Figure 9-47 gives an overhead view of the path taken by a 0.165 kgcue ball as it bounces from a rail of a pool table. The ball’s initial speed is 2.00 m/s, and the angle ${\mathit{\theta }}_{\mathbf{1}}$is$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$. The bounce reverses the y component of the ball’s velocity but does not alter the x component. What are (a) angle ${\mathit{\theta }}_{\mathbf{2}}$and (b) the change in the ball’s linear momentum in unit-vector notation? (The fact that the ball rolls is irrelevant to the problem.)

The mass of a cue ball,m=0.165 kg .

The initial velocity of a ball,$v_1=2.00\mathrm{m}/\mathrm{s}$ .

The angle given is,${\theta }_1=30°$.

Here, we need to consider the vector nature of velocity. So, we need to calculate the components of velocity before using the equation of momentum by using the formula given below.

$$\begin{gathered} \mathrm{P}=\mathrm{mv} \\ ∆\mathrm{P}=\mathrm{m}\left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \end{gathered}$$

As velocity is a vector quantity, we consider the components of velocity.

So, the initial velocity components are:

$$\begin{gathered} {\mathrm{v}}_{1\mathrm{x}}={\mathrm{v}}_1\sin \left(30°\right) \\ =\left(2.00\mathrm{m}/\mathrm{s}\right)\times \frac{1}{2} \\ =1.00\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{1\mathrm{y}}={\mathrm{v}}_1\cos \left(30°\right) \\ =\left(2.00\mathrm{m}/\mathrm{s}\right)\times \frac{\sqrt{3}}{2} \\ =1.73\mathrm{m}/\mathrm{s} \end{gathered}$$

We have given that; the x component of velocity is conserved.

So,$v_{1x}=v_{2x}$and hence

$v_{1}sin{\theta }_1=v_{1}sin{\theta }_2$

This implies that,${\theta }_1={\theta }_2=30.0°$.

As the velocity along the x-axis is conserved, the momentum along x is also conserved.

So, we get,$P_{x=0}$

So, we need to consider the change in momentum only in the y direction.

Also, it is given that the bounce reverses the y component of velocity. So,

$$\begin{gathered} {\mathrm{v}}_{2\mathrm{y}}=-1.73\mathrm{m}/\mathrm{s} \\ {\mathrm{P}}_{\mathrm{y}}=\mathrm{m}\times \left({\mathrm{v}}_{2\mathrm{y}}-{\mathrm{v}}_{1\mathrm{y}}\right) \\ =0.165\mathrm{kg}\times \left(-1.73\mathrm{m}/\mathrm{s}-1.73\mathrm{m}/\mathrm{s}\right) \\ =-0.572\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the change in momentum can be expressed in component form as,

$\mathrm{P}=(-0.572\hspace{0.33em}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s})\stackrel{⏜}{\mathrm{j}}$