A ball of mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence `45^0`. The ball is reflected by the plane at an equal angle of reflection with the same speed. Calculate (a). the magnitude of the change in momentum of the ball (b). the change in the magnitude of the mometum of the ball.

Explanation

To solve the problem step by step, we will calculate the change in momentum of the ball before and after it strikes the plane surface.

### Given Data:
- Mass of the ball, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Initial speed of the ball, \( v = 2.0 \, \text{m/s} \)
- Angle of incidence (and reflection), \( \theta = 45^\circ \)

### Step 1: Calculate the initial momentum of the ball
The initial velocity can be broken down into its components:
- \( v_{ix} = v \cdot \cos(45^\circ) = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \, \text{m/s} \)
- \( v_{iy} = -v \cdot \sin(45^\circ) = -2 \cdot \frac{1}{\sqrt{2}} = -\sqrt{2} \, \text{m/s} \)

Thus, the initial momentum \( \vec{p_i} \) can be calculated as:
\[ \vec{p_i} = m \cdot \vec{v_i} = 0.05 \cdot (\sqrt{2} \hat{i} - \sqrt{2} \hat{j}) = 0.05\sqrt{2} \hat{i} - 0.05\sqrt{2} \hat{j} \]

### Step 2: Calculate the final momentum of the ball
After striking the plane surface, the ball is reflected with the same speed but the direction of the y-component of the velocity changes. Therefore:
- \( v_{fx} = -v_{ix} = -\sqrt{2} \, \text{m/s} \)
- \( v_{fy} = v_{iy} = -\sqrt{2} \, \text{m/s} \)

Thus, the final momentum \( \vec{p_f} \) can be calculated as:
\[ \vec{p_f} = m \cdot \vec{v_f} = 0.05 \cdot (-\sqrt{2} \hat{i} - \sqrt{2} \hat{j}) = -0.05\sqrt{2} \hat{i} - 0.05\sqrt{2} \hat{j} \]

### Step 3: Calculate the change in momentum
The change in momentum \( \Delta \vec{p} \) is given by:
\[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} \]
Substituting the values:
\[ \Delta \vec{p} = \left(-0.05\sqrt{2} \hat{i} - 0.05\sqrt{2} \hat{j}\right) - \left(0.05\sqrt{2} \hat{i} - 0.05\sqrt{2} \hat{j}\right) \]
\[ = -0.05\sqrt{2} \hat{i} - 0.05\sqrt{2} \hat{j} - 0.05\sqrt{2} \hat{i} + 0.05\sqrt{2} \hat{j} \]
\[ = -0.1\sqrt{2} \hat{i} \]

### Step 4: Calculate the magnitude of the change in momentum
The magnitude of the change in momentum is:
\[ |\Delta \vec{p}| = |-0.1\sqrt{2}| = 0.1\sqrt{2} \approx 0.1414 \, \text{kg m/s} \]

### Step 5: Change in the magnitude of the momentum
The magnitude of the initial momentum \( |\vec{p_i}| \) is:
\[ |\vec{p_i}| = \sqrt{(0.05\sqrt{2})^2 + (-0.05\sqrt{2})^2} = \sqrt{0.0025 + 0.0025} = \sqrt{0.005} = 0.0707 \, \text{kg m/s} \]

The magnitude of the final momentum \( |\vec{p_f}| \) is the same:
\[ |\vec{p_f}| = \sqrt{(-0.05\sqrt{2})^2 + (-0.05\sqrt{2})^2} = 0.0707 \, \text{kg m/s} \]

Thus, the change in the magnitude of the momentum is:
\[ \Delta |\vec{p}| = |\vec{p_f}| - |\vec{p_i}| = 0.0707 - 0.0707 = 0 \]

### Final Answers:
(a) The magnitude of the change in momentum of the ball is approximately \( 0.1414 \, \text{kg m/s} \).

(b) The change in the magnitude of the momentum of the ball is \( 0 \).