Question #db371

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Question #db371

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Answer 1 · 2016-05-05T17:47:14

56.25%

Explanation:

Momentum is defined as mass times velocity, or $p = m v$ $p=mv$

Kinetic energy is given by $K E = \frac{1}{2} m v^{2}$ $KE=1/2mv^2$

We can also write KE as
$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \frac{{(m v)}^{2}}{m} = \frac{1}{2} \frac{p^{2}}{m}$ $KE=1/2mv^2=1/2(mv)^2/m=1/2p^2/m$

So if we take the above equations as the position before the increase, then increase momentum by 25%:

$p = 1.25 m v$ $p=1.25mv$
$K E = \frac{1}{2} \frac{{(1.25 p)}^{2}}{m} = \frac{1}{2} \cdot \frac{p^{2}}{m} \cdot 1.5625$ $KE=1/2(1.25p)^2/m=1/2*p^2/m*1.5625$

So KE will increase by 56.25%

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Question #db371

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