Question 8: Average Velocity of an Oscillating Plate
1. Understanding the Motion Cycles
The problem describes a repetitive process. We assume the upward direction is positive ($+\hat{j}$).
Key Parameters:
- Initial ball velocity: $u$ (upwards)
- Plate speed: $v_P$ (constant magnitude)
- Condition: The plate reverses direction immediately after every collision.
2. Analyzing the Collision Intervals
Cycle 1: Plate moves UP ($+v_P$)
The relative acceleration is $g$ (downward). The time of flight $t_1$ is found using the relative velocity of separation $u_{rel} = u – v_P$.
$$ t_1 = \frac{2(u – v_P)}{g} $$
Plate displacement during Cycle 1:
$$ \Delta y_1 = v_P t_1 = \frac{2v_P(u – v_P)}{g} $$
Derivation: Rebound Velocity Reset
We need to prove that after the first collision, the ball’s velocity relative to the ground is $u$ (upwards).
Step A: Velocity of ball just before collision ($v_{b1}$)
Using the equation of motion $v = u + at$ for the ball:
$$ v_{b1} = u – g t_1 $$
Substituting $t_1 = \frac{2(u – v_P)}{g}$:
$$ v_{b1} = u – g \left[ \frac{2(u – v_P)}{g} \right] $$
$$ v_{b1} = u – 2(u – v_P) = u – 2u + 2v_P $$
$$ v_{b1} = 2v_P – u $$
Step B: Elastic Collision with the Plate
For an elastic collision with a massive wall (the plate), the velocity of separation equals the velocity of approach.
$$ v_{sep} = v_{app} $$
$$ v_{ball, after} – v_{plate} = -(v_{ball, before} – v_{plate}) $$
At the instant of collision, the plate is still moving upwards ($v_{plate} = v_P$).
$$ v_{ball, after} – v_P = -[ (2v_P – u) – v_P ] $$
$$ v_{ball, after} – v_P = -[ v_P – u ] $$
$$ v_{ball, after} – v_P = u – v_P $$
$$ v_{ball, after} = u $$
Conclusion: The ball leaves the plate with velocity $u$ upwards. Immediately after this interaction, the plate reverses its direction to downwards.
Cycle 2: Plate moves DOWN ($-v_P$)
Now, the ball moves up with $u$, and the plate moves down with $v_P$. The relative velocity of separation is $u_{rel} = u – (-v_P) = u + v_P$.
Time of flight for Cycle 2:
$$ t_2 = \frac{2(u + v_P)}{g} $$
Plate displacement during Cycle 2:
$$ \Delta y_2 = (-v_P) t_2 = -\frac{2v_P(u + v_P)}{g} $$
3. Calculation of Average Velocity
Total Time ($T$):
$$ T = t_1 + t_2 = \frac{2}{g} (u – v_P + u + v_P) = \frac{4u}{g} $$
Net Displacement ($\Delta Y$):
$$ \Delta Y = \Delta y_1 + \Delta y_2 = \frac{2v_P}{g}(u – v_P) – \frac{2v_P}{g}(u + v_P) $$
$$ \Delta Y = \frac{2v_P}{g} [ (u – v_P) – (u + v_P) ] = -\frac{4v_P^2}{g} $$
Average Velocity ($\langle v \rangle$):
$$ \langle v \rangle = \frac{\Delta Y}{T} = \frac{-4v_P^2 / g}{4u / g} = -\frac{v_P^2}{u} $$
Average Velocity:
$$ \frac{v_P^2}{u} \quad \text{(Downwards)} $$
1. Understanding the Motion Cycles
The problem describes a repetitive process. We assume the upward direction is positive ($+\hat{j}$).
Key Parameters:
- Initial ball velocity: $u$ (upwards)
- Plate speed: $v_P$ (constant magnitude)
- Condition: The plate reverses direction immediately after every collision.
2. Analyzing the Collision Intervals
Cycle 1: Plate moves UP ($+v_P$)
The relative acceleration is $g$ (downward). The time of flight $t_1$ is found using the relative velocity of separation $u_{rel} = u – v_P$.
$$ t_1 = \frac{2(u – v_P)}{g} $$Plate displacement during Cycle 1:
$$ \Delta y_1 = v_P t_1 = \frac{2v_P(u – v_P)}{g} $$Derivation: Rebound Velocity Reset
We need to prove that after the first collision, the ball’s velocity relative to the ground is $u$ (upwards).
Step A: Velocity of ball just before collision ($v_{b1}$)
Using the equation of motion $v = u + at$ for the ball:
Substituting $t_1 = \frac{2(u – v_P)}{g}$:
$$ v_{b1} = u – g \left[ \frac{2(u – v_P)}{g} \right] $$ $$ v_{b1} = u – 2(u – v_P) = u – 2u + 2v_P $$ $$ v_{b1} = 2v_P – u $$Step B: Elastic Collision with the Plate
For an elastic collision with a massive wall (the plate), the velocity of separation equals the velocity of approach.
At the instant of collision, the plate is still moving upwards ($v_{plate} = v_P$).
$$ v_{ball, after} – v_P = -[ (2v_P – u) – v_P ] $$ $$ v_{ball, after} – v_P = -[ v_P – u ] $$ $$ v_{ball, after} – v_P = u – v_P $$ $$ v_{ball, after} = u $$Conclusion: The ball leaves the plate with velocity $u$ upwards. Immediately after this interaction, the plate reverses its direction to downwards.
Cycle 2: Plate moves DOWN ($-v_P$)
Now, the ball moves up with $u$, and the plate moves down with $v_P$. The relative velocity of separation is $u_{rel} = u – (-v_P) = u + v_P$.
Time of flight for Cycle 2:
$$ t_2 = \frac{2(u + v_P)}{g} $$Plate displacement during Cycle 2:
$$ \Delta y_2 = (-v_P) t_2 = -\frac{2v_P(u + v_P)}{g} $$3. Calculation of Average Velocity
Total Time ($T$):
$$ T = t_1 + t_2 = \frac{2}{g} (u – v_P + u + v_P) = \frac{4u}{g} $$Net Displacement ($\Delta Y$):
$$ \Delta Y = \Delta y_1 + \Delta y_2 = \frac{2v_P}{g}(u – v_P) – \frac{2v_P}{g}(u + v_P) $$ $$ \Delta Y = \frac{2v_P}{g} [ (u – v_P) – (u + v_P) ] = -\frac{4v_P^2}{g} $$Average Velocity ($\langle v \rangle$):
$$ \langle v \rangle = \frac{\Delta Y}{T} = \frac{-4v_P^2 / g}{4u / g} = -\frac{v_P^2}{u} $$$$ \frac{v_P^2}{u} \quad \text{(Downwards)} $$