Question

Momentum of a body increased by 50% its kinetic energy will increase by

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%