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 decrease by 36%

Moderate

Solution

K is decreased by 36%.
Therefore, new value of K is 64% of its original value or
$K^{'} = 0.64 K$
$p^{'} = \sqrt{2 K^{'} m}$
$= \sqrt{2 (0.64 K) m}$
$= 0.8 \sqrt{2 K m}$
Now, percentage change in momentum
$= (\frac{Final value}{Initial value} - 1) \times 100$
$= (\frac{0.8 \sqrt{2 K m}}{\sqrt{2 K m}} - 1) \times 100$
= -20%
Therefore, momentum will decrease by 20%.

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