50% confusing

If the momentum of a body is increased by 50% ,what would be the percentage increase in its kinetic energy?

80 100 125 75 50

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1 solution

Let initial momentum be p 0 p_0 , final momentum be p 1 p_1 , mass be m m , initial velocity be u u , final velocity be v v , initial kinetic energy be K E 0 KE_0 and final kinetic energy be K E 1 KE_1 .

p 0 = m u p_0 = mu

p 1 = m v = 1.5 p 0 = 1.5 m u p_1 = m v = 1.5 p_0 = 1.5 mu

m v = 1.5 m u \implies m v=1.5 m u

v = 1.5 u \implies v=1.5 u

K E 0 = 1 2 m u 2 = 0.5 m u 2 KE_0 = \dfrac{1}{2} m u^2 = 0.5 m u^2

K E 1 = 1 2 m v 2 = 1 2 m ( 1.5 u ) 2 = 1 2 m × 2.25 u 2 = 1.125 m u 2 KE_1 = \dfrac{1}{2} m v^2 = \dfrac{1}{2} m (1.5u)^2 = \dfrac{1}{2} m \times2.25 u^2 = 1.125 m u^2

Increase in K E = K E 1 K E 0 K E 1 × 100 % KE = \dfrac{KE_1 - KE_0}{KE_1} \times 100\%

\implies Increase in K E = 1.125 m u 2 0.5 m u 2 0.5 m u 2 × 100 % = 0.625 m u 2 0.5 m u 2 × 100 % = 1.25 m u 2 m u 2 × 100 % KE = \dfrac{1.125 m u^2 -0.5 m u^2}{0.5 m u^2} \times 100\%=\dfrac{0.625 m u^2}{0.5 m u^2} \times 100\%=\dfrac{1.25 m u^2}{ m u^2} \times 100\%

\implies Increase in K E = 1.25 × 100 % = 125 % KE = 1.25 \times 100\% = \textcolor{#3D99F6}{\boxed{125\%}}

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