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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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When kinetic energy of body is increased by 300 % momentum of body is increased by

\[(a)\;200\;\% \quad (b)\;100\;\% \quad (c)\;50\;\% \quad (d)\;150\;\%\]
Can you answer this question?
 
 

1 Answer

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$P_1,P_2$ momentum before and after let orginal K.E be 100 J
K.E increased by $300 \%$
new $KE=400 J$
$\large\frac{{P_2}^2}{{P_1}^2}=\frac{E_2}{E_1}=\frac{400}{100}$$=4$
$\large\frac{P_2}{P_1}$$=2=>P_2=2P_1$
Increase % $=\large\frac{P_2-P_1}{P_1} $$\times 100$
$\qquad=\large\frac{2P_1 -P_1}{P_1} $$ \times 100$
$\qquad=100 \;\%$
Hence b is the correct answer.

 

answered Jul 22, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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