Explanation

(a) $Since,\ F = \frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}$
$or\ \ \mathrm{\Delta}p = F\ \mathrm{\Delta}t$
We can say that momentum between 0 to 7 s is equal to the vector area enclosed by the force-time graph from 0 to 7 s. So, Change in linear momentum
= vector area of triangle OAB + vector area of square BCDE+ vector area of triangle EFG + vector area of square GHIJ + vector area of triangle JKL
$= \left\lbrack \frac{1}{2} \times 1 \times ( - 1) \right\rbrack + \lbrack 2 \times 2\rbrack + \left\lbrack \frac{1}{2} \times 2 \times ( - 2) \right\rbrack + \lbrack 1 \times 1\rbrack + \left\lbrack \frac{1}{2} \times 1 \times ( - 1) \right\rbrack$
$= - \frac{1}{2} + 4 - 2 + 1 - \frac{1}{2} = 2\ Ns$