Question

A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes. If the club and ball are in contact for 2.00 ms, what average force acts on the ball? *

Answer 1

Answer:

Average force, F = 562.5 N

Explanation:

Mass of the golf ball, m = 0.045 kg

Initially, it is at rest, u = 0

Final speed of the ball, v = 25 m/s

The club and the ball are in contact for, $t=2\ ms=2\times 10^{-3}\ s$

We need to find the average force acting on the ball. It can be calculated using the formula as :

$F\times t=m(v-u)$

$F=m\dfrac{v-u}{t}$

$F=0.045\times \dfrac{25}{2\times 10^{-3}}$  

F = 562.5 N

So, the average force acting on the ball is 562.5 N. Hence, this is the required solution.