Question #249271
Two practically elastic spheres collide. If their velocities before impact have same magnitude and opposite directions and one has a mass 3 times the other, what are their velocities after impact?
1
Expert's answer
2021-10-10T18:11:21-0400

Let the velocity of the spheres before collision u

And the mass of them is m and 3m

Let there speed after collision be v1v_{1} and v2v_{2}

Hence,

Velocity of approach = velocity of separation

u1u2=v2v1u(u)=v2v1v2v1=2uv2=2u+v1...(i)u_{1}-u_{2}=v_{2}-v_{1} \\ u-(-u)=v_{2}-v_{1} \\ v_{2}-v_{1}=2u \\ v_{2}=2u+v_{1}...(i)

Now from the conservation of momentum

m1u1+m2u2=m1v1+m2v2mu+3m(u)=mv1+3mv2mu3mu=mv1+3mv22mu=mv1+3mv2v1+3v2=2u...(ii)m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2} \\ mu+3m(-u)=mv_{1}+3mv_{2} \\ mu-3mu=mv_{1}+3mv_{2} \\ -2mu=mv_{1}+3mv_{2} \\ v_{1}+3v_{2}=-2u... (ii)

Solve (i) and (ii)

v1+3(2u+v1)=2uv1+6u+3v1=2uv1+3v1=2u6u4v1=8uv1=2uv_{1}+3(2u+v_{1})=-2u \\ v_{1}+6u+3v_{1}=-2u \\ v_{1}+3v_{1}=-2u-6u \\ 4v_{1}=-8u \\ v_{1}=-2u

Putting this in (i)

v2=2u+(2u)v2=0v_{2}=2u+(-2u) \\ v_{2}=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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