Figure shows a0.300 kgbaseball just before and just after it collides with a bat. Just before, the ball has velocity of magnitude 12.0 m/sand angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{°}$. Just after, it is traveling directly upward with velocity of magnitude $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. The duration of the collision is 2.00 ms. What is the (a) magnitude and direction (relative to the positive direction of the x axis) of the impulse on the ball from the bat? What is the (c) magnitude and (d) direction of the average force on the ball from the bat?

1. The mass of the baseball is 0.300 kg .
2. The velocity of the ball before collision,$v_1\mathrm{is}12.0\mathrm{m}/\mathrm{s}$ .
3. The angle of collision,${\theta }_1=35.0°$ .
4. The velocity f the ball after collision,${\mathrm{v}}_2\mathrm{is}10.0\mathrm{m}/\mathrm{s}$.
5. The duration of collision,$t\mathrm{is}2.00\times {10}^{-3}\mathrm{s}$ .

The collision between the ball and bat imparts an impulse on the ball as well asthebat. Using the impulse-linear momentum theorem, it can be determined the impulse and the average force on the ball and bat which is given below.

$$\begin{gathered} \overrightarrow{\mathbf{J}}\mathbf{=}\mathbf{∆}\overrightarrow{\mathbf{p}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{m}\mathbf{∆}\overrightarrow{\mathbf{v}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathbf{F}}_{\mathbf{avg}}\mathbf{∆}\mathbf{t} \end{gathered}$$

(a)

The collision between the baseball and the bat is 2-dimensional. Hence, we need to write the velocity vectors intheunit vector notation. Then, we will determine the impulse along each direction to get the magnitude and the direction of the impulse vector.

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_1={\mathrm{v}}_1\mathrm{cos\theta }\overrightarrow{\mathrm{i}}-{\mathrm{v}}_1\mathrm{sin\theta }\hat{\mathrm{j}} \\ =12\cos 35.0\hat{\mathrm{i}}-12\sin 35.0\hat{\mathrm{j}} \\ =9.83\hat{\mathrm{i}}-6.88\hat{\mathrm{j}} \end{gathered}$$

As the velocity after collision is directed upwards i.e., along +Y axis, we write

${\overrightarrow{\mathrm{v}}}_2=+{\mathrm{v}}_2\hat{\mathrm{j}}=10.0\hat{\mathrm{j}}$

Now, the impulse components can be calculated as

$$\begin{gathered} J_x=m\left(v_{2x}-v_{1x}\right) \\ =0.300\times \left(0-9.83\right) \\ =-2.95 \end{gathered}$$

$$\begin{gathered} and{J}_{y}m(v_{2y}-V_{1y}) \\ =0.300\times (10.0+6.88) \\ =5.064 \end{gathered}$$

Now,$$\begin{gathered} \mathrm{J}=\sqrt{{\mathrm{J}}_{\mathrm{x}}^2+{\mathrm{J}}_{\mathrm{y}}^2} \\ =\sqrt{{\left(-2.95\right)}^2+\left(5.064\right)} \\ =5.86\mathrm{N}.\mathrm{s} \end{gathered}$$

(b)

The direction of the impulse vector can be calculated as

$$\begin{gathered} \tan \varphi =\frac{J_y}{J_x} \\ =\frac{5.064}{-2.95} \\ \varphi ={\tan }^{-1}(-1.717) \\ =-59.8° \end{gathered}$$

The direction of impulse J is $59.8°$ measured counterclockwise from the +x axis.

(c)

The magnitude of the average force can be calculated as

$$\begin{gathered} J=F_{avg}∆t \\ {F}_{avg}=\frac{J}{t} \\ =\frac{5.86}{2\times {10}^{-3}} \\ =2.93\times {10}^3\mathrm{N} \\ =2.93\mathrm{kN} \end{gathered}$$

(d)

The average force $\overrightarrow{F_{avg}}$and the impulse $\overrightarrow{J}$are in the same direction. Hence, the angle of the average force and impulse is $59.8°$measured counterclockwise from the +x axis.