A body of mass `0.5 kg` is projected under the gravity with a speed of `98 m//s` at an angle of `30^(@)` with the horizontal. The change in momentum (in magnitude) of the body when it strikes the ground is

Options

A. 24.5 N-s
B. 49.0 N-s
C. 98.0 N-s
D. 50.0 N-s

Explanation

To find the change in momentum of a body projected under gravity, we can follow these steps:

### Step 1: Identify the given values
- Mass of the body, \( m = 0.5 \, \text{kg} \)
- Initial speed, \( u = 98 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)

### Step 2: Break down the initial velocity into components
The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos(30^\circ) \)
- Vertical component: \( u_y = u \sin(30^\circ) \)

Calculating these components:
- \( u_x = 98 \cos(30^\circ) = 98 \cdot \frac{\sqrt{3}}{2} \)
- \( u_y = 98 \sin(30^\circ) = 98 \cdot \frac{1}{2} = 49 \, \text{m/s} \)

### Step 3: Determine the final velocity components just before striking the ground
At the point of impact (when the body strikes the ground), the horizontal component of the velocity remains unchanged, while the vertical component will be reversed in direction:
- Final horizontal component: \( v_x = u_x = 98 \cos(30^\circ) \)
- Final vertical component: \( v_y = -u_y = -49 \, \text{m/s} \)

### Step 4: Calculate the initial and final momentum
The initial momentum \( \vec{P_i} \) and final momentum \( \vec{P_f} \) can be expressed as:
- Initial momentum:
\[ \vec{P_i} = m \cdot (u_x \hat{i} + u_y \hat{j}) = 0.5 \cdot \left(98 \cos(30^\circ) \hat{i} + 49 \hat{j}\right) \]
- Final momentum:
\[ \vec{P_f} = m \cdot (v_x \hat{i} + v_y \hat{j}) = 0.5 \cdot \left(98 \cos(30^\circ) \hat{i} - 49 \hat{j}\right) \]

### Step 5: 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:
\[ \Delta \vec{P} = 0.5 \cdot \left(98 \cos(30^\circ) \hat{i} - 49 \hat{j}\right) - 0.5 \cdot \left(98 \cos(30^\circ) \hat{i} + 49 \hat{j}\right) \]
This simplifies to:
\[ \Delta \vec{P} = 0.5 \cdot \left(-49 \hat{j} - 49 \hat{j}\right) = -49 \hat{j} \, \text{kg m/s} \]

### Step 6: Calculate the magnitude of the change in momentum
The magnitude of the change in momentum is:
\[ |\Delta \vec{P}| = 49 \, \text{kg m/s} \]

### Final Answer
The change in momentum (in magnitude) of the body when it strikes the ground is \( 49 \, \text{kg m/s} \).

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