# Question #e1ca3

Feb 22, 2017

You got it correct.

#### Explanation:

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

Using equation (1) we get
$K {E}_{\text{2kg}} = \frac{1}{2} \left(2\right) {v}^{2} = {v}^{2}$ .....(2)
and $K {E}_{\text{1kg}} = \frac{1}{2} \left(1\right) {\left(2 v\right)}^{2} = 2 {v}^{2}$ ......(3)

Dividing (2) with (3)
$\left(K {E}_{\text{2kg")/(KE_"1kg}}\right) = {v}^{2} / \left(2 {v}^{2}\right)$
$\implies \left(K {E}_{\text{2kg")/(KE_"1kg}}\right) = \frac{1}{2}$

$\implies$ Compared to the 1kg ball, the 2kg ball has half the kinetic energy.