A ball of mass 0.2 kg travelling in a straight line with a speed of `m//s` along negative x- axis id deflected by a bat `15 m//s` along negative x - axis is deflected by a bat at an angle of `30^(@)` If the speed of the ball after deflection is `10 m//s`, find the impulse on the ball.

Options

A. `4000 N`
B. `5000 N`
C. `3000 N`
D. `2500 N`

Explanation

To find the impulse on the ball, we will follow these steps:

### Step 1: Understand the initial and final conditions
The ball has an initial mass \( m = 0.2 \, \text{kg} \) and is traveling along the negative x-axis with a speed of \( 15 \, \text{m/s} \). After being deflected by the bat at an angle of \( 30^\circ \), the speed of the ball becomes \( 10 \, \text{m/s} \).

### Step 2: Calculate the initial momentum
The initial momentum \( p_i \) of the ball can be calculated using the formula:
\[ p_i = m \cdot v_i \]
where \( v_i = -15 \, \text{m/s} \) (negative because it is along the negative x-axis).

\[ p_i = 0.2 \, \text{kg} \cdot (-15 \, \text{m/s}) = -3 \, \text{kg m/s} \]

### Step 3: Calculate the final momentum
After deflection, we need to resolve the final velocity into its components. The final speed is \( 10 \, \text{m/s} \) at an angle of \( 30^\circ \).

The components of the final velocity \( v_f \) are:
- \( v_{fx} = 10 \cos(30^\circ) \)
- \( v_{fy} = 10 \sin(30^\circ) \)

Calculating these:
\[ v_{fx} = 10 \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]
\[ v_{fy} = 10 \cdot \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]

Now, calculate the final momentum \( p_f \):
\[ p_f = m \cdot v_f = m \cdot (v_{fx}, v_{fy}) = (0.2 \cdot v_{fx}, 0.2 \cdot v_{fy}) = (0.2 \cdot 5\sqrt{3}, 0.2 \cdot 5) \]
\[ p_f = (1\sqrt{3}, 1) \, \text{kg m/s} \]

### Step 4: Calculate the impulse
Impulse \( J \) is defined as the change in momentum:
\[ J = p_f - p_i \]
Calculating the x and y components:
- For the x-component:
\[ J_x = 1\sqrt{3} - (-3) = 1\sqrt{3} + 3 \]
- For the y-component:
\[ J_y = 1 - 0 = 1 \]

### Step 5: Calculate the magnitude of the impulse
The magnitude of the impulse can be calculated using the Pythagorean theorem:
\[ |J| = \sqrt{J_x^2 + J_y^2} \]
Substituting the values:
\[ |J| = \sqrt{(1\sqrt{3} + 3)^2 + 1^2} \]

Calculating \( (1\sqrt{3} + 3)^2 \):
\[ = (1.732 + 3)^2 = (4.732)^2 \approx 22.3 \]
Thus,
\[ |J| = \sqrt{22.3 + 1} = \sqrt{23.3} \approx 4.83 \, \text{N·s} \]

### Final Answer
The impulse on the ball is approximately \( 4.83 \, \text{N·s} \).

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