A force of $20\text{N}$ acts on a body of mass $40\text{kg}$ for $20$ seconds. Change in its momentum is

The correct option is B $400\text{kg m/sec}$
Applying Newton's second law
$\Sigma F=ma$
or, $F=ma$
$\Rightarrow 20=40\times a\Rightarrow a=0.5{\text{m/s}}^2$
Since, acceleration is constant
$\therefore$ Applying equation of motion
$v=u+at$
$v=0+0.5\times 20\Rightarrow v=10\text{m/s}$
So, change in momentum is given by
$\Delta P=m(v-u)=40(10-0)=400\text{kgm/s}$

Alternate solution:

Impulse $J=$ Force $\times$ time
$\Rightarrow 20\times 20=400\text{Ns}$
Impulse, $J=\Delta P=400\text{kg m/s}$