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If kinetic energy of a body is increased...

If kinetic energy of a body is increased by 300%, then percentage change in momentum will be

A

`100%`

B

`200%`

C

`sqrt300%`

D

`400%`

Text Solution

Verified by Experts

The correct Answer is:
A
Let Initial KE= `100 alpha`
Final KE `=(100 + 300) alpha= 400alpha`
`(p_(1)^(2))/(p_(2)^(2))= (2mk_(1))/(2mk_(2))= (100 alpha)/(400 alpha)= (1)/(4)`
or `(p_(2))/(p_(1)) =(2)/(1) rArr p_(2) = 2p_(1)`
Percentage increament in `p= (p_(2)- p_(1))/(p_(1)) xx 100`
`=("Change in momentum")/("Original momentum") xx 100`
`(2p_(2)-p_(1))/(p_(1)) xx 100= 100%`
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