The momentum of a body is increased by 50 %. | vedclass

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Question

Kinetic energy of body $=\frac{p^2}{2 m}$

Initial kinetic energy $=\frac{p_1^2}{2 m }$

Final kinetic energy $=\frac{p_f^2}{2 m }=\frac{\left(1.5

Vedclass · Accepted Answer

$125$

Vedclass · Answer

Kinetic energy of body $=\frac{p^2}{2 m}$

Initial kinetic energy $=\frac{p_1^2}{2 m }$

Final kinetic energy $=\frac{p_f^2}{2 m }=\frac{\left(1.5 p _{ i }ight)^2}{2 m }$

$=\frac{2.25 p _{ i }^2}{2 m }$

$\%$ increase in $KE =\frac{2.25 \frac{ p _{ i }^2}{2 m }-\frac{ p _{ i }^2}{2 m }}{\frac{ p _{ i }^2}{2 m }} 	imes 100=125 \%$

The momentum of a body is increased by $50 \%$. The percentage increase in the kinetic energy of the body is $...........\,\%$

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