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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)


There is an increase in 4 % of k.e.

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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%
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