Question

16. If a momentum of a particle is increased by 100%. Then percentage increase in kinetic energy of the
particle is:
A) 400%
B) 300%
C) 100%
D) 200%

Please help me out guys ​

Answer 1

Answer:

it will be B

velocity becomes doubled so kinetic energy is 4 times

Explanation:

Answer 2

Answer:

300%

p = mv

p2/p1=2

==》V2/v1=2

KE2/KE1= Square of (v2/v1) =4

percentage increase is (4-1)/1×100= 300%