Question

A ball of mass $4\text{kg}$ is moving in $3$ dimensional plane and the velocity of ball w.r.t an observer is ${\overrightarrow{V}}_1=-3\hat{i}+4\hat{j}+5\hat{k}$ and the velocity of observer w.r.t ground is given as ${\overrightarrow{V}}_2=2\hat{i}+2\hat{j}-4\hat{k}$. Find out kinetic energy of ball w.r.t ground.

• A. $100\text{J}$
• B. $38\text{J}$
• C. $76\text{J}$
• D. $19\text{J}$

Answer 1

The correct option is C $76\text{J}$

Velocity of ball w.r.t ground,

${\overrightarrow{V}}_{bg}={\overrightarrow{V}}_1+{\overrightarrow{V}}_2$

$\Rightarrow {\overrightarrow{V}}_{bg}=(-3\hat{i}+4\hat{j}+5\hat{k})+(2\hat{i}+2\hat{j}-4\hat{k})$

$\Rightarrow {\overrightarrow{V}}_{bg}=-\hat{i}+6\hat{j}+\hat{k}$

Magnitude of ${\overrightarrow{V}}_{bg}:$

$|{\overrightarrow{V}}_{bg}|=\sqrt{(-1{)}^2+6^2+1^2}=\sqrt{38}\text{m/s}$

Kinetic energy of the ball:

$=\frac{1}{2}m|{\overrightarrow{V}}_{bg}{|}^2$

$=\frac{1}{2}\times 4\times (\sqrt{38}{)}^2$

$=2\times 38$

$=76\text{J}$

Hence, option $\left(c\right)$ is the correct answer.