Question

Rank the following velocities according to the kinetic energy a particle will have with each

velocity, greatest first: (i) $\overrightarrow{v}=4\hat{i}+3\hat{j}$ (ii) $\overrightarrow{v}=-4\hat{i}+3\hat{j}$ (iii) $\overrightarrow{v}=-3\hat{i}+4\hat{j}$ (iv) $\overrightarrow{v}=3\hat{i}-4\hat{j}$ (v) $\overrightarrow{v}=5\hat{i}$ (vi) $\overrightarrow{v}$ = $5\frac{m}{s}$ at ${30}^{\circ }$ to the horizontal.

• A. (i) > (ii) > (iii) > (iv) > (v) > (vi)
• B. (i) < (ii) < (iii) < (iv) < (v) < (vi)
• C. (i) > (ii) = (iii) = (iv) > (v) > (vi)
• D. (i) = (ii) = (iii) = (iv) = (v) = (vi)

Answer 1

The correct option is D

(i) = (ii) = (iii) = (iv) = (v) = (vi)

So let's check one by one if particle has the particle has the given velocities then what will be its corresponding

kinetic energies.

(i) $\overrightarrow{v}=4\hat{i}+3\hat{j}$

$|v|=5$ $(\because \sqrt{4^2+3^2}=5)$

$K.E.=\frac{1}{2}m{v}^2=\frac{25}{2}mjoules$

(ii) $\overrightarrow{v}=4\hat{i}+3\hat{j}$

$|v|=5$
$\therefore$ K.E = $\frac{1}{2}m{v}^2=\frac{25}{2}mjoules$
(iii) $\overrightarrow{V}=-3\hat{i}+4\hat{j}V=\sqrt{9+16}=5$
$\therefore$ K.E = $\frac{1}{2}m{v}^2=\frac{25}{2}mjoules$
(iv) $\overrightarrow{V}=-3\hat{i}-4\hat{j}V=\sqrt{9+16}=5$
$\therefore$ K.E = $\frac{1}{2}m{v}^2=\frac{25}{2}mjoules$
(v) V = 5
K.E = $\frac{25}{2}$ mJ
$\therefore$ K.E = $\frac{1}{2}m{v}^2=\frac{25}{2}mjoules$
(vi) V = 5
K.E = $\frac{25}{2}$ mJ
$\therefore$ K.E = $\frac{1}{2}m{v}^2=\frac{25}{2}mjoules$