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Question

If the kinetic energy of a body is increased by 300% then determine the percentage increase in the  momentum 


A
100 %
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B
150 %
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C
300
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D
175 %
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Solution

The correct option is A 100 %
When the kinetic energy increases by $$300$$%,new kinetic energy will be
$$K'=K+300$$%K
$$K'=K+3K$$
$$K'=4K$$.............(i)
We know that,
K.E, $$K = \frac{{{P^2}}}{{2m}}$$ 9$$m$$ is the mass of object0
Therefore, eq.(i) become
$$\begin{array}{l} \frac { { P{ '^{ 2 } } } }{ { 2m } } =4\frac { { { p^{ 2 } } } }{ { 2m } }  \\ p{ '^{ 2 } }=4{ p^{ 2 } } \\ p'=\sqrt { 4 } { p^{ 2 } } \\ p'=2p...........\left( { ii } \right)  \end{array}$$
% change in momentum $$ = \frac{{p' - p}}{p} \times 100$$
$$ = \frac{{2p - p}}{p} \times 100$$%$$=1 \times 100=100$$%
Therefore, increases in momentum is $$100$$%.

Physics

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