A cue stick strikes a stationary pool ball, with an average force of 32 N over a time of 14 ms. If the ball has mass 0.20 kg, what speed does it have just after impact?

i) Force, F = 32N

ii) Time,$t=14\mathrm{s}=14\times {10}^{-3}\mathrm{s}$

iii) Mass of the ball,$m=0.20\mathrm{kg}$

iv) Initial speed of the ball,$v_i=0\frac{\mathrm{m}}{\mathrm{s}}$

Using the formula of impulse in terms of force and time, find the impulse of the ball. Impulse is defined as the change in momentum. So, using the formula of impulse in terms of change in momentum, find the speed of the ball after the impact.

Formulae are as follow:

$$\begin{gathered} J=F\times t \\ J=P_f-P_i=m{v}_f-m{v}_i \end{gathered}$$

Here, m is mass, v is velocity, t is time, J is impulse, P is momentum and F is force.

For impulse in terms of force and time,

$$\begin{gathered} \mathrm{J}=\mathrm{F}\times \mathrm{t} \\ \mathrm{J}=32\mathrm{N}\times 14\times {10}^{-3}\mathrm{s} \\ \mathrm{J}=0.448\mathrm{Ns} \end{gathered}$$

Now, for impulse in terms of change in momentum,

$$\begin{gathered} J=P_F-P_I=m{v}_f-m{v}_i \\ J=m{v}_f-m{v}_i \\ J=m\left(v_f-v_i\right) \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} 0.448\mathrm{Ns}=\left(0.20\mathrm{kg}\right)\left({\mathrm{v}}_{\mathrm{f}}-0\frac{\mathrm{m}}{\mathrm{s}}\right) \\ \frac{0.448\mathrm{Ns}}{0.20\mathrm{kg}}={\mathrm{v}}_{\mathrm{f}} \\ {\mathrm{v}}_{\mathrm{f}}=2.24\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, thespeed of the ball after the impact is$2.24\frac{\mathrm{m}}{\mathrm{s}}$

Therefore, the speed of the ball after the impact can be found, using the formula of impulse in terms of change in momentum.