homework help 1677

The kinetic energy increases by 125 %.

Let initial momentum be ##p## and kinetic energy be ##K##.

##p = mv##

If ##p## increases by 50% (0.50##p##)

the new momentum ##p’ = p + 0.50p = 1.50p##

The initial ##K = 1/2 mv^2 = (mv)^2/(2m) = p^2/(2m)##

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

So ##(K’)/K = (p’^2)/ p^2##

##(K’)/K = (1.50p)^2/p^2 = 1.50^2/1 = 2.25##

##K’ = 2.25 K##

% change = ##(K’ – K)/K × 100 % = (2.25 K – K)/K × 100 % = 125 %##

% increase in kinetic energy = 125%