Question

Relation between kinetic energy K and momentum p of a particle is
$p = \sqrt{2 K m}$
Find the percentage of change in p if K is increased by 1%

Moderate

Solution

$p \propto \sqrt{K} [m = constant] or p \propto K^{\frac{1}{2}}$
As % change is small. Hence,
$(% change in p) = \frac{1}{2}$ (%change in K)
$= \frac{1}{2} (+ 1 %) = + 0.5 %$
Therefore, momentum will increase by 0.5%.

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