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# if the kinetic energy of a body is increased by 100% ,then the change in momentum of the body is: pls. Explain.

10 years ago

300 percent

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as we know the relation b/w kinetic energy & momentum as KINETIC ENERGY=P2/2M,

WHERE,,, p=momentum

now substitute in the relation u will get the answer as 300 percent

10 years ago

As KE incr. 100%, it bcms 2 times.

As KE direct. prop. to v2, so v bcms √2 times.

As v direct. prop. to p, so p bcms √2 times, so change in p = √2E - E / E Χ 100

= 100(√2 - 1) %

10 years ago

Let k1 be the initial kinetic energy,k2 be the final kinetic energy.

let p1 be intial momentum & p2 be final momentum.

acc. to question,

k2=k1+100%k1

k2=k1+k1=2k1

1\2(mV2)2=2(1\2mV12)

V2=√2V1

on mutiplying by mass m,

mV2=√2mV1

p2=√2p1.         now,

Δp=p2-p1             =>p1(√2-1)

%age of change in momentum=100(Δp\p1)

=(1.414-1)100

=41.4%

I Hope u understand