Question

What happens to K.E. of a body when 3/4th of its mass is removed and velocity is
doubled?​

Answer 1

Answer:

Solution

Explanation:

You see, the formula of K.E. is

$ke = \frac{1}{2} mv^{2}$

Let the mass be x and velocity v

Then

$ke = \frac{1}{2} \times \frac{1}{4} m \times {2v}^{2} \\ \\ = \frac{1}{2} \times mv {}^{2} \\ \\ therefore \: there \: is \: no \: change \: at \: all$

Answer 2

Given : Kinetic Energy of the body.

To Find : K.E. of body when 3/4th of its mass is removed and velocity is

doubled

Solution :

Let the mass be m, and velocity v

Then,

Kinetic Energy = $\[\frac{1}{2}m{v^2}\]$

When 3/4th of its mass is removed then

Mass of new body = $\[m - \frac{3}{4}m\]$ = $\[\frac{1}{4}m\]$

When velocity is doubled ,

New velocity will be = 2v

Then,

New Kinetic Energy = $\[\frac{1}{2} \times \frac{1}{4}m \times {(2v)^2}\]$

New Kinetic Energy = $\[\frac{1}{2} \times \frac{1}{4}m \times {(2v)^2}\]$

New Kinetic Energy =  $\[\frac{1}{2}m{v^2}\]$

Thus, Kinetic Energy = New Kinetic Energy

Hence, Kinetic Energy will not changed when 3/4th of its mass is removed and velocity is  doubled.