Question

the momentum of a body is increased by 50% when is the % change in kinetic energy 3 marks

Answer 1

hello friend..!!

p = mv , if this increases by 50 % ( 0.50 p ) 

the new momentum p' = p + 0.50 p = 1.50 p 

the intial kinetic energy (k)  = 1/2 mv²  =  (mv)²/ 2m  =  p²/ 2m 

the new kinetic energy (k')  = (p')² / 2m .

therefore ,

          k' / k  = (p')² / p²

         k'/k    =  ( 1.50 p) ² / p²

        k'/k     = 2.25

         k' =  2.25 ( k)

       k'  =  2.25 k

therefore the percentage change is ,

(k' - k ) / k x 100 = ( 2.25 k - k ) / k x 100 = 125 % .

therefore the % change in the kinetic energy is 125 % 

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hope it is useful...!!

Answer 2

Answer:

125%

Explanation:

pf = pi + 50/100 pi

pf = 150/100 pi

change in K.E = pf^2/2m - pi^2/2m

= 5pi^2/8m

% change = 5/4 *100 = 125%