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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Kinetic Energy
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tinchubenz (5)

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When the K.E. of a body is increased by 300%, the momentum of the body is increased by:
a.20%
b.50%
c.100%
d.200%
    
harsha_27 (1343)

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200% because momentum (MV)=sqr.rt.[2m(k.e.)]...





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rooney (894)

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Nope. Its 100%.

Velocity is doubled. So, momentum is doubled. So, increase is of 100%.

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spideyunlimited (3493)

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KE = 1/2 m v^2

KE is increased by 300% means that new KE = 4 x original KE
[ original KE + (300 / 100) * original KE ]

so new KE = 4 * 1/2 * m * v^2
= 1/2 * m * (2v)^2
so velocity has doubled.

so if original momentum was mv
new momentum is m(2v)
= twice of original momentum
means momentum has increased by 100% [ mv + (100/100)mv ]

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karthik2007 (3375)

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The basis for such problems is simply this:

Kinetic Energy = (Momentum)2/2 x mass

Will nip in at times to solve problems :)
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