A ball of mass `0.2kg` strikes an obstacle and moves at `60^(@)` to its original direction If its speed also changes from `20m//s` to `10m//s`, the magnitude of the impulse received by the ball is .

Options

A. `2sqrt7 Ns`
B. `2sqrt3Ns`
C. `2sqrt5Ns`
D. `3sqrt2Ns`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a ball of mass \( m = 0.2 \, \text{kg} \) that strikes an obstacle and changes its direction and speed. The initial speed is \( u = 20 \, \text{m/s} \) and the final speed is \( v = 10 \, \text{m/s} \) at an angle of \( 60^\circ \) to its original direction.

### Step 2: Calculate Initial Momentum
The initial momentum \( P_i \) of the ball can be calculated using the formula:
\[ P_i = m \cdot u \]
Substituting the values:
\[ P_i = 0.2 \, \text{kg} \cdot 20 \, \text{m/s} = 4 \, \text{kg m/s} \]

### Step 3: Calculate Final Momentum
The final momentum \( P_f \) needs to be calculated in both x and y components since the direction has changed.

#### Step 3.1: Calculate the x-component of Final Momentum
The x-component of the final velocity \( v_x \) is:
\[ v_x = v \cdot \cos(60^\circ) = 10 \, \text{m/s} \cdot \cos(60^\circ) = 10 \, \text{m/s} \cdot \frac{1}{2} = 5 \, \text{m/s} \]
Thus, the x-component of final momentum \( P_{fx} \) is:
\[ P_{fx} = m \cdot v_x = 0.2 \, \text{kg} \cdot 5 \, \text{m/s} = 1 \, \text{kg m/s} \]

#### Step 3.2: Calculate the y-component of Final Momentum
The y-component of the final velocity \( v_y \) is:
\[ v_y = v \cdot \sin(60^\circ) = 10 \, \text{m/s} \cdot \sin(60^\circ) = 10 \, \text{m/s} \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]
Thus, the y-component of final momentum \( P_{fy} \) is:
\[ P_{fy} = m \cdot v_y = 0.2 \, \text{kg} \cdot 5\sqrt{3} \, \text{m/s} = \sqrt{3} \, \text{kg m/s} \]

### Step 4: Calculate Change in Momentum
Now we can find the change in momentum in both x and y directions.

#### Step 4.1: Change in x-direction Momentum
\[ \Delta P_x = P_{fx} - P_i = 1 \, \text{kg m/s} - 4 \, \text{kg m/s} = -3 \, \text{kg m/s} \]

#### Step 4.2: Change in y-direction Momentum
Since the initial momentum in the y-direction was zero (the ball was moving horizontally):
\[ \Delta P_y = P_{fy} - 0 = \sqrt{3} \, \text{kg m/s} \]

### Step 5: Calculate the Magnitude of the Impulse
The impulse \( J \) is equal to the change in momentum:
\[ J = \sqrt{(\Delta P_x)^2 + (\Delta P_y)^2} \]
Substituting the values:
\[ J = \sqrt{(-3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \, \text{N s} \]

### Final Answer
The magnitude of the impulse received by the ball is:
\[ \boxed{2\sqrt{3} \, \text{N s}} \]