(a) Explain why, a cricket player moves his hands backwards while catching a fast cricket ball
(b) A 150 g ball, travelling at 30 m/s, strikes the palm of a player's hand and is stopped in 0.05 second. Find the force exerted by the ball on the hand.

Answer

### Solution

**(a)** A cricket player moves his hands backward while catching a fast cricket ball to prevent injury. This action is based on the principle of momentum and the relationship between force, momentum, and time.

1. **Understanding Momentum**: Momentum (p) is defined as the product of mass (m) and velocity (v). When a fast-moving ball strikes the player's hand, it has a certain momentum.

\[ p = m \times v \]

2. **Change in Momentum**: When the ball is caught, its momentum changes from its initial value to zero (since it comes to a stop). The change in momentum (Δp) can be expressed as:

\[ \Delta p = p_{final} - p_{initial} = 0 - (m \times v) = - (m \times v) \]

3. **Force and Time Relation**: The force (F) exerted on the hand can be calculated using the formula:

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

where Δt is the time taken to stop the ball.

4. **Increasing Time**: By moving his hands backward, the player increases the time (Δt) over which the ball's momentum is brought to zero. According to the impulse-momentum theorem, if the time taken to stop the ball increases, the force exerted on the hand decreases.

5. **Conclusion**: Thus, moving the hands backward reduces the force experienced by the player's hand, minimizing the risk of injury.

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**(b)** To find the force exerted by the ball on the player's hand, we can use the following steps:

1. **Convert Mass to Kilograms**: The mass of the ball is given as 150 grams. We need to convert this to kilograms.

\[ m = 150 \, \text{g} = \frac{150}{1000} \, \text{kg} = 0.15 \, \text{kg} \]

2. **Calculate Initial Momentum**: The initial velocity of the ball is 30 m/s. The initial momentum (p_initial) is:

\[ p_{initial} = m \times v = 0.15 \, \text{kg} \times 30 \, \text{m/s} = 4.5 \, \text{kg m/s} \]

3. **Final Momentum**: When the ball stops, its final momentum (p_final) is:

\[ p_{final} = 0 \, \text{kg m/s} \]

4. **Change in Momentum**: The change in momentum (Δp) is:

\[ \Delta p = p_{final} - p_{initial} = 0 - 4.5 = -4.5 \, \text{kg m/s} \]

5. **Time Taken to Stop**: The time taken to stop the ball is given as 0.05 seconds.

6. **Calculate Force**: Now, we can calculate the force (F) exerted by the ball on the hand using the formula:

\[ F = \frac{\Delta p}{\Delta t} = \frac{-4.5 \, \text{kg m/s}}{0.05 \, \text{s}} = -90 \, \text{N} \]

The negative sign indicates that the force exerted by the ball is in the opposite direction to the motion of the ball.

7. **Final Answer**: The magnitude of the force exerted by the ball on the player's hand is:

\[ F = 90 \, \text{N} \]

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