A body of mass 2kg is projected from the ground with a velocity `20ms^(-1)` at an angle `30^(@)` with the vertical. If `t_(1)` is the time in seconds at which the body is projected and `t_(2)` is the time in seconds at which it reaches the ground, the change in momentum in `kgms^(-1)` during the time `(t_(2)-t_(1))` is

Options

A. 40
B. `40sqrt3`
C. `50sqrt3`
D. 60

Answer

To solve the problem, we need to determine the change in momentum of a body projected at an angle with respect to the vertical. Here are the steps to find the solution:

### Step 1: Identify the given values
- Mass of the body, \( m = 2 \, \text{kg} \)
- Initial velocity, \( v = 20 \, \text{m/s} \)
- Angle of projection with the vertical, \( \theta = 30^\circ \)

### Step 2: Resolve the initial velocity into vertical and horizontal components
Since the angle is given with respect to the vertical, we can find the vertical component of the velocity using:
\[ v_y = v \cdot \cos(\theta) \]
Substituting the values:
\[ v_y = 20 \cdot \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 3: Determine the final vertical velocity just before hitting the ground
When the body reaches the ground, its vertical velocity will be equal in magnitude but opposite in direction to the initial vertical velocity (ignoring air resistance). Therefore, the final vertical velocity \( v_{y_f} \) is:
\[ v_{y_f} = -v_y = -10\sqrt{3} \, \text{m/s} \]

### Step 4: Calculate the change in vertical velocity
The change in vertical velocity \( \Delta v_y \) can be calculated as:
\[ \Delta v_y = v_{y_f} - v_y = -10\sqrt{3} - 10\sqrt{3} = -20\sqrt{3} \, \text{m/s} \]

### Step 5: Calculate the change in momentum
The change in momentum \( \Delta p \) is given by the formula:
\[ \Delta p = m \cdot \Delta v_y \]
Substituting the values:
\[ \Delta p = 2 \cdot (-20\sqrt{3}) = -40\sqrt{3} \, \text{kg m/s} \]

### Step 6: Present the final answer
The magnitude of the change in momentum is:
\[ |\Delta p| = 40\sqrt{3} \, \text{kg m/s} \]

### Final Answer
The change in momentum during the time \( (t_2 - t_1) \) is \( 40\sqrt{3} \, \text{kg m/s} \).

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