Question

A golf club exerts and average force of 1000N on a 0.045-kg golf ball which is initially at rest. The club is in contact with the ball for 1.8ms. What is the speed of the golf ball it leaves the tee?
A batter applies an average force if 8000N to a baseball for 1.10ms. What is the magnitude of the impulse delivered to the baseball?

Answer 1

Explanation:

Given that,

Average force exerting on the golf club, F = 1000 N

Mass of the ball, m = 0.045 kg

Initial speed of the ball, u = 0

The time of contact between the ball and the club, $t=1.8\ ms=1.8\times 10^{-3}\ s$

1. Let v is the speed of the golf ball as it leaves the tee. The product of force and time is equal to the change in its momentum as :

$Ft=m(v-u)$

$Ft=mv$

$v=\dfrac{Ft}{m}$

$v=\dfrac{1000\times 1.8\times 10^{-3}}{0.045}$

v = 40 m/s

2. Force applied by the batter, F = 8000 N

Time, $t=1.1\ ms=1.1\times 10^{-3}\ s$

Let J is the magnitude of the impulse delivered to the baseball. It is equal to the product of force and time as :

$J=F\times t$

$J=8000\ N\times 1.1\times 10^{-3}\ s$

J = 8.8 kg-m/s