Cricket (the sport)!

Use Newton’s second law or the Impulse Equation to relate force, time of impact, and change in momentum.

The relationship between force and linear momentum is given by Newton’s Second Law in the form:

\begin{equation*}
\vec{F}=\frac{d\vec{p}}{dt},
\end{equation*}

where \(d\vec{p}\) or \(dt\) as finite changes can be written as \(\Delta \vec{p}\) and \( \Delta t \) respectively. Then

\begin{equation*}
\vec{F}=\frac{m \vec{v}_f – m \vec{v}_i}{dt}.
\end{equation*}

Plugging in numerical values, we have:

\begin{equation*}
\vec{F}\approx 44.75\,\text{N}\,\hat{\textbf{i}}+24.75\,\text{N}\,\hat{\textbf{j}}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

In order to find the force that the bat exerts on the ball, we need to find the change in the linear momentum of the ball during the 0.2 seconds that it’s in contact with the bat. Since we know the ball’s mass \(m_b\), the initial velocity, and the final velocity, we have everything that we need to find the change in linear momentum and the force \(\vec{F}\) exerted on the ball. In particular, the relation between force and linear momentum is given by Newton’s second law in the form:

\begin{equation}
\vec{F}=\frac{d\vec{p}}{dt},
\end{equation}

where \(\frac{d\vec{p}}{dt}\) is the derivative of linear momentum \(\vec{p}\) with respect to time. In the case that the force is applied in a very short time, the expression above can be approximated to

\begin{equation}
\vec{F}\approx\frac{\Delta \vec{p}}{\Delta t},
\end{equation}

where \(\Delta \vec{p}\) is the change in linear momentum and \(\Delta t\) is the interval of time in which the force is exerted. Explicitly writing the change in linear momentum, we have

\begin{equation}
\label{impulse1}
\vec{F}\approx \frac{\vec{p}_f – \vec{p}_i}{\Delta t},
\end{equation}

where \(\vec{p}_f\) is the final linear momentum and \(\vec{p}_i\) the initial linear momentum. Since the linear momentum is given by \( \vec{p}=m \vec{v} \), we then know that \eqref{impulse1} becomes

\begin{equation}
\label{impulse2}
\vec{F}= \frac{m\vec{v}_f – m\vec{v}_i}{\Delta t},
\end{equation}

where \(\vec{v}_f\) is the velocity after Emily hits the ball with the bat and \(\vec{v}_i\) is the velocity of the ball before she hits it.

Figure 1: On the left, we see the ball’s velocity before the collision with Emily’s bat. On the right, we see the velocity of the ball after the collision with Emily’s bat. Notice that after the collision the velocity has components along the X and Y axis.

The initial velocity, according the coordinate system shown in figure 1 is given by

\begin{equation}
\label{vi}
\vec{v}_i=-20\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation}

The velocity of the ball after Emily hits it can be decomposed in the X and Y axis with the use of the trigonometric functions \(\sin\) and \(\cos\), namely

\begin{equation}
\label{vf}
\vec{v}_f=(35\,\text{m/s})\cos(45^{\circ})\,\hat{\textbf{i}}+(35\,\text{m/s})\sin(45^{\circ})\,\hat{\textbf{j}}.
\end{equation}

Using the given numerical values and the explicit expression for the velocities given by equations \eqref{vi} and \eqref{vf} in equation \eqref{impulse2}, we obtain

\begin{equation}
\vec{F}\approx \frac{(0.2\,\text{kg})((35\,\text{m/s})\cos(45^{\circ})\,\hat{\textbf{i}}+(35\,\text{m/s})\sin(45^{\circ})\,\hat{\textbf{j}})-(0.2\,\text{kg})(-20\,\text{m/s}\,\hat{\textbf{i}})}{0.2\,\text{s}},
\end{equation}

which is

\begin{equation}
\vec{F}\approx 44.75\,\text{N}\,\hat{\textbf{i}}+24.75\,\text{N}\,\hat{\textbf{j}}.
\end{equation}

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