Question

The velocity of the centre of mass of a system changes form $\overrightarrow{V_1}=8\hat{i}\text{m/s}$ to $\overrightarrow{V_2}=6\hat{j}\text{m/s}$ during time interval $△t=3\text{s}.$ If the mass of the system is $m=6\text{kg}$, the constant force acting on the system is

• A. $25\text{N}$
• B. $12.5\text{N}$
• C. $10\text{N}$
• D. $20\text{N}$

Answer 1

The correct option is D $20\text{N}$

Given velocites, $\overrightarrow{V_1}=8\hat{i}\text{m/s}$ and
$\overrightarrow{V_2}=6\hat{j}\text{m/s}$

Change in velocity of COM $\left(△{\overrightarrow{V}}_{CM}\right)=(6\hat{j}-8\hat{i})\text{m/s}$

Rate of change of momentum gives the force acting on the system
$\overrightarrow{F}=m\frac{△{\overrightarrow{V}}_{cm}}{△t}=\frac{6(6\hat{j}-8\hat{i})}{3}=2(6\hat{j}-8\hat{i})$
$|\overrightarrow{F}|=2\sqrt{8^2+6^2}=2\times 10=20\text{N}$

Therefore, constant force acting on the system is $20\text{N}$.