Question

If the linear momentum of a body is increased by by 50% then by what percentage will its kinetic energy increase?​

Answer 1

Let initial momentum be p and kinetic energy be K

p=mv

If p increases by 50%, the new momentum

p'=p+p/2= 3p/2

The initial K=mv^2/2=(mv) ^2/2m=p^2/2m

The new kinetic energy K'=(p')^2/2m

So K'/K=(p')^2/p^2

K'/K= 9/4

K'=9K/4

% change = [(K'–K)×100%]/K=(5×100%)/4=125%

So the percentage increase in kinetic energy is 125%

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Comments Report

abhishek665 Ace

Hey mate

ciao!!

Let initial momentum be p and kinetic energy be K

p=mv

If p increases by 50%, the new momentum

p'=p+p/2= 3p/2

The initial K=mv^2/2=(mv) ^2/2m=p^2/2m

The new kinetic energy K'=(p')^2/2m

So K'/K=(p')^2/p^2

K'/K= 9/4

K'=9K/4

% change = [(K'–K)×100%]/K=(5×100%)/4=125%

So the percentage increase in kinetic energy is 125%

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Answer 2

Answer:

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