The figure above gives an overhead view of the path taken by a cue ball with mass m as it bounces from a rail of a pool table. The ball's initial speed is v; and the angle of impact is 01. The bounce reverses the y component of the ball's velocity but does not alter the x component. NOTE: Express your answers in terms of the given variables (m, v;, 0 1).Enclose arguments of functions in parentheses. For example, sin(2x). (a) What is 02? 02 T – 01 -

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Chapter1: Units, Trigonometry. And Vectors
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The figure above illustrates an overhead view of the path taken by a cue ball with mass \( m \) as it bounces off a rail of a pool table. The ball's initial speed is \( v_i \) and the angle of impact is \( \theta_1 \). The bounce reverses the \( y \)-component of the ball’s velocity but does not alter the \( x \)-component.

**Note:** Express your answers in terms of the given variables (\( m, v_i, \theta_1 \)). Enclose arguments of functions in parentheses. For example, \(\sin(2x)\).

**(a) What is \( \theta_2 \)?**

\[
\theta_2 = \pi - \theta_1 \quad \textcolor{red}{\text{✘ Incorrect}}
\]

**(b) What is the change in the ball’s linear momentum in unit-vector notation?** (The fact that the ball rolls is irrelevant to the problem.)

\[
\Delta \vec{p} = 0 \, \hat{\imath} + (-2 m v_i \cos(\theta_1)) \, \hat{\jmath} \quad \textcolor{green}{\text{✔ Correct}}
\]

**Explanation:**

- **Diagram:** It shows a cue ball approaching a rail at an angle \( \theta_1 \) and bouncing off at an angle \( \theta_2 \), with the direction labeled along the \( x \)-axis.
- The angle \( \theta_2 \) should be calculated correctly by considering the properties of the bounce and symmetry.
- The change in momentum tensor indicates that the \( x \)-component remains constant, whereas the \( y \)-component reverses, hence the negative sign in the momentum expression.
Transcribed Image Text:The figure above illustrates an overhead view of the path taken by a cue ball with mass \( m \) as it bounces off a rail of a pool table. The ball's initial speed is \( v_i \) and the angle of impact is \( \theta_1 \). The bounce reverses the \( y \)-component of the ball’s velocity but does not alter the \( x \)-component. **Note:** Express your answers in terms of the given variables (\( m, v_i, \theta_1 \)). Enclose arguments of functions in parentheses. For example, \(\sin(2x)\). **(a) What is \( \theta_2 \)?** \[ \theta_2 = \pi - \theta_1 \quad \textcolor{red}{\text{✘ Incorrect}} \] **(b) What is the change in the ball’s linear momentum in unit-vector notation?** (The fact that the ball rolls is irrelevant to the problem.) \[ \Delta \vec{p} = 0 \, \hat{\imath} + (-2 m v_i \cos(\theta_1)) \, \hat{\jmath} \quad \textcolor{green}{\text{✔ Correct}} \] **Explanation:** - **Diagram:** It shows a cue ball approaching a rail at an angle \( \theta_1 \) and bouncing off at an angle \( \theta_2 \), with the direction labeled along the \( x \)-axis. - The angle \( \theta_2 \) should be calculated correctly by considering the properties of the bounce and symmetry. - The change in momentum tensor indicates that the \( x \)-component remains constant, whereas the \( y \)-component reverses, hence the negative sign in the momentum expression.
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