Question

The club was in contact with a ball, initially at rest, for about 0.0045 s. If the ball has a mass of 55 g and leaves the head of the club with a speed of 1.4 ✕ 102 ft/s, find the average force (in kN) exerted on the ball by the club

Answer 1

Answer:

0.38 kN

Explanation:

From newton's second law of motion,

F = m(v-u)/t........................ Equation 1

Where F = Average force exerted on the ball, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact.

Given: m = 55 g = 55/1000 = 0.055 kg, v = 102 ft/s = (102×0.3048) m/s = 31.0896 m/s, u = 0 m/s (initially at rest), t = 0.0045 s

Substitute into equation 1

F = 0.055(31.0896-0)/0.0045

F = 1.709928/0.0045

F = 379.984 N

F ≈ 0.38 kN

Answer 2

Answer:

The average force exerted on the ball is 0.522 kN

Explanation:

Given:

t = 0.0045 s

m = 55 g = 0.055 kg

vi = 0

v = 1.4x10²ft/s = 42.672 m/s

The average force is equal to:

$F_{average} =\frac{m(v-v_{i}) }{t} =\frac{0.055*(42.672-0)}{0.0045} =521.55N=0.522kN$