# Question 786a6

Dec 21, 2017

The ratio of their $K E$ is $= \frac{3}{2}$

#### Explanation:

The momentum is $P = m v$

The masses are ${m}_{1} = 10 k g$ and ${m}_{2} = 15 k g$

As their momentum are identical

$p = {m}_{1} {v}_{1} = {m}_{2} {v}_{2}$

So,

$10 \cdot {v}_{1} = 15 \cdot {v}_{2}$

${v}_{2} = \frac{10}{15} {v}_{1} = \frac{2}{3} {v}_{1}$

Therefore,

Their kinetic energies are

$K {E}_{1} = \frac{1}{2} {m}_{1} {v}_{1}^{2} = \frac{1}{2} \cdot 10 \cdot {v}_{1}^{2} = 5 {v}_{1}^{2}$

$K {E}_{2} = \frac{1}{2} {m}_{2} {v}_{2}^{2} = \frac{1}{2} \cdot 15 \cdot {v}_{2}^{2} = \frac{15}{2} \cdot {\left(\frac{2}{3} {v}_{1}\right)}^{2}$

$= \frac{15}{2} \cdot \frac{4}{9} \cdot {v}_{1}^{2} = \frac{10}{3} {v}_{1}^{2}$

Therefore,

$\left(\frac{K {E}_{1}}{K {E}_{2}}\right) = \frac{5 {v}_{1}^{2}}{\frac{10}{3} {v}_{1}^{2}} = \frac{3}{2}$

Dec 21, 2017

The first objects KE is $\frac{3}{2}$ times more than the second objects KE.

#### Explanation:

We know that,

Momentum,color(blue)(p=mv#

The mass of the first object,${m}_{1} = 10 k g$

The mass of the second object,${m}_{2} = 15 k g$

$\therefore$The momentum of the first object$, {p}_{1} = {m}_{1} {v}_{1} = 10 {v}_{1}$

$\therefore$the momentum of the second object$, {p}_{2} = {m}_{2} {v}_{2} = 15 {v}_{2}$
As,you mentioned${p}_{1} = {p}_{2}$
Hence,${m}_{1} {v}_{1} = {m}_{2} {v}_{2}$
$\mathmr{and} , 10 {v}_{1} = 15 {v}_{2}$

$\therefore {v}_{1} / {v}_{2} = \frac{15}{10} = \frac{3}{2}$[As it is a ratio, there will remain no unit.]
We know,
$\textcolor{g r e e n}{k E = \frac{1}{2} \left(m {v}^{2}\right)}$
Then,$\frac{K {E}_{1}}{K {E}_{2}} = \frac{\cancel{\frac{1}{2}} {m}_{1} {v}_{1}^{2}}{\cancel{\frac{1}{2}} {m}_{2} {v}_{2}^{2}} = {m}_{1} / \left({m}_{2}\right) . {\left({v}_{1} / {v}_{2}\right)}^{2} = \frac{10 \cancel{k g}}{15 \cancel{k g}} . {\left(\frac{3}{2}\right)}^{2} = \frac{2}{3.} \left(\frac{9}{4}\right) = \frac{3}{2} \textcolor{b r o w n}{\left(A n s .\right)}$