Question

if the kinetic energy of the body is increased by 200% by what percent will the momentum increase​

Answer 1

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%

Answer 2

Answer:

let p1, p2 and k1,k2 be initial & final momentum and K.E .

Explanation:

let initial K.E, k1 = 100J

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 !

#MukeshBhardwaj