Question #e1ca3

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Question #e1ca3

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Answer 1 · 2017-02-22T11:41:48

You got it correct.

Explanation:

Let the speed of 2kg ball be $= v$ $=v$ $m s^{- 1}$ $ms^-1$
1kg ball has speed $= 2 v$ $=2v$ $m s^{- 1}$ $ms^-1$
Kinetic energy is given as $= \frac{1}{2} m v^{2}$ $=1/2mv^2$ ......(1)

Using equation (1) we get
$K E_{2kg} = \frac{1}{2} (2) v^{2} = v^{2}$ $KE_"2kg"=1/2 (2)v^2=v^2$ .....(2)
and $K E_{1kg} = \frac{1}{2} (1) {(2 v)}^{2} = 2 v^{2}$ $KE_"1kg"=1/2(1)(2v)^2=2v^2$ ......(3)

Dividing (2) with (3)
$\frac{K E_{2kg}}{K E_{1kg}} = \frac{v^{2}}{2 v^{2}}$ $(KE_"2kg")/(KE_"1kg")=v^2/(2v^2)$
$\Rightarrow \frac{K E_{2kg}}{K E_{1kg}} = \frac{1}{2}$ $=>(KE_"2kg")/(KE_"1kg")=1/(2)$

$\Rightarrow$ $=>$ Compared to the 1kg ball, the 2kg ball has half the kinetic energy.

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