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A particle of mass m has half the kinetic energy of another particle of mass m/2. If the speed of the heavier particle is increades by 2m/s, its new kinetic energy becomes equal to the original kinetic energy of the lighter particle. The ratio of the original speeds of the lighter and heavier particle is ????

8 years ago

147 Points

Dear ankit Shukla

let original speed is V1 and V2

given

1/2 mv12 =1/2  *1/2(m/2)v22

or  v12 =1/4 v22

or v1 =1/2 v2 ..............1

and 1/2 m(v1+2)2 =1/2(m/2)v22

or v1+2 =v2/ √2..................2

solve equation 1 and 2

v1=2/( √2-1)

v2=4/(√2-1)

Regards

8 years ago
alina !
33 Points

I dont understand why the 2nd statement of increment of kinetic energy is given......my ans acc to me is-

Soln-

Let the velocity of 1st particle is v and dat of lighter particle be v'.

Now,B/Q

1/2 m.v2= 1/2{1/2(m/2).v'2}

=>2 m.v2=mv'2/2

=>4mv2=v'2

=>v':v=2:1

ie ratio of speeds of lighter particle to heavier..

7 years ago
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