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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 ????
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
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..
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..
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