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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 ????

147 Points
12 years ago

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

alina !
33 Points
11 years ago

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..