The force-time (F-t) curve of a particle exec | vedclass

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(D) The change in momentum $(\Delta p)$ of a particle is equal to the impulse,which is given by the area under the force-time $(F-t)$ graph.$\Delta p

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$0$

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(D) The change in momentum $(\Delta p)$ of a particle is equal to the impulse,which is given by the area under the force-time $(F-t)$ graph.$\Delta p = \int F \, dt = 	ext{Area under } F-t 	ext{ curve}$.In the given graph,the area from $t = 2 \ s$ to $t = 6 \ s$ is above the time axis,representing a positive impulse.The area from $t = 0 \ s$ to $t = 2 \ s$ and $t = 6 \ s$ to $t = 8 \ s$ is below the time axis,representing a negative impulse.Due to the symmetry of the curve,the positive area (between $2 \ s$ and $6 \ s$) is equal in magnitude to the total negative area (from $0$ to $2 \ s$ and $6$ to $8 \ s$).Therefore,the net area is $0$,which implies the net change in momentum acquired by the particle in the interval from $0$ to $8 \ s$ is $0 \ N-s$.

The force-time $(F-t)$ curve of a particle executing linear motion is as shown in the figure. The momentum acquired by the particle in the time interval from $0$ to $8 \ s$ will be ........... $N-s$.

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