A) \[2\,N-s\]
B) \[3\,N-s\]
C) \[4\,N-s\]
D) \[5\,N-s\]
Correct Answer: A
Solution :
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 \[=\left[ \frac{1}{2}\times 1\times (-1) \right]+[2\times 2]+\left[ \frac{1}{2}\times 2\times (-2) \right]\] \[+\,[1\times 1]+\left[ \frac{1}{2}\times 1\times (-1) \right]\] \[=-\frac{1}{2}+4-2+1-\frac{1}{2}=2N-s\]
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