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.

Answer 2

The average force acts on the ball, when the club strikes to the golf ball for 2.00 ms is 562.5 N.

What is average force?

Average force is the force applied by a body which is in the motion and with a particular velocity and for a particular time.

The average force acting on a body can be given as,

$F=m\dfrac{\Delta v}{\Delta t}$

Here, (m) is the mass of the body, $\Delta v$ is the change in speed and $\Delta t$ is the change in time.

As the Initial speed of the golf ball is 0 and the final speed of the golf balls is 25.0 m/s when a club strikes. Thus the change of speed is,

$\Delta v={25-0}\\\Delta v=25\rm m/s$

The mass of the golf ball is 0.0450 kg and the time of contact between the club and ball is 2.00 ms.

Thus, put the values in the above formula to find the average force acts on the ball as,

$F=0.045\times\dfrac{25}{2\times10^{-3}}\\F=562.5\rm N$

Thus the average force acts on the ball, when the club strikes to the ball for 2.00 ms is 562.5 N.

Learn more about the average force here;

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