Question #13b3d

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Answer 1 · 2014-03-06T18:52:09

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%