Question

When momentum of a body increases by 200%, its ke increases by (1) 200% (2) 300% (3) 400% (4) 800%?

Answer 1

p=√(2mK)……,where K is the Kinetic Energy.

Let K1 and K2 be the initial and final K.E.

Let p1 and p2 be the initial and final momentum.

According to ques.,

K2 = K1+K1*300%

K2 = K1+3*K1

K2 = 4*K1

p1 = √(2mK1)

p2 = √(2mK2)

p2 = √(2m*4*K1)

p2 = 2√(2mK1)

p2 = 2*p1

Now,

% increase in momentum=(p2-p1)/p1*100%

= (2*p1-p1)/p1*100%

= 100%

Answer 2

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