Question

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.)

Short answer

a) The value of the angle ${\theta }_2$is,${\theta }_2=30.0°$

b) The change in the ball’s linear momentum,

$\mathrm{P}=\left(-0.571\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{j}}$.

Step-by-step solution

Understanding the given information

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,${\theta }_1=30°$.

Concept and Formula used for the given question

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.

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

(a) Calculation for the angle θ2 

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,${\theta }_1={\theta }_2=30.0°$.

(b) Calculation for the change in the ball’s linear momentum in unit-vector notation

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,

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