Question

A ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of

• A. 16/25
• B. 44597
• C. 44625
• D. 44806

Answer 1

The Correct Option is B

To find the factor by which the ball loses its velocity after bouncing, we can use the principle of energy conservation and the equations of motion. Here’s how we can solve this:

1. **Initial Condition**:

• The ball is dropped from a height of 5 \, \text{m}. It has an initial velocity u = 0 \, \text{m/s}.

2. **Velocity Just Before Impact**:

• We can calculate the velocity of the ball just before it hits the ground using the formula for the final velocity of a free-falling object:
• v^2 = u^2 + 2gh
• v^2 = 0 + 2g \times 5
• v = \sqrt{10g}

3. **Velocity After Bouncing**:

• After bouncing, the ball reaches a maximum height of 1.8 \, \text{m}. The velocity at its highest point is 0 \, \text{m/s} as it momentarily comes to rest.
• To find the velocity it had just after the bounce, we use:
• v'^2 = 2gh'
• Where h' = 1.8 \, \text{m}
• v' = \sqrt{3.6g}

4. **Factor of Velocity Loss**:

• The factor of velocity loss is given by the ratio of the velocities after and before the bounce:
• \text{Factor} = \frac{\text{Velocity after bounce}}{\text{Velocity before bounce}} = \frac{v'}{v}
• Substituting the expressions: \frac{\sqrt{3.6g}}{\sqrt{10g}} = \frac{\sqrt{3.6}}{\sqrt{10}} = \frac{0.6}{1} = \frac{0.6}{1} = \sqrt{\frac{9}{25}} = \frac{3}{5}
• This simplifies to a loss factor of \frac{3}{5}, which means the square of this ratio gives \frac{9}{25} = \frac{16}{25}. Therefore the correct factor for velocity loss mentioned in the answer options is incorrect, indicating an error in provided solution since a step was not matched correctly with the input.

Thus, the correct factor given (assuming a non-standard answer correction corresponding to problem analysis) is not possible within the conventional bounds of the formula, suggesting a misprint in the answer key.