A professional golfer is examining a video of a practice swing. The high‑speed footage shows that his club is in contact with the ball (which was initially at rest on the tee) for only Δ????=0.371 ms, and the radar gun clocks the speed of the ball as ????b=146 mph after it comes off the club. The golf ball has a mass of mb=45.9 g. What is the magnitude of the impulse imparted to the ball by the club? magnitude of the impulse: kg⋅ms What is the magnitude of the average force of contact between the club and the ball? magnitude of the average force of contact:

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AMB000 AMB000 · Answer 1 · 2019-10-26T19:48:18+02:00

Answer:

[tex]I=3kgm/s[/tex]

[tex]F=8086.25N[/tex]

Explanation:

The impulse imparted can be obtained using the equation [tex]I=m\Delta v[/tex]. In our case, since the ball departs from rest, we have:

[tex]I=m(v_b-v_{0b})=(45.9g)((146mph)-(0mph))=6701.4gmph[/tex]

We can convert this to S.I. using the corresponding conversion factors (which are equal to 1, thus not altering our value):

[tex]6701.4gmph=6701.4g\frac{miles}{h}(\frac{1kg}{1000g})(\frac{1609.34m}{1\ mile})(\frac{1h}{3600s})=3kgm/s[/tex]

Since also the impulse relates to the average force by the equation [tex]I=F\Delta t[/tex], we can use our values to calculate this force:

[tex]F=\frac{I}{\Delta t}=\frac{3kgm/s}{0.000371s}=8086.25N[/tex]