Question

A force of 10N acts on a body of mass 2kg for three seconds, initially at rest. Calculate the

(1) The velocity acquired by a body

(2) Change in the momentum of the body

Answer 1

Step1: Given data

A force of 10N acts on a body of mass 2kg for three seconds, initially at rest.

Step2: Formula used

Newton's second law, $F=ma[whereF=force,m=mass,a=acceleration]$

First kinematic equation, $v=u+at[whereu=initialvelocity,v=finalvelocity,t=time,a=acceleration]$

Momentum, $p=mv[wherem=mass,v=velocity]$

Step3: Calculating the velocity of the body

We can determine the acceleration of the body using Newton’s second law of motion as follows, $a=\frac{F}{m}$
Substituting $F=10N$ and $m=2kg$
In the above equation, we get, $a=\frac{10}{2}=5m/s^2$

Let’s determine the final velocity of the body using the first kinematic equation as follows, $v=u+at$
Substituting $u=0$, $a=5m/s^2$and $t=3s$
In the above equation, we get, $v=0+\left(5\right)\left(3\right)=15m/s$

Step4: Calculating the momentum of the body

We can express the change in the momentum of the body as,
$$\begin{gathered} \Delta p=mv-mu \\ \Delta p=m(v-u) \end{gathered}$$

Substituting $m=2kg$ , $v=15m/s$ and $u=0$
in the above equation, we get, $\Delta p=\left(2\right)(15-0)=30kgm/s$.

(1) Therefore, the final velocity of the body is 15 m/s.

(2) Therefore, the change in the momentum of the body is 30 kg m/s.