Question

When the momentum of a body increases by 100%, its K.E. increases by

• A. 20%
• B. 40%
• C. 100%
• D. 300%

Answer 1

Answer: (D)

300%

K.E${}_1$ = $\frac{{\rho }^2}{2m}$ ($\rho$ = momentum and mass = mass)

Momentum is increased by 100% then,

K.E${}_2$ = $\frac{(2\rho {)}^2}{2m}$ = $\frac{4{\rho }^2}{2m}$

Therefore percentage increase in kinetic energy,

= ($\frac{4{\rho }^2}{2m}$ - $\frac{{\rho }^2}{2m}$} / {$\frac{{\rho }^2}{2m}$} = 300%