Question

Now it's your turn

6
A golfer hits a ball of mass 45 g at a speed of 40 m 5-1 (Figure 4.15). The golf club is
in contact with the ball for 3.0 ms. Calculate the average force exerted by the club on
the ball.

Answer 1

The average force the golf club exerts on the ball is 600 N

Step-by-step explanation:

Newton's second law of motion states that force, F, is directly proportional to the rate of change of momentum produced

F = m× (v₂ - v₁)/(Δt)

The given parameters of the motion of the ball are;

The mass of the ball, m = 45 g = 0.045 kg

The initial velocity of the ball, v₁ = 0 m/s

The speed with which the ball was hit by the golfer, v₂ = 40 m/s

The duration of contact between the golf club and the ball, Δt = 3 ms = 0.003 seconds (s)

By Newton's law of motion, the average force, 'F', which the golf club exerts on the ball is therefore, given as follows;

F = 0.045 kg × (40 m/s - 0 m/s)/(0.003 s) = 600 N

The average force the golf club exerts on the ball = F = 600 N.