A player catches a ball of 200 gm moving with a speed of 20 m/s. if the time taken to complete the cathc is 0.5s, the force exerted on the player, hand is :-

Options

A. 8 N
B. 4N
C. 2N
D. 0N

Explanation

To solve the problem of finding the force exerted on the player's hand when catching a ball, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the mass of the ball:**
The mass of the ball is given as 200 grams. We need to convert this into kilograms for our calculations.
\[ \text{Mass} (m) = 200 \, \text{grams} = 0.2 \, \text{kg} \]

2. **Identify the initial speed of the ball:**
The speed of the ball before being caught is given as 20 m/s.
\[ \text{Initial speed} (u) = 20 \, \text{m/s} \]

3. **Determine the final speed of the ball:**
When the player catches the ball, the final speed (v) of the ball becomes 0 m/s since it comes to rest in the player's hand.
\[ \text{Final speed} (v) = 0 \, \text{m/s} \]

4. **Calculate the change in velocity:**
The change in velocity (Δv) can be calculated as:
\[ \Delta v = v - u = 0 - 20 = -20 \, \text{m/s} \]

5. **Determine the time taken to catch the ball:**
The time taken to complete the catch is given as 0.5 seconds.
\[ \text{Time} (t) = 0.5 \, \text{s} \]

6. **Calculate the acceleration:**
Acceleration (a) can be calculated using the formula:
\[ a = \frac{\Delta v}{t} = \frac{-20 \, \text{m/s}}{0.5 \, \text{s}} = -40 \, \text{m/s}^2 \]
(The negative sign indicates that the ball is decelerating.)

7. **Calculate the force exerted on the player's hand:**
Using Newton's second law of motion, \( F = m \cdot a \):
\[ F = 0.2 \, \text{kg} \times (-40 \, \text{m/s}^2) = -8 \, \text{N} \]
The negative sign indicates that the force is in the opposite direction of the ball's initial motion. However, we are interested in the magnitude of the force, which is:
\[ F = 8 \, \text{N} \]

### Final Answer:
The force exerted on the player's hand is **8 Newtons**.

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