Question

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

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

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+3KK'=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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