Question

The momentum of a body is increased by 50%, its k.E. Will increase by

Answer 1

We know,

K.E = 1/2 mv^2

P = m.v

So, If

Let initial momentum be possible and kinetic energy be K

If p= mv

If possible increases by 50% the new momentum will be

p' = p + p/2 = 3p/2

The initial K = mv^2= (mv)^2/2Mg

The new Kinetic energy K'= (p')^2/2 mm

So,K'/' =(p')^2/p^2 mm

K'/K = 9/4K'oK/4

% Change = (K'-K)×100%/K

= (5 × 100%)/4 = 125%

So, percentage increase is 125 %.

Hope it helps....

Answer 2

m =mv

m increased by 50%=1/2+mv =1+2mv/2

ke =1/2m ×v^2

ke=1/2 m×v×v

ke=1/2 ×(1+2mv) /2 ×v

new ke=v(1+2mv)/4