? m/s practice it

Explanation:

Step1: Analyze the relationship between force, impulse and time

We know that \(F_{av}=\frac{\Delta p}{\Delta t}\), and \(\Delta p = m\Delta v\) (where \(m\) is the mass of the ball, \(\Delta v\) is the change in velocity). Also, \(\Delta t=\frac{\Delta x}{v_{av}}\) (where \(\Delta x\) is the displacement, \(v_{av}\) is the average velocity).

Let the initial average velocity be \(v_{av1}\), the new average velocity be \(v_{av2}\). The mass \(m = 5.0\times10^{- 2}\text{kg}\), \(\Delta x=2.0\times10^{-2}\text{m}\)

The initial impulse \(I=\Delta p=(5.0\times10^{-2}\text{kg})\times44\text{m/s}\), initial \(\Delta t_1=\frac{\Delta x}{v_{av1}}\), \(F_{av1}=\frac{\Delta p}{\Delta t_1}=\frac{\Delta p\cdot v_{av1}}{\Delta x}\)

If \(F_{av2} = 2F_{av1}\), since \(F_{av}=\frac{\Delta p\cdot v_{av}}{\Delta x}\) ( \(\Delta p = m\Delta v\), and assuming \(\Delta v\) is proportional to \(v_{club}\), and \(v_{av}=\frac{v_{club}}{2}\) in the original formula for \(\Delta t\)), and \(\Delta p\) and \(\Delta x\) are constant (mass of ball and radius of ball don't change)

We have \(F_{av}\propto v_{av}\) (because \(F_{av}=\frac{\Delta p\cdot v_{av}}{\Delta x}\), \(\Delta p\) and \(\Delta x\) are fixed)

Step2: Calculate the new average velocity

Since \(F_{av}\propto v_{av}\), if \(F_{av2} = 2F_{av1}\), then \(v_{av2}=2v_{av1}\)

The original \(v_{av1}=\frac{44\text{m/s}}{2}=22\text{m/s}\) (because average velocity \(v_{av}=\frac{v_{final}}{2}\) in the original problem's context for calculating \(\Delta t\)), so \(v_{av2} = 44\text{m/s}\)

Since \(v_{av}=\frac{v_{club}}{2}\), then \(v_{club}=2v_{av}\)

Answer:

\(88\)