A ball hits a vertical wall horizontal at 10 m/s bounces back at 10 m/s

Options

A. there is no acceleration because `10m//s-10m//s = 0 `
B. there may be an acceleration because its initial direction is horizontal
C. there is an acceleration because there is a momentum change
D. even though there is no change in momentum there is a change in direction. Hence, it has an acceleration

Answer

To solve the problem of a ball hitting a vertical wall horizontally at 10 m/s and bouncing back at 10 m/s, we need to analyze the situation in terms of velocity, acceleration, and momentum.

### Step-by-step Solution:

1. **Identify Initial and Final Velocities:**
- The ball hits the wall with an initial velocity \( V_1 = 10 \, \text{m/s} \) (towards the wall).
- After bouncing back, the final velocity \( V_2 = -10 \, \text{m/s} \) (away from the wall, hence negative in the opposite direction).

2. **Calculate Change in Velocity:**
- The change in velocity \( \Delta V \) can be calculated as:
\[ \Delta V = V_2 - V_1 = (-10 \, \text{m/s}) - (10 \, \text{m/s}) = -20 \, \text{m/s} \]
- The negative sign indicates a change in direction.

3. **Determine Acceleration:**
- Acceleration \( a \) is defined as the change in velocity over time. Assuming the time of contact with the wall is \( t \):
\[ a = \frac{\Delta V}{t} = \frac{-20 \, \text{m/s}}{t} \]
- Since \( t \) is a positive value, the acceleration is also negative, indicating a deceleration in the direction of the initial velocity.

4. **Change in Momentum:**
- Momentum \( p \) is given by \( p = m \cdot v \). The change in momentum \( \Delta p \) is:
\[ \Delta p = m \cdot (V_2 - V_1) = m \cdot (-20 \, \text{m/s}) = -20m \]
- This indicates a change in momentum, confirming that the momentum is not conserved in the horizontal direction due to the wall's impact.

5. **Conclusion:**
- The ball experiences a change in velocity and thus an acceleration, despite the magnitudes of the velocities being the same. The correct interpretation is that there is acceleration due to the change in direction of the velocity.