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Q.
If the momentum of a body increases by $50\%$, its kinetic energy will increase by
Chhattisgarh PMTChhattisgarh PMT 2005
Solution:
Relation between kinetic energy (KE) and momentum (p) is,
$K E=\frac{p^{2}}{2 m}$ Initial kinetic energy
$K E_{1}=\frac{\vec{p}_{2}^{2}}{2 m}$,Similarly, final kinetic energy
$K E_{2}=\frac{\vec{p}_{1}^{2}}{2 m}$
$\therefore \frac{E_{2}-E_{1}}{E_{1}}=\frac{p_{2}^{2} / 2 m-p_{1}^{2} / 2 m}{p_{1}^{2} / 2 m}$
ie. Percentage increase in $KE =\frac{p_{2}^{2}-p_{1}^{2}}{p_{1}^{2}} \times 100$
Now, let, $p_{1}=100$ then $p_{2}=150$
So, $\%$ increase in KE $=\frac{(50)^{2}-(100)^{2}}{(100)^{2}} \times 100$
or $\%$ increase in $E=125$