physics question

Question

A high-speed photograph of a club hitting a golf ball is shown in the figure below. The club was in contact with a ball, initially at rest, for about 0.0029 s. If the ball has a mass of 55 g and leaves the head of the club with a speed of 3.3 102 ft/s, find the average force exerted on the ball by the club.

Answer 1

here we would use newton's 2nd law in the following form:

F=m(v2-v1)/(t2-t1) 

where we know v1=0 and t2-t1= .0029 

then we just plug in using a calculator

Answer 2

The force is the change in momentum divided by the time interval.  Sine the ball started from rest,

 

Δp = m(v_f - v_i)

 

m = 55 g = 0.055 kg

 

v_i = 0  [started from rest]

 

v_f = 3.3 x 10² ft/s = (3.3 x 10² ft/s)(0.3048 m/ft) = 100.58 m/s

 

Δp = (0.055 kg)(100.58 m/s - 0) = 5.532 kg·m/s

 

t = 0.0029 s

 

F = Δp/t = (5.532/0.0029) N ≅ 1908 N