Question

If the kinetic energy of a body is increased by 300%, its momentum will increase by
A. 300%
B. 30%
C. 10%
D. 100%

Answer 1

When we say that kinetic energy of the particle is increased by 300%, it means that new kinetic energy is equal to 4 times the initial kinetic energy. We know the relation between kinetic energy and momentum. The momentum formula is

$\rho {\text{ }} = mv$

And kinetic energy formula is

$K.E. = \dfrac{1}{2}m{v^2}$

Complete step by step solution:
The relation between kinetic energy and momentum formula is given by

$K.E. = \dfrac{{{\rho ^2}}}{{2m}} \to (1)$

Now, when the KE is increased is increased by 300% is taken as ‘KE’ and the new kinetic energy is taken as KE’

$KE' = KE + 300KE$
$KE' = KE + 3KE$
$KE' = 4KE \to (2)$

Therefore, the equation (1) becomes,

$\Rightarrow KE' = \dfrac{{\rho {'^2}}}{{2m}}$
$\Rightarrow (2) \Rightarrow \dfrac{{\rho {'^2}}}{{2m}} = \dfrac{{4{\rho ^2}}}{{2m}}$
$\Rightarrow \rho {'^2} = 4{\rho ^2}$
$\Rightarrow p' = \sqrt {4{p^2}}$
$\Rightarrow p' = 2p \to (3)$

% change in momentum = $\dfrac{{\rho ' - \rho }}{\rho } \times 100\%$
$= \dfrac{{2\rho - \rho }}{\rho } \times 100\%$
$= \dfrac{\rho }{\rho } \times 100\%$
$= 100\%$

Therefore, increase in momentum is 100%.

Note: This means that the new momentum of the particle is two times the initial momentum, which means that the momentum of the particle increased by 100%. Momentum is a vector quantity and kinetic energy is a scalar quantity. Kinetic energy has a velocity that means velocity has both magnitude and direction.
${\text{The change in percentage =}}\dfrac{{{\text{Final initial value}}}}{{{\text{Initial value}}}} \times 100\%$