A cricket ball of mass `160g` is moving horizontally directly towards a batsman. Its speed just before it hits the bat is `30m//s`. It leaves the bat at `40 m//s` at `90^(@)` to its original direction. Find the magnitude of the impulse (in N`-`s) imparted to the ball.

Options

A. `4.8`
B. `8`
C. `6.4`
D. `10`

Explanation

To solve the problem of finding the magnitude of the impulse imparted to the cricket ball, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass of the Ball:**
The mass of the cricket ball is given as \( m = 160 \, \text{g} \).
Convert this to kilograms:
\[ m = \frac{160}{1000} = 0.16 \, \text{kg} \]

2. **Identify the Initial and Final Velocities:**
The initial velocity of the ball before it hits the bat is \( v_i = 30 \, \text{m/s} \) (horizontally).
The final velocity after it leaves the bat is \( v_f = 40 \, \text{m/s} \) (perpendicular to the original direction).

3. **Calculate the Initial Momentum:**
The initial momentum \( p_i \) is given by:
\[ p_i = m \cdot v_i = 0.16 \, \text{kg} \cdot 30 \, \text{m/s} = 4.8 \, \text{kg m/s} \]

4. **Calculate the Final Momentum:**
The final momentum \( p_f \) is given by:
\[ p_f = m \cdot v_f = 0.16 \, \text{kg} \cdot 40 \, \text{m/s} = 6.4 \, \text{kg m/s} \]

5. **Determine the Change in Momentum:**
The change in momentum \( \Delta p \) is given by:
\[ \Delta p = p_f - p_i \]
Since the initial and final velocities are perpendicular to each other, we can treat them as vectors and use the Pythagorean theorem to find the magnitude of the change in momentum:
\[ \Delta p = \sqrt{(p_f)^2 + (p_i)^2} = \sqrt{(6.4)^2 + (4.8)^2} \]
Calculate:
\[ \Delta p = \sqrt{40.96 + 23.04} = \sqrt{64} = 8 \, \text{kg m/s} \]

6. **Impulse Imparted to the Ball:**
The impulse \( J \) imparted to the ball is equal to the change in momentum:
\[ J = \Delta p = 8 \, \text{N s} \]

### Final Answer:
The magnitude of the impulse imparted to the ball is \( 8 \, \text{N s} \).