Question

A force-time graph for a linear motion of a body is shown in the figure. The change in linear momentum between 0 and 7 s is :

• A. $2N-s$
• B. $3N-s$
• C. $4N-s$
• D. $5N-s$

Answer 1

Answer: (A)

Since, $F=\frac{\Delta p}{\Delta t}$ or $\Delta p=F\Delta t$ we can say that momentum between 0 to 7 s isequal to the vector area enclosed by theforce-time graph from 0 to 7s. So, Change in linear momentum = vector area of triangle OAB+vectorarea ofsquare BCDE + vector area of triangle $EFG+$ vector area of square $GHIJ+$ vector area oftriangle JKL $=\left[\frac{1}{2}\times 1\times (-1)\right]+[2\times 2]+\left[\frac{1}{2}\times 2\times (-2)\right]$ $+[1\times 1]+\left[\frac{1}{2}\times 1\times (-1)\right]$ $=-\frac{1}{2}+4-2+1-\frac{1}{2}=2N-s$