(c) A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 60° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

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Chapter1: Units, Trigonometry. And Vectors
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### Problem Statement

(c) A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 60° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

### Analysis and Solution

**Step 1: Understand the Collision and Given Data**
- Initial speed of the projectile proton: \( v_i = 500 \, \text{m/s} \)
- Target proton is initially at rest: \( v_{ti} = 0 \, \text{m/s} \)
- Angle between projectile proton's path and original direction after collision: \( 60^\circ \)

**Step 2: Conservation Laws Applicable**
- **Conservation of momentum** in the x and y directions.
- **Conservation of kinetic energy** as the collision is elastic.

**Step 3: Setup the Momentum Equations**
Let:
- \( v_{tf} \) be the final speed of the target proton.
- \( v_{pf} \) be the final speed of the projectile proton (different from initial speed).

**Momentum Conservation in the X-direction:**
\[ m \cdot v_i = m \cdot v_{tf} \cos(90^\circ) + m \cdot v_{pf} \cos(60^\circ) \]
\[ 500 \, \text{m/s} = 0 + v_{pf} \cdot \frac{1}{2} \]
\[ v_{pf} = 1000 \, \text{m/s} \rightarrow (1) \]

**Momentum Conservation in the Y-direction:**
\[ 0 = m \cdot v_{tf} + m \cdot v_{pf} \sin(60^\circ) \]
\[ 0 = v_{tf} + 1000 \cdot \frac{\sqrt{3}}{2} \]
\[ v_{tf} = - 500\cdot\sqrt{3} \rightarrow (2) \]

### Conclusion
The solved values for the final speeds are:

- **(a) The speed of the target proton**: \( v_{tf} = 500 \cdot \sqrt{3} \)
- **(b) The speed of the projectile proton
Transcribed Image Text:### Problem Statement (c) A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 60° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton? ### Analysis and Solution **Step 1: Understand the Collision and Given Data** - Initial speed of the projectile proton: \( v_i = 500 \, \text{m/s} \) - Target proton is initially at rest: \( v_{ti} = 0 \, \text{m/s} \) - Angle between projectile proton's path and original direction after collision: \( 60^\circ \) **Step 2: Conservation Laws Applicable** - **Conservation of momentum** in the x and y directions. - **Conservation of kinetic energy** as the collision is elastic. **Step 3: Setup the Momentum Equations** Let: - \( v_{tf} \) be the final speed of the target proton. - \( v_{pf} \) be the final speed of the projectile proton (different from initial speed). **Momentum Conservation in the X-direction:** \[ m \cdot v_i = m \cdot v_{tf} \cos(90^\circ) + m \cdot v_{pf} \cos(60^\circ) \] \[ 500 \, \text{m/s} = 0 + v_{pf} \cdot \frac{1}{2} \] \[ v_{pf} = 1000 \, \text{m/s} \rightarrow (1) \] **Momentum Conservation in the Y-direction:** \[ 0 = m \cdot v_{tf} + m \cdot v_{pf} \sin(60^\circ) \] \[ 0 = v_{tf} + 1000 \cdot \frac{\sqrt{3}}{2} \] \[ v_{tf} = - 500\cdot\sqrt{3} \rightarrow (2) \] ### Conclusion The solved values for the final speeds are: - **(a) The speed of the target proton**: \( v_{tf} = 500 \cdot \sqrt{3} \) - **(b) The speed of the projectile proton
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