Question

The momentum of a body is 50 %, the percentage increase in its kinetic energy is

Answer 1

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$\underline\bold\red{\huge{ANSWER : }}$

⭐ If p is momentum of a body and m is mass, then,

⭐ Kinetic energy = p²/2m.

⭐ New momentum (p') =
$p + \frac{50}{100} p$
$= \frac{3}{2} p$
⭐ If E' is the new kinetic energy, then :

E'
$= ( \frac{3p}{2} )^{2} \div 2m$

$= \frac{9}{4} \times \frac{ {p}^{2} }{2m}$

$= \frac{9}{4} \times kinetic \: energy$

⭐ Increase in kinetic energy = E'-E

= 9/4E - E
= 9E-4E/4
= 5E/4

⭐ PERCENTAGE INCREASE = ((5E/4)/E)×100

= 5/4×100
= 125%

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