Question

the momentum of a body is increased by 200% what is the percentage change in kinetic energy

Answer 1

Answer:

4 times or 400%

Explanation:

P₁=mv--------1

when increasd by 200%,

P₂=2mv----------2

KE₁=1/2 mv²                                                             [derived from 1]

   =mv²/2

KE₂=1/2× m×(2v)²                                                         [derived from 2]

    =2mv²

KE₁ :KE₂=mv²/2 : 2mv²

             =1:4

the kinetic energy is increased by 400% or in other words, 4 times its original value.

hope it helped and pls mark as brainliest:)