In the overhead view of Figure, a 300 g ball with a speed v of 6.0 m/sstrikes a wall at an angle $\mathit{\theta }$of$\mathbf{30}\mathbf{°}$and then rebounds with the same speed and angle. It is in contact with the wall for10 ms. In unit vector notation, What are (a) the impulse on the ball from the wall and (b) The average force on the wall from the ball?

1. The mass of the ball $m=300\mathrm{gm}=0.300\mathrm{kg}$.
2. The initial and final velocity of the ball $v_1=v_2=6.0\mathrm{m}/\mathrm{s}$.
3. The angle of collision $\mathrm{\theta }\mathrm{is}30.0°$.
4. The duration of collision $t=10.0\mathrm{ms}=10.0\times {10}^{-3}\mathrm{s}$.

The collision between the ball and wall imparts an impulse on the ball. The impulse-linear momentum theorem helps you to determine the impulse and the average force on the ball and the wall and is given as follows.

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

We write the velocity vectors in unit vector notation. Then, we determine the impulse along each direction to get the magnitude and the direction of the resultant impulse vector.

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_1={\mathrm{v}}_1\cos \mathrm{\theta }\hat{\mathrm{i}}={\mathrm{v}}_1\mathrm{sin\theta }\hat{\mathrm{j}} \\ \therefore {\overrightarrow{\mathrm{v}}}_1=6.0\cos 30\hat{\mathrm{i}}-6.0\sin 30\hat{\mathrm{j}} \\ =5.19\hat{\mathrm{i}}-3.0\hat{\mathrm{j}} \end{gathered}$$

Similarly,

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_2={\mathrm{v}}_2\cos \mathrm{\theta }\hat{\mathrm{i}}+{\mathrm{v}}_2\mathrm{sin\theta }\hat{\mathrm{j}} \\ {\overrightarrow{\mathrm{v}}}_1=6.0\cos 30\hat{\mathrm{i}}+6.0\sin 30\hat{\mathrm{j}} \\ {\overrightarrow{\mathrm{v}}}_1=5.19\hat{\mathrm{i}}+3.0\hat{\mathrm{j}} \end{gathered}$$

Now, the impulse components can be calculated as

role="math" localid="1661242697207" $$\begin{gathered} J_x=m(v_{2x}-v_{1x}) \\ {J}_x=0.300\times (5.19-5.19) \\ =0 \\ and{J}_y=m(v_{2y}-v_{1y}) \\ {J}_y=0.300\times (3.0+3.0) \\ =(1.8\mathrm{N}.\mathrm{s})\hat{\mathrm{j}} \\ =1.8\mathrm{N}.\mathrm{s} \\ Now\mathit{,}\mathit{}J\mathit{=}\sqrt{{\mathit{J}}_{\mathit{x}}^{\mathit{2}}\mathit{+}{\mathit{J}}_{\mathit{y}}^{\mathit{2}}}\mathit{=}\sqrt{{\left(0\right)}^2+{(1.8)}^2} \\ \overrightarrow{\mathit{J}}\mathit{=}(1.8\mathrm{N}.\mathrm{s})\hat{\mathrm{j}} \\ =1.8N.s \end{gathered}$$

The magnitude of the average force can be calculated as

$$\begin{gathered} J=F_{avg}∆t\hat{\mathrm{j}} \\ {F}_{avg}=\frac{J}{t} \\ =\frac{1.8}{10\times {10}^{-3}} \\ =(1.8\times {10}^{2}N)\hat{\mathrm{j}} \\ =(180N)\hat{\mathrm{j}} \end{gathered}$$