Question

A particle of mass $70g$, moving at $50cm/s$, is acted upon by a variable force opposite to its direction of motion. The force $F$ is shown as a function of time $t$. Then:

• A. its speed will be $50cm/s$ after the force stops acting
• B. its direction of motion will reverse
• C. its average acceleration will be $1m/s^2$ during the interval in which the force acts
• D. its average acceleration will be $10m/s^2$ during the interval in which the force acts

Answer 1

Answer: (A), (B)

The correct options are
A its direction of motion will reverse
B its speed will be $50cm/s$ after the force stops acting

Initial momentum of the particle $=\left(0.07\right)\left(0.5\right)=0.035kgm/s$
The impulse given to the particle is equal to the area under the $F-t$ graph $=0.07kgm/s=2\times$ initial momentum of the particle. Hence, the particle will reverse direction and move with its initial speed.