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S1: If momentum of a body increases by 50%, its KE will increase by 125%. S2: KE is proportional to square of velocity.

$\begin{array}{1 1} \text{(a) S1 is true, S2 is true; S2 is correct explanation of S1.} \\ \text{(b) S1 is true, S2 is true; S2 is not correct explanation of S1.} \\\text{(c) S1 is false; S2 is true.} \\ \text{(d) S1 is true; S2 is false.} \end{array}$
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Ans : (a)
Momentum $p_2 = p_1+50 \% \: of\: p_1$
$ \Rightarrow p_2 = \large\frac{3}{2} \times p_1$
Since, $KE \sim v^2 \sim p^2$
Hence, $KE_2 = \large\frac{9}{4} KE_1$
So, $ \% $ increase in $KE$ is
$\large\frac{KE_2 – KE_1}{KE_1} \times100 =( { \large\frac{9p_1}{4} – p_1} ) \times100 = 125 \%$

 

answered Dec 30, 2013 by thanvigandhi_1
edited Nov 7, 2014 by thagee.vedartham
 

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