Question #13b3d

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Answer 1 · 2014-03-06T18:52:09

The kinetic energy increases by 125 %.

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

$p = m v$ $p = mv$

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

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%