Answer to Question #249271 in Electric Circuits for Jayson

Question

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?

Answer 1

Let the velocity of the spheres before collision u

And the mass of them is m and 3m

Let there speed after collision be $v_{1}$ and $v_{2}$

Hence,

Velocity of approach = velocity of separation

$u_{1}-u_{2}=v_{2}-v_{1} \\ u-(-u)=v_{2}-v_{1} \\ v_{2}-v_{1}=2u \\ v_{2}=2u+v_{1}...(i)$

Now from the conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2} \\ mu+3m(-u)=mv_{1}+3mv_{2} \\ mu-3mu=mv_{1}+3mv_{2} \\ -2mu=mv_{1}+3mv_{2} \\ v_{1}+3v_{2}=-2u... (ii)$

Solve (i) and (ii)

$v_{1}+3(2u+v_{1})=-2u \\ v_{1}+6u+3v_{1}=-2u \\ v_{1}+3v_{1}=-2u-6u \\ 4v_{1}=-8u \\ v_{1}=-2u$

Putting this in (i)

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

hence the 1st sphere will come back with a speed twice of its initial and the 2nd sphere will attains at rest.