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Cool goIITian

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20 Apr 2009 16:11:01 IST

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kinetic energy of a body is increased by 300%.What is the percentage increase in its momentum ??

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kinetic energy of a body is increased by 300%.What is the percentage increase in its momentum ??

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~goiit user~

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20 Apr 2009 16:14:57 IST

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K=P^2/2m

so differentiating

so 600%

Gaurav |spideyunlimited| Ragtah

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20 Apr 2009 16:37:30 IST

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Kinetic energy = 1/2 m v^2

Since it has increased by 300% , it means the KE has become thrice.. - specifically speaking, the new velocity is sqrt(3) times the original velocity.

So the new momentum (mv) is sqrt(3) times the original momentum

in terms of %, this is equal to 100.rt(3) or approx 173%

~goiit user~

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20 Apr 2009 16:48:10 IST

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@Gaurav

I dont think that is correct

When it is said that KE is increased it can be only mass, or only velocity, or both which are increased

dhyan sourav

Cool goIITian

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21 Apr 2009 05:23:32 IST

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Answer is 100%.But how please explain ??how ??

Gaurav |spideyunlimited| Ragtah

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21 Apr 2009 10:57:26 IST

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Ohhh, I got what you're talking about.

The question says INCREASED by 300%, so the new KE is actually 400% of original KE. Which means, the new velocity is 2 times the original velocity (KE = 1/2m (2v)^2 = 4*1/2 mv^2)

So the new momentum = m(2v) = 2mv, which is 200% of the original momentum, that means the momentum increased by 100%

dhyan sourav

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24 Apr 2009 14:38:45 IST

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Gaurav is right.......!!

BALGANESH

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24 Apr 2009 14:50:40 IST

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go for usual method instead...

K=P^2/2m...

after the increse in KE 4K=p'^2/2m

divide the equations

u get p/p'=1/2 p'=2p

so % increase in p is p'-p/p* *100 = 2p-p/p*100 = p/p*100 = 100%

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