Question

If the kinetic energy of a body is increased by300%, what will be the percentage change in momentum of body

Answer 1

Answer:

$\boxed{\mathfrak{Percentage \ change \ in \ momentum \ of \ body = 100\%}}$

Given:

Kinetic energy is increased by 300%

Explanation:

Formula of momentum (P):

$\bold{P = mv}$

Formula of Kinetic Energy (K):

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

m → Mass

v → Velocity

Relation between kinetic energy and momentum:

$\bold{K = \dfrac{P^2}{2m}}$

Initial kinetic energy: = K

Final kinetic energy = K'

= 100%K + 300%K

= 400%K

= 4K

Initial momentum (P):

$\rm P ^{2} = 2mK \\ \rm P= \sqrt{2mK}$

Final momentum (P'):

$\rm P'^{2} = 2mK' \\ \rm P' = \sqrt{2mK'} \\ \rm P' = \sqrt{2m \times 4K} \\ \rm P' = 2\sqrt{2m K} \\ \rm P' = 2P$

$\therefore$ Percentage change in momentum of body:

$\rm \implies \dfrac{P' - P}{P} × 100 \\ \\ \rm \implies \dfrac{2P - P}{P} × 100 \\ \\ \rm \implies \cancel{\dfrac{P}{P}} × 100 \\ \\ \rm \implies 100\%$