# Conservation of momentum in Newton’s cradle

collisionenergy-conservationmomentumnewtonian-mechanics

Imagine a Newton's cradle with 5 balls with mass of each ball is $$m$$. In a case where two balls are dropped against three balls, if we write an equation considering that momentum is conserved,
$$2mu=2mv_1+3mv_2$$
where $$u$$ is the initial velocity of the two balls dropped, $$v_1$$ is the final velocity of the two balls dropped, $$v_2$$ is the velocity of the three balls that were stationary.

And an equation for the kinetic energy conserved (assuming the collision is elastic),
$$\frac{1}{2}\left(2mu^2\right)=\frac{1}{2}\left(2mv_1^2\right)+\frac{1}{2}\left(3mv_2^2\right)$$
And solving the two equations, $$v_1$$ doesn't equate to $$0$$ and $$v_2$$ doesn't equate to $$u$$. But in a Newton's cradle, a number of balls equal to the dropped number of balls always should go up from the other side with the same velocity. My question is why can't we use conservation laws considering the collision as between two systems, one with mass $$2m$$ and the other with mass $$3m$$?

Here are the balls' velocities at each step: $$\begin{eqnarray*} (u, u, 0, 0, 0)\\ (u, 0, u, 0, 0)\\ (0, u, 0, u, 0)\\ (0, 0, u, 0, u)\\ (0, 0, 0, u, u) \end{eqnarray*}$$