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Question

The kinetic energy of a body increases by 300% due to change in its velocity. The momentum increases by 


A
100%
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B
200%
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C
300%
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D
400%
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Solution

The correct option is A $$100 \%$$
when the kinetic energy increasing by $$300$%, new kinetic energy t will be
$$K' = k + 300$$% k
$$K'=K+3K$$
$$K'=4K...........(i)$$
We know that $$K.E=K= \dfrac{{{p^2}}}{{2m}}$$($$m$$ is the mass of object)
$$\therefore$$ ,equation (i) become
$$\begin{array}{l} =\dfrac { { p{ '^{ 2 } } } }{ { 2m } } =4\dfrac { { { p^{ 2 } } } }{ { 2m } }  \\ =p{ '^{ 2 } }=4{ p^{ 2 } } \\ p'=\sqrt { 4 } { p^{ 2 } } \\ p'=2p..........\left( { ii } \right)  \end{array}$$
% change in momentum $$ = \frac{{p' - p}}{p} \times 100$$
$$ = \dfrac{{2p - p}}{p} \times 100$$$=$$1 \times 100=100$$
$$\therefore $$ increases in momentum is $$100$$%.

Physics

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