In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.  Text Solution Solution : Let the mass of the body be m . <br> Initial momentum = `p_1` , hence , final momentum after a 100 % increase, `p_2=2p_1`. <br> If `K_1 and K_2` are initial and final kinetic energies, <br> `K_1=(p_1^2)/(2m) and K_2 =(p_2^2)/(2m)` <br> `therefore (K_2)/(K_1)=(p_2^2)/(p_1^2)=((2p_1)^2)/(p_1^2) =4 =4/1` <br> `or, (K_2-K_1)/(K_1)xx100=(4-1)/1xx100=300` <br> So, the kinetic energy increases by 300% . 1) 150% 2) 200% 3) 225% 4) 300% Answer: 4) 300% Solution: Kinetic Energy (K.E) = p2/2m ( p = momentum, m = mass) Momentum is increased by 100% then, Kinetic Energy (K.E) = (2p)2/2m = 4p2/2m The percentage increase in kinetic energy = [4p2/2m – p2/2m]/p2/2m = 300 %
No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses |
When the KE of body is increased by 100% the momentum of the body is increased by?
zusammenhängende Posts
Urheberrechte © © 2024 frojeostern Inc.
