Question

A 20 kg body is pushed with 7N for 1.5 sec then with a force of 5N for 1.7 sec and finally with a force of 10N for 3 sec, the total impulse applied to the body and change in velocity will be

• A. $49Ns,12.5m{s}^{-1}$
• B. $49Ns,2.45m{s}^{-1}$
• C. $98Ns,4.9m{s}^{-1}$
• D. $4.9Ns,2.45m{s}^{-1}$

Answer 1

Answer: (B)

$49Ns,2.45m{s}^{-1}$

Impulse of a force which is defined as force times its duration of action is equal to the change in the momentum of the body.
Therefore total impulse will be summation of all the individual ones.
$J=F_1\times t_1+F_2\times t_2+F_3\times t_3$
$J=49Ns$
Change in velocity will be equal to the change in momentum divided by its mass.
$\delta V=49/20$
$\delta V=2.45m{s}^{-1}$