Question

If momentum of a particle is increased by $60$$, then percentage increase in kinetic energy is:

• A. $125$
• B. $148$
• C. $156$
• D. $200$

Answer 1

Answer: (C)

$156$

We know kinetic energy $E_K=\frac{1}{2}m{v}^2$ and momentum of an object $P=mv$

Now multiplying and dividing$E_K$ by mass we get

$E_K=\frac{1}{2m}{m}^2{v}^2=\frac{p^2}{2m}$------(1)

Now when the momentum is increased by 60% $P_{new}=1.6P$

$E_{K_{new}}=\frac{1}{2m}{m}^2{v}^2=\frac{{1.6p}^2}{2m}$------(2)

Solving 1 and 2 we get ,

$\frac{E_K}{E_{Knew}}=\frac{\frac{P^2}{2m}}{\frac{1.6P^2}{2m}}$

$E_{Knew}=2.56E_K$

Thus percentage increase in its kinetic energy=$156$