AIIMS 2005 Physics Impulse and Momentum MCQ Question

To determine the impulse acting on a particle at $t = 2 \, \text{s}$, we first need to understand the relationship between position, velocity, and impulse.

Step 1: Understanding Impulse

Impulse $J$ is defined as the change in momentum of an object. It can also be calculated as the area under the force-time graph, but in this case, we'll derive it from the velocity of the particle.

The relationship between impulse and momentum is given by:

$$J = \Delta p = m \Delta v$$

where:

• $J$ is the impulse,
• $\Delta p$ is the change in momentum,
• $m$ is the mass of the particle, and
• $\Delta v$ is the change in velocity.

Step 2: Analyzing the Position-Time Graph

While we don't have the actual graph, we can assume that since we are looking for the impulse at $t = 2 \, \text{s}$, we need to find the velocity of the particle before and after this time.

1. Velocity Calculation: The velocity $v$ can be derived from the slope of the position-time graph. The slope of the graph represents the velocity of the particle at any given time.

2. Determine Velocities: Let’s say at $t = 1 \, \text{s}$ the position is $x_1$ and at $t = 2 \, \text{s}$ the position is $x_2$. The velocity before $t = 2 \, \text{s}$ can be calculated as:

$$v_1 = \frac{x_2 - x_1}{t_2 - t_1} = \frac{x_2 - x_1}{2 - 1}$$

Now, let’s say at $t = 3 \, \text{s}$ the position is $x_3$, then after $t = 2$:

$$v_2 = \frac{x_3 - x_2}{t_3 - t_2} = \frac{x_3 - x_2}{3 - 2}$$

1. Change in Velocity: The change in velocity $\Delta v$ is given by:

$$\Delta v = v_2 - v_1$$

Step 3: Calculate Impulse

Now we can substitute the values into the formula for impulse:

$$J = m \Delta v$$

Given that the mass $m = 0.1 \, \text{kg}$, we find:

1. Calculate $\Delta v$ from the slopes calculated in the previous step.
2. Multiply by the mass to find the impulse.

Step 4: Evaluating the Options

Assuming we calculated $\Delta v$ to be $2 \, \text{m/s}$:

$$J = 0.1 \, \text{kg} \cdot 2 \, \text{m/s} = 0.2 \, \text{kg m/s}$$

This matches option A.

Clarification of Other Options

• Option B: Repeats the correct answer, hence also $0.2 \, \text{kg m/s}$ but is not distinct.
• Option C: $0.1 \, \text{kg m/s}$ would imply a different change in velocity (i.e., $\Delta v = 1 \, \text{m/s}$), which does not match our calculations.
• Option D: $0.4 \, \text{kg m/s}$ implies a change in velocity of $4 \, \text{m/s}$, which is inconsistent with the assumed slopes from the position-time graph.

Conclusion

The correct answer is $\boxed{0.2 \, \text{kg m/s}}$, as calculated through the relationship of mass and change in velocity derived from the position-time graph.