Example on Impulsive Cricketers

A cricket ball weighing \(\text{156}\) \(\text{g}\) is moving at \(\text{54}\) \(\text{km·hr$^{-1}$}\) towards a batsman. It is hit by the batsman back towards the bowler at \(\text{36}\) \(\text{km·hr$^{-1}$}\). Calculate

1. the ball's impulse, and

2. the average force exerted by the bat if the ball is in contact with the bat for \(\text{0.13}\) \(\text{s}\).

Step 1: Identify what information is given and what is asked for

The question explicitly gives

• the ball's mass,

• the ball's initial velocity,

• the ball's final velocity, and

• the time of contact between bat and ball

We are asked to calculate the impulse:

\[\text{Impulse}=\Delta \vec{p}=\vec{F}_{\text{net}}\Delta t\]

Since we do not have the force exerted by the bat on the ball ( \(\vec{F}_{\text{net}}\)), we have to calculate the impulse from the change in momentum of the ball. Now, since

\begin{align*} \Delta \vec{p}& = \vec{p}_{f}-\vec{p}_{i} \\& = m\vec{v}_{f}-m\vec{v}_{i},\end{align*}

we need the ball's mass, initial velocity and final velocity, which we are given.

Step 2: Convert to S.I. units

Firstly let us change units for the mass

\begin{align*} \text{1 000} \text{g}& = 1 \text{kg} \\ \text{So.} 1 \text{g}& = \frac{1}{\text{1 000}} \text{kg} \\ \therefore 156\times 1 \text{g}& = 156\times \frac{1}{\text{1 000}} \text{kg} \\& = \text{0.156} \text{kg}\end{align*}

Next we change units for the velocity

\begin{align*} 1 \text{km}·{\text{h}}^{-1}& = \frac{\text{1 000} \text{m}}{\text{3 600} \text{s}} \\ \therefore 54\times 1 \text{km}·{\text{h}}^{-1}& = 54\times \frac{\text{1 000} \text{m}}{\text{3 600} \text{s}} \\& = 15 \text{m·s$^{-1}$}\end{align*}

Similarly, \(\text{36}\) \(\text{km·hr$^{-1}$}\) = \(\text{10}\) \(\text{m·s$^{-1}$}\).

Step 3: Choose a frame of reference

Let us choose the direction from the batsman to the bowler as the positive direction. Then the initial velocity of the ball is \(\vec{v}_{i}=-\text{15}\text{ m·s$^{-1}$}\), while the final velocity of the ball is \(\vec{v}_{f}=+\text{10}\text{ m·s$^{-1}$}\).

Step 4: Calculate the momentum

Now we calculate the change in momentum,

\begin{align*} \Delta \vec{p}& = \vec{p}_{f}-\vec{p}_{i} \\& = m\vec{v}_{f}-m\vec{v}_{i} \\& = m\left(\vec{v}_{f}-\vec{v}_{i}\right) \\& = \left(\text{0.156}\right)\left(\left(+10 \right)-\left(-15 \right)\right) \\& = \text{+3.9} \\& = \text{3.9}\text{ kg·m·s$^{-1}$}~\text{in the direction from the batsman to the bowler}\end{align*}

Step 5: Determine the impulse

Finally since impulse is just the change in momentum of the ball,

\begin{align*} \text{Impulse}& = \Delta \vec{p} \\& =\text{3.9}\text{ kg·m·s$^{-1}$}~\text{in the direction from the batsman to the bowler}\end{align*}

Step 6: Determine the average force exerted by the bat

\(\text{Impulse}=\vec{F}_{net}\Delta t=\Delta \vec{p}\)

We are given \(\Delta t\) and we have calculated the impulse of the ball.

\begin{align*} \vec{F}_{net}\Delta t& = \text{Impulse} \\ \vec{F}_{net}\left(\text{0.13}\right)& = \text{+3.9} \\ \vec{F}_{net}& = \frac{\text{+3.9}}{\text{0.13}} \\& = +30 \\& = \text{30}\text{ N}~\text{in the direction from the batsman to the bowler}\end{align*}