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5.Work, Energy, Power and Collision
normal
When the momentum of a body increases by $100\%$, its $KE$ increases by .............. $\%$
A
$400$
B
$100$
C
$300$
D
none
Solution
$\mathrm{K.E}_{1}=\frac{\rho^{2}}{2 m}(\rho=\text { momentum and mass }=\text { mass })$
Momentum is increased by $100 \%$ then,
$\mathrm{KE}_{2}=\frac{(2 \rho)^{2}}{2 m}=\frac{4 \rho^{2}}{2 m}$
Therefore percentage increase in kinetic energy,
$=\left(\frac{4 \rho^{2}}{2 m}-\frac{\rho^{2}}{2 m}\right) /\left(\frac{\rho^{2}}{2 m}\right)=300 \%$
Standard 11
Physics
