# Particles velocities directly after an inelastic impact with a smooth frictionless surface

Let us consider the following situation where a dumbbell falls onto a smooth and frictionless surface.

The dumbbell is formed of 2 particles, each of mass $m$, linked by a massless rod of length $l$. Before impacting with the surface, $\theta=45^{\circ}$ and $\dot\theta=0\,rad.s^{-1}$. Since the surface is smooth, the collision is considered inelastic and there is no rebound.

Now, the only impulse applied to particle 1 upon impact should be vertical since the surface is frictionless. Also, because the collision is inelastic, particle 1 should slide away horizontally from its initial position. Since there is no horizontal impulse, the dumbbell center of mass horizontal linear momentum is equal to zero.

I tried to imagine what would happen directly after the collision (drawn in light gray) and since there is no horizontal impulse, particles 1 and 2 would move horizontally away from their initial positions with the same speed. The center of mass will obviously continue to fall down in a strictly vertically. So, if we take particle 1 as a reference point $P$, particle 2 velocity directly after the impact would be

$$\mathbf{v}_2 = l\dot\theta\vec{e_\theta} = l\dot\theta \left( -\frac{\sqrt{2}}{2}\vec{i} + \frac{\sqrt{2}}{2}\vec{j} \right)$$

where $\vec{e_\theta}$ is perpendicular to the radial unit vector; $\vec{i}$ and $\vec{j}$ are unit vectors of the $(x,y)$ frame.

Now, since the two particles are moving horizontally away from their respective initial positions and the center of mass is not, I assume that their horizontal velocities must be equal and opposite. But since the particle 1 already has a velocity after impact, I am not sure if the equation above can hold. Also, since it is an inelastic contact, there should be some velocity loss. All of these factors are confusing to me.

How am I supposed to be calculating the velocities of these two particles?

I finally found a way to obtain these velocities with calculus. I am sure that one can always try to deduce the particles velocities by mere logic; in fact, more than once, I "thought" about the answer that I finally found. But I was not comfortable without a calculus development.

Finding the velocity of particle 1

We can first calculate the velocity of the system center of mass by taking particle 1 as a reference.

$$\vec{v}_c = \vec{v}_1 + \dot\theta\vec{k} \times \frac{l}{2}\vec{e}_r$$

where $\vec{v}_c$ is the system center of mass velocity; $\vec{v}_1$ is the total velocity of particle 1; $\dot\theta\vec{k}$ is the angular velocity around an axis perpendicular to the $(x,y)$ plane; $\times$ is the cross product operator ;and $\vec{e}_r$ is the radial vector along the massless rod.

We know that particle 1 cannot have a vertical velocity because of the assumption of an inelastic collision. So the equation above becomes:

$$\vec{v}_c = \vec{v}_{1x} + \dot\theta\vec{k} \times \frac{l}{2}\vec{e}_r$$

where $\vec{v}_{1x}$ is the horizontal velocity of particle 1. By developing the above equation, we end up with:

$$\vec{v}_c = \vec{v}_{1x} + \frac{l\dot\theta\sqrt{2}}{4} \left( -\vec{i} + \vec{j} \right)$$

Now, if the system center of mass does not move horizontally, that means that the horizontal component of its velocity has to be zero. Consequently, in the above equation:

$$\vec{v}_{1x} - \frac{l\dot\theta\sqrt{2}}{4}\vec{i} = 0$$

and since the vertical velocity of particle 1 is simply zero, then:

$$\vec{v}_{1} = \frac{l\dot\theta\sqrt{2}}{4}\vec{i}$$

Finding the velocity of particle 2

The velocity of particle 2 can now be simply calculated as:

$$\vec{v}_2 = \vec{v}_1 + \dot\theta\vec{k} \times l\vec{e}_r$$

and by developing the above equation :

$$\vec{v}_2 = \frac{l\dot\theta\sqrt{2}}{2} \left( -\frac{1}{2}\vec{i} + \vec{j} \right)$$