A ball falls from 20 m height on floor and rebounds to 5 m. Time of contact is 0.02 s. Find acceleration during impact.

Options

A. `1200 ms^(-2)`
B. `1000 ms^(-2)`
C. `2000 ms^(-2)`
D. `1500 ms^(-2)`

Explanation

To solve the problem of finding the acceleration during the impact of a ball that falls from a height of 20 m and rebounds to a height of 5 m, we can follow these steps:

### Step 1: Calculate the velocity just before impact
We can use the equation of motion to find the velocity of the ball just before it hits the ground. The equation is:

\[ v^2 = u^2 + 2gh \]

Where:
- \( v \) = final velocity just before impact
- \( u \) = initial velocity (0 m/s, since it starts from rest)
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( h \) = height (20 m)

Substituting the values:

\[ v^2 = 0 + 2 \times 10 \times 20 \]
\[ v^2 = 400 \]
\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Step 2: Calculate the velocity just after impact
Next, we calculate the velocity of the ball just after it rebounds to a height of 5 m. Again, we use the equation of motion:

\[ v^2 = u^2 - 2gh \]

Where:
- \( v \) = final velocity (0 m/s at the highest point)
- \( u \) = initial velocity just after impact
- \( g \) = 10 m/s²
- \( h \) = 5 m

Rearranging gives:

\[ 0 = u^2 - 2 \times 10 \times 5 \]
\[ u^2 = 100 \]
\[ u = \sqrt{100} = 10 \, \text{m/s} \]

### Step 3: Calculate the change in momentum
The change in momentum (\( \Delta p \)) during the impact can be calculated as:

\[ \Delta p = p_{\text{final}} - p_{\text{initial}} \]

Where:
- \( p_{\text{final}} = m \times v_{\text{final}} \) (after impact, going upwards)
- \( p_{\text{initial}} = m \times v_{\text{initial}} \) (before impact, going downwards)

Substituting the values:

- \( v_{\text{final}} = -10 \, \text{m/s} \) (upward direction is negative)
- \( v_{\text{initial}} = 20 \, \text{m/s} \)

Thus:

\[ \Delta p = m \times (-10) - m \times (20) \]
\[ \Delta p = -10m - 20m \]
\[ \Delta p = -30m \]

### Step 4: Calculate the force during impact
Using the impulse-momentum theorem, we know that:

\[ F = \frac{\Delta p}{\Delta t} \]

Where:
- \( \Delta t = 0.02 \, \text{s} \)

Substituting the values:

\[ F = \frac{-30m}{0.02} \]
\[ F = -1500m \, \text{N} \]

### Step 5: Calculate the acceleration during impact
Using Newton's second law, \( F = ma \):

\[ -1500m = ma \]

Dividing both sides by \( m \) (assuming \( m \neq 0 \)):

\[ a = -1500 \, \text{m/s}^2 \]

### Final Answer
The acceleration during impact is \( -1500 \, \text{m/s}^2 \) (the negative sign indicates that the acceleration is in the upward direction).

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