A block of mass `m = 3kg` resting on a horizontal frictionless flooe is horizontally struck by a 9N force that acts for 0.02 sec. After 3 sec it receivers a second blow of force 9 N but in opposite direction which acts for 0.01 sec. The speed of the body after 30 sec is

Options

A. 0
B. 3 cm/sec
C. 90 cm/sec
D. 30 cm/sec

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and calculate the resulting velocities after each force is applied.

### Step 1: Calculate the initial velocity after the first force is applied

1. **Given Data**:
- Mass of the block, \( m = 3 \, \text{kg} \)
- Force applied, \( F_1 = 9 \, \text{N} \)
- Time duration of the first force, \( t_1 = 0.02 \, \text{s} \)

2. **Calculate the impulse from the first force**:
\[ \text{Impulse} (J_1) = F_1 \cdot t_1 = 9 \, \text{N} \cdot 0.02 \, \text{s} = 0.18 \, \text{Ns} \]

3. **Using the impulse-momentum theorem**:
\[ J_1 = m \cdot \Delta V_1 \]
where \( \Delta V_1 \) is the change in velocity due to the first force. Since the block starts from rest, we have:
\[ 0.18 = 3 \cdot (V_1 - 0) \]
\[ V_1 = \frac{0.18}{3} = 0.06 \, \text{m/s} \]

### Step 2: Determine the velocity after 3 seconds

After the first force is applied for 0.02 seconds, the block continues to move at the same velocity \( V_1 \) for the remaining time until the second force is applied (3 seconds total).

- Thus, after 3 seconds, the velocity remains:
\[ V = 0.06 \, \text{m/s} \]

### Step 3: Calculate the effect of the second force

1. **Given Data for the second force**:
- Force applied, \( F_2 = -9 \, \text{N} \) (opposite direction)
- Time duration of the second force, \( t_2 = 0.01 \, \text{s} \)

2. **Calculate the impulse from the second force**:
\[ \text{Impulse} (J_2) = F_2 \cdot t_2 = -9 \, \text{N} \cdot 0.01 \, \text{s} = -0.09 \, \text{Ns} \]

3. **Using the impulse-momentum theorem for the second force**:
\[ J_2 = m \cdot \Delta V_2 \]
where \( \Delta V_2 \) is the change in velocity due to the second force. The initial velocity before the second force is applied is \( V_1 = 0.06 \, \text{m/s} \):
\[ -0.09 = 3 \cdot (V_2 - 0.06) \]
\[ -0.09 = 3V_2 - 0.18 \]
\[ 3V_2 = 0.09 \implies V_2 = \frac{0.09}{3} = 0.03 \, \text{m/s} \]

### Final Step: Determine the speed after 30 seconds

After the second force is applied, the block will continue to move at the new velocity \( V_2 = 0.03 \, \text{m/s} \) for the remaining time (27 seconds) since there are no other forces acting on it.

### Conclusion

The speed of the body after 30 seconds is:
\[ \text{Speed} = 0.03 \, \text{m/s} \quad \text{(or 3 cm/s)} \]