Question

A force of $10N$ acts on a body of mass $2kg$ for $3s$, initially at rest. Calculate: (i) the velocity acquired by the body, and (ii) change in momentum of the body.

• A. $15$ $m$ $s^{-1}$ $30$ $kg$ $m$ $s^{-1}$
• B. $15$ $m$ $s^{-1}$ $20$ $kg$ $m$ $s^{-1}$
• C. $5$ $m$ $s^{-1}$ $30$ $kg$ $m$ $s^{-1}$
• D. $25$ $m$ $s^{-1}$ $30$ $kg$ $m$ $s^{-1}$

Answer 1

The correct option is A $15$ $m$ $s^{-1}$ $30$ $kg$ $m$ $s^{-1}$

$F=10N,m=2kg,\Delta t=3s,u=0$
(i) $F=ma$

Putting the values, $10N=2kg\times \left(a\right)$

Therefore, acceleration, $a=5m/s^2$

Acceleration, $a=\frac{\Delta v}{\Delta t}$

Change in velocity, $\Delta v=a\times \Delta t=5m/s^2\times 3s=15m/s$

$v-u=15m/s$

Therefore, velocity acquired by the body, $v=15m/s$
(ii) Force can be also written as, $F=\frac{\Delta p}{\Delta t}$

Where, $p$ is momentum.

Therefore, change in momentum, $\Delta p=F\times \Delta t=10N\times 3s=30kgm/s$