A bowler tbrows a ball horizontally along east direction with speed of `144"km"//"hr".` The batsman hits the ball such that it deviates from its initial direction of motion by `74^(@)` north of east direction, without changing its speed. If mass of the ball is `1/3 kg` and time of contact between bat and ball is 0.02 s. Average force applied by batsman on ball is:

A. 800 N, `53^(@)`East of North
B. 800N, `53^(@)` North of East
C. 800 N, `53^(@)` North of West
D. 800N, `53^(@)` West of North

Options

A. 800 N, `53^(@)`East of North
B. 800N, `53^(@)` North of East
C. 800 N, `53^(@)` North of West
D. 800N, `53^(@)` West of North

Answer

To solve the problem, we need to find the average force applied by the batsman on the ball. We will follow these steps:

### Step 1: Convert the speed of the ball from km/hr to m/s
The speed of the ball is given as 144 km/hr. To convert this to m/s, we use the conversion factor \( \frac{5}{18} \).

\[ V_1 = 144 \times \frac{5}{18} = 40 \, \text{m/s} \]

### Step 2: Determine the final velocity vector after the batsman hits the ball
The batsman hits the ball such that it deviates by \( 74^\circ \) north of east. The final velocity \( V_2 \) remains the same in magnitude as \( V_1 \), which is 40 m/s.

### Step 3: Calculate the change in velocity
We need to find the change in velocity \( \Delta V \). The initial velocity vector \( V_1 \) can be represented as:

\[ V_1 = 40 \hat{i} \, \text{(east direction)} \]

The final velocity vector \( V_2 \) can be represented using trigonometric components:

\[ V_2 = 40 \cos(74^\circ) \hat{i} + 40 \sin(74^\circ) \hat{j} \]

Calculating the components:

\[ V_2 = 40 \cos(74^\circ) \hat{i} + 40 \sin(74^\circ) \hat{j} \]

Using \( \cos(74^\circ) \approx 0.2756 \) and \( \sin(74^\circ) \approx 0.9613 \):

\[ V_2 \approx 40 \times 0.2756 \hat{i} + 40 \times 0.9613 \hat{j} \approx 11.024 \hat{i} + 38.452 \hat{j} \]

### Step 4: Calculate the change in velocity vector
The change in velocity \( \Delta V \) is given by:

\[ \Delta V = V_2 - V_1 = (11.024 \hat{i} + 38.452 \hat{j}) - (40 \hat{i}) = (-28.976 \hat{i} + 38.452 \hat{j}) \]

### Step 5: Calculate the magnitude of the change in velocity
The magnitude of the change in velocity is given by:

\[ |\Delta V| = \sqrt{(-28.976)^2 + (38.452)^2} \]

Calculating this:

\[ |\Delta V| = \sqrt{839.4 + 1474.5} = \sqrt{2313.9} \approx 48.1 \, \text{m/s} \]

### Step 6: Calculate the change in momentum
The change in momentum \( \Delta p \) is given by:

\[ \Delta p = m \cdot \Delta V \]

Given that the mass \( m = \frac{1}{3} \, \text{kg} \):

\[ \Delta p = \frac{1}{3} \cdot 48.1 \approx 16.03 \, \text{kg m/s} \]

### Step 7: Calculate the average force
The average force \( F \) is given by:

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

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

\[ F = \frac{16.03}{0.02} \approx 801.5 \, \text{N} \approx 800 \, \text{N} \]

### Step 8: Determine the direction of the force
To find the direction of the force, we can use the components of the change in velocity:

\[ \tan(\theta) = \frac{38.452}{28.976} \]

Calculating \( \theta \):

\[ \theta = \tan^{-1}\left(\frac{38.452}{28.976}\right) \approx 53^\circ \]

Thus, the direction of the average force is \( 53^\circ \) north of west.

### Final Answer
The average force applied by the batsman on the ball is approximately \( 800 \, \text{N} \) at \( 53^\circ \) north of west.

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