Question

the kinetic energy of the body increased by 200% find the percentage increase in momentum​

Answer 1

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Momentum,p = mv

P increases by 200%

P1 = 200/100P + P = 3P

m1 = m

v1 = 3v

Increase in K.E = {1/2m(3v)^2 - 1/2mv^2} / 1/2mv^2 x 100%

= 9 - 1 x 100%

= 800%

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Answer 2

Answer:

let initial K.E k1 = 100J

Explanation:

increased K.E

k2 = 100J + 200% of 100J = 300J

we know

p2^2/p1^2 = k2/k1

p2^2/p1^2 = 300J/100J = 3

p2/p1 = √3 = 1.73

p2 = 1.73p1

increase in momentum

p2 - p1 = 1.73p1 - p1 = 0.73p1

increase% = 0.73p1/p1 *100

= 73% Answer