JEE Main Physics question
If the momentum of a body is increased by 100%, then the percentage increase in the kinetic energy is
Answer: (D)
Solution
E = p²/2m. With p' = 2p: E' = 4E, so the percentage increase is (4E−E)/E × 100 = 300%.
If the momentum of a body is increased by 100%, then the percentage increase in the kinetic energy is
E = p²/2m. With p' = 2p: E' = 4E, so the percentage increase is (4E−E)/E × 100 = 300%.