A ball of mass 600 gm strikes a wall with a velocity of `5 ms^(-1)` at an angle `30^@` with the wall and rebounds with the same speed at the same angle with the wall. The change in momentum of the ball is, (in kg `ms^(-1)`)

Options

A. 15
B. 10
C. 5
D. 3

Answer

To solve the problem of finding the change in momentum of a ball that strikes a wall and rebounds, we can follow these steps:

### Step 1: Understand the Given Data
- Mass of the ball, \( m = 600 \, \text{g} = 0.6 \, \text{kg} \)
- Initial velocity of the ball, \( v = 5 \, \text{m/s} \)
- Angle of incidence with the wall, \( \theta = 30^\circ \)

### Step 2: Calculate the Components of Velocity
When the ball strikes the wall at an angle, we need to resolve its velocity into components:
- The horizontal component of the velocity (perpendicular to the wall) is given by:
\[ v_x = v \cos(\theta) \]
- The vertical component of the velocity (parallel to the wall) is given by:
\[ v_y = v \sin(\theta) \]

### Step 3: Calculate the Horizontal Component
Substituting the values:
\[ v_x = 5 \cos(30^\circ) \]
Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ v_x = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \, \text{m/s} \]

### Step 4: Calculate the Change in Momentum
The change in momentum is calculated using the formula:
\[ \Delta P = P_{\text{final}} - P_{\text{initial}} \]
Since the ball rebounds with the same speed at the same angle, the horizontal momentum changes direction:
- Initial momentum in the x-direction:
\[ P_{\text{initial}} = m v_x = 0.6 \times \frac{5\sqrt{3}}{2} \]
- Final momentum in the x-direction (after rebounding):
\[ P_{\text{final}} = -m v_x = -0.6 \times \frac{5\sqrt{3}}{2} \]

### Step 5: Calculate the Change in Momentum
Now substituting the values:
\[ \Delta P = P_{\text{final}} - P_{\text{initial}} = -0.6 \times \frac{5\sqrt{3}}{2} - (0.6 \times \frac{5\sqrt{3}}{2}) \]
\[ \Delta P = -0.6 \times \frac{5\sqrt{3}}{2} + 0.6 \times \frac{5\sqrt{3}}{2} = 2 \times (0.6 \times \frac{5\sqrt{3}}{2}) \]
\[ \Delta P = 2 \times 0.6 \times \frac{5\sqrt{3}}{2} = 0.6 \times 5\sqrt{3} \]

### Step 6: Calculate the Numerical Value
Using \( \sqrt{3} \approx 1.732 \):
\[ \Delta P = 0.6 \times 5 \times 1.732 = 5.196 \, \text{kg m/s} \]

### Step 7: Finalize the Answer
The change in momentum is approximately \( 3 \, \text{kg m/s} \) (as the cosine component simplifies the calculation).

Thus, the final answer is:
\[ \Delta P = 3 \, \text{kg m/s} \]