Question

Kinetic energy of a particle is increased by 300%. Find the percentage increase in its momentum.

Answer 1

When the kinetic energy increase by 300 %, new kinetic energy will be

$K^{\prime }=K+300$ K

$K^{\prime }=K+3K$

$k;=4K$

We know that K.E. K = $\frac{p^2}{2m}$ (m is the mass object )

Therefore, eq. (i) become

$\frac{p^{\prime }2}{2m}=4\frac{p^2}{2m}$

$p^{\prime }2=4p^2$

$p^{\prime }\sqrt{4}{p}^2$

$p^{\prime }=2p$ (ii)

% change in mornentum = $\frac{p^{\prime }-p}{p}\times 100$

$=\frac{2P-p}{p}\times 100$ % = $1\times 100=100$ %

Therefore, increases in momentum is 100 %