Q109 / 180 · AIIMS 2005 · Physics
In the figure given, the position-time graph of a particle of mass $0.1 \mathrm{~kg}$ is shown.
The impulse at $t=2 \mathrm{sec}$ is
- Correct answerA$0.2 \mathrm{~kg} \mathrm{~m} \mathrm{sec}^{-1}$
- B$-0.2 \mathrm{~kg} \mathrm{~m} \mathrm{sec}{ }^{-1}$
- C$0.1 \mathrm{~kg} \mathrm{~m} \mathrm{sec}{ }^{-1}$
- D$-0.4 \mathrm{~kg} \mathrm{~m} \mathrm{sec}{ }^{-1}$
Solution
Impulse $=$ change in momentum
$$
=m \Delta v=\frac{m \Delta x}{\Delta t}=0.1 \times \frac{4-0}{2-0}=0.2 \mathrm{~kg} \mathrm{msec}^{-1} \text {. }
$$
Hence, the correct answer is option: 'A'.