Question

if the mass of the body is halved and its kinetic energy becomes eight times ,then velocity of the body will became?​

Answer 1

Answer:

4 times

Explanation:

K.E=1/2mv^2

=1/2(1/2)(4v)^2 ( Do trial and error method by keeping v=4

=1/2(8)mv^2

therefore velocity of the body will became 4 times

Answer 2

To find:

The mass of the body is halved and its kinetic energy becomes eight times ,then velocity of the body will became?

Calculation:

Let initial mass be m , initial velocity be v and initial kinetic energy be KE.

$\therefore \: KE = \dfrac{1}{2} m {v}^{2} \: \: \: \: ......(1)$

Now , new mass be m/2 , new kinetic energy be 8(KE) and let new velocity be v_(2):

$\therefore \: 8(KE )= \dfrac{1}{2} (\dfrac{m}{2}) {v_{2}}^{2} \: \: \: \: ......(2)$

Dividing the equations ( 2 ÷ 1 ) :

$\therefore \: 8 = \dfrac{ \frac{1}{2} {(v_{2})}^{2} }{ {v}^{2} }$

$\implies \: \dfrac{ {(v_{2})}^{2} }{ {v}^{2} } = 8 \times 2$

$\implies \: \dfrac{ {(v_{2})}^{2} }{ {v}^{2} } = 16$

$\implies \: {(v_{2})}^{2} = 16 {v}^{2}$

$\implies \: v_{2} = 4v$

So, velocity becomes 4 times than that of initial value.