A ball of mass 3 kg collides with a wall with velocity 10 m/sec at an angle of 30° and after collision reflects at the same angle with the same speed. The change in momentum of ball in MKS unit is-

Options

A. 20
B. 30
C. 15
D. 45

Answer

To solve the problem of the change in momentum of a ball colliding with a wall, we can follow these steps:

### Step 1: Understand the Initial Conditions
- The mass of the ball (m) = 3 kg
- The initial velocity (V) = 10 m/s
- The angle of incidence (θ) = 30°

### Step 2: Resolve the Initial Velocity into Components
The initial velocity can be resolved into two components:
- The horizontal component (Vx) = V * cos(θ)
- The vertical component (Vy) = V * sin(θ)

Calculating these components:
- Vx = 10 * cos(30°) = 10 * (√3/2) = 5√3 m/s
- Vy = 10 * sin(30°) = 10 * (1/2) = 5 m/s

### Step 3: Determine the Initial Momentum
The initial momentum (P_initial) can be expressed as a vector:
\[ P_{\text{initial}} = m \cdot V = m \cdot (V_x \hat{i} + V_y \hat{j}) \]
\[ P_{\text{initial}} = 3 \cdot (5\sqrt{3} \hat{i} + 5 \hat{j}) \]
\[ P_{\text{initial}} = 15\sqrt{3} \hat{i} + 15 \hat{j} \, \text{kg m/s} \]

### Step 4: Determine the Final Velocity After Collision
After the collision, the ball reflects off the wall at the same angle and speed. The horizontal component of the velocity will change direction (becomes negative), while the vertical component remains the same:
- Final horizontal component (Vx') = -V * cos(30°) = -5√3 m/s
- Final vertical component (Vy') = V * sin(30°) = 5 m/s

### Step 5: Determine the Final Momentum
The final momentum (P_final) can be expressed as:
\[ P_{\text{final}} = m \cdot V' = m \cdot (V_x' \hat{i} + V_y' \hat{j}) \]
\[ P_{\text{final}} = 3 \cdot (-5\sqrt{3} \hat{i} + 5 \hat{j}) \]
\[ P_{\text{final}} = -15\sqrt{3} \hat{i} + 15 \hat{j} \, \text{kg m/s} \]

### Step 6: Calculate the Change in Momentum
The change in momentum (ΔP) is given by:
\[ \Delta P = P_{\text{final}} - P_{\text{initial}} \]
\[ \Delta P = (-15\sqrt{3} \hat{i} + 15 \hat{j}) - (15\sqrt{3} \hat{i} + 15 \hat{j}) \]
\[ \Delta P = (-15\sqrt{3} - 15\sqrt{3}) \hat{i} + (15 - 15) \hat{j} \]
\[ \Delta P = -30\sqrt{3} \hat{i} + 0 \hat{j} \]

### Step 7: Calculate the Magnitude of Change in Momentum
The magnitude of the change in momentum is:
\[ |\Delta P| = \sqrt{(-30\sqrt{3})^2 + 0^2} = 30\sqrt{3} \, \text{kg m/s} \]

### Step 8: Final Answer
Thus, the change in momentum of the ball is:
\[ |\Delta P| = 30\sqrt{3} \, \text{kg m/s} \]