Question

If K.E. of a particle increases by $125$ % then what is the % increase in its momentum?

Answer 1

We can relate kinetic energy with momentum of an object by the given relation

$2m{E}_K=p^2$

Let the initial kinetic energy of the object $E_{K1}=\frac{p_1^2}{2m}$------(A)

Now as mentioned in the question final kinetic energy is 125% more than initial kinetic energy.

$E_{K2}=E_{K1}+1.25E_{K1}=2.25E_{K1}$-----(1)

Again we know$E_{K2}=\frac{p_2^2}{2m}$------(B)

Now as we can observe from the equation that mass remains constant thus we can relate A and B by the given equation,

$\frac{E_{K1}}{E_{K2}}=\frac{p_1^2}{p_2^2}$

$p_2^2=p_1^2\frac{E_{K2}}{E_{K1}}=2.25p_1^2=1.5p_1$---from 1

Now percentage increase in momentum $=\frac{p_2-p_2}{p_1}\times 100=\frac{1.5p_1-p_1}{p_1}\times 100=50$%