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 ????

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

Badiuddin

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..