Question

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?=0.871 ms, and the radar gun clocks the speed of the ball as ?b=161 mph after it comes off the club. The golf ball has a mass of ?b=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: N

Answer 1

We know that change in the momentum is called impulse  

I=P=P₂ -P₁

Here P₂ =m_bv_b  P₁=0

I=m_bv_b   here v_b=161mph=71.967m/s (1mph=0.447m/s), m_b=45.9g=0.0459 kg

I=0.0459*71.967

I=3.3032853 kgm/s

Average force

F _average=P/t here P=3.3032853 kgm/s, t=0.871ms=0.000871 s

F _average =3.3032853/0.000871

F _average=3792.52 N