Question

If the momentumof the body is incresed by 20% what will be the incresed in k e of the body

Answer 1

Solution :-

Kinetic energy = K

momentum = P = mv

where, v = velocity

m = mass

$k = \frac{1}{2} \times {mv}^{2} = \frac{ {p}^{2} }{2m}$

of momentum is increased by 20%

$(p + \frac{ \frac{20}{100} }{2m} ) = \frac{ {p}^{2} ( {1.2}^{2}) }{2m}$

= 1.44k ( present kinetic energy)

∆k = 1.44k - k =0.44

∆k/k ×100 = 44%

kinetic energy will increased by 44% is momentum is by increased by 20%

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@GauravSaxena01