91Ó°ÊÓ

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 $\mathrm{θ}\mathrm{is}30.0°$.
4. The duration of collision $t=10.0\mathrm{ms}=10.0×{10}^{-3}\mathrm{s}$.

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.

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

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.

$$\begin{gathered} {\stackrel{→}{\mathrm{v}}}_1={\mathrm{v}}_1\cos \mathrm{θ}\hat{\mathrm{i}}={\mathrm{v}}_1\mathrm{²õ¾\pm ²Ôθ}\hat{\mathrm{j}} \\ ∴{\stackrel{→}{\mathrm{v}}}_1=6.0\cos 30\hat{\mathrm{i}}-6.0\sin 30\hat{\mathrm{j}} \\ =5.19\hat{\mathrm{i}}-3.0\hat{\mathrm{j}} \end{gathered}$$

Similarly,

$$\begin{gathered} {\stackrel{→}{\mathrm{v}}}_2={\mathrm{v}}_2\cos \mathrm{θ}\hat{\mathrm{i}}+{\mathrm{v}}_2\mathrm{²õ¾\pm ²Ôθ}\hat{\mathrm{j}} \\ {\stackrel{→}{\mathrm{v}}}_1=6.0\cos 30\hat{\mathrm{i}}+6.0\sin 30\hat{\mathrm{j}} \\ {\stackrel{→}{\mathrm{v}}}_1=5.19\hat{\mathrm{i}}+3.0\hat{\mathrm{j}} \end{gathered}$$

Now, the impulse components can be calculated as

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

The magnitude of the average force can be calculated as

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