Question #26e1f

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Question #26e1f

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Answer 1 · 2017-05-12T04:29:49

We know that KE ( $E$ $E$ ) and momentum ( $p$ $p$ ) of a body of mass $m$ $m$ is related as follows

$E = \frac{p^{2}}{2 m}$ $E =p^2/(2m)$

If $E_{i} and p_{i}$ $E_i and p_i$ are respectively initial KE and momentum and $E_{f} and p_{f}$ $E_fand p_f$ are respectively final KE and momentum then

$E_{i} = \frac{p_{i}^{2}}{2 m}$ $E_i =p_i^2/(2m)$

and

$E_{f} = \frac{p_{f}^{2}}{2 m}$ $E_f =p_f^2/(2m)$

combining these two we have

$\frac{E_{f}}{E_{i}} = \frac{p_{f}^{2}}{p_{i}^{2}}$ $E_f/E_i=p_f^2/p_i^2$

By the given condition $p_{f} = 50 % o f p_{i} = \frac{p_{i}}{2}$ $p_f=50%of p_i=p_i/2$

so

$\frac{E_{f}}{E_{i}} = \frac{p_{f}^{2}}{{(2 p_{f})}_{f}^{2}} = \frac{1}{4}$ $E_f/E_i=p_f^2/(2p_f)^2=1/4$

$\Rightarrow \frac{E_{f} - E_{i}}{E_{i}} = - \frac{3}{4} = - 75 %$ $=>(E_f-E_i)/E_i=-3/4=-75%$

So change in KE is 75% , here negative sign denotes the decrease in KE

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Question #26e1f

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