Question

If the momentum of the body increases by 20% what will be the increase in the K.E. of the body?

CLASS - XI PHYSICS (Work, Energy and Power)

Answer 1

there is an increase in 4 % of k.e.

 

Answer 2

momentum(p) = mv
KE = 1/2(mv²)

thus KE = $\frac{ p^{2} }{2m}$

p₁ = p
p₂ is 20% more than p₁.
p₂ = 1.2p

ΔKE = KE₂ - KE₁ = $\frac{ (1.2p)^{2} }{2m} - \frac{ p^{2} }{2m}$ 

⇒ΔKE = $\frac{ (1.44p^{2} }{2m} - \frac{ p^{2} }{2m}$ 

⇒ΔKE = $\frac{ 0.44p^{2} }{2m}$

% increase = (ΔKE/KE₁)×100 = 0.44×100 = 44%