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If momentum of an object is increased by 10%,then its kinetic energy will b increased by--?

using relation:

E=P 2/2m

let pi= initial momentum

then according to question,final momentum pf=10% of pi+pi

=11pi/10

initial kinetic energy Ei=Pi2/2m

 

final kinetic energy Ef=Pf2/2m=(11pi/10)2/2m


 

=121pi2/200m

 

so change in kinetic energy = Ef – Ei


 

= (121Pi2/200m) - (Pi2 /2m)


 

= 21Pi2/200m


 

= 21/200(Pi2/2m)


 

= 21% of Ei


 

so, the kinetic energy increases by 21% when momentum increases by 10%

 

 

@Anushree....Very nicely done, thanks for the help!

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momentum = mass X velocity..........(mass always remains constant, and increase in momentum means increase in the velocity only)

let initial velocity = v  m/s, kinetic energy = 1/2 mv2...............(I)

hence, increase in velocity = 10 % i.e., 10/100 X v = v/10.

now, final velocity = v+(v/10) = 11v/10.

final kinetic energy = 1/2 m (11v/10)2.....................(II)

now, change in kinetic energy = (II) - (I) = 1/2 m [ (11v/10) + v ] [ (11v/10) - v ]

  = 1/2m ( 21v/10* v/10) = 1/2m* 21v2/100 ................(III)

% increase = change/initial * 100

= [1/2m* 21v2/100] / 1/2 mv2*100

= 21 %.

hope this helped.....

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