Let the velocity of the spheres before collision u
And the mass of them is m and 3m
Let there speed after collision be v1 and v2
Hence,
Velocity of approach = velocity of separation
u1−u2=v2−v1u−(−u)=v2−v1v2−v1=2uv2=2u+v1...(i)
Now from the conservation of momentum
m1u1+m2u2=m1v1+m2v2mu+3m(−u)=mv1+3mv2mu−3mu=mv1+3mv2−2mu=mv1+3mv2v1+3v2=−2u...(ii)
Solve (i) and (ii)
v1+3(2u+v1)=−2uv1+6u+3v1=−2uv1+3v1=−2u−6u4v1=−8uv1=−2u
Putting this in (i)
v2=2u+(−2u)v2=0
hence the 1st sphere will come back with a speed twice of its initial and the 2nd sphere will attains at rest.
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