8.5: Collisions
An important application of the law of momentum conservation is the dynamic analysis of collisions between two objects, like the collision between two billiard balls (see Figure 8.3). We note that methods from previous chapters often cannot be easily applied because the forces and accelerations during a collision are often not accurately known. This inaccuracy is caused by the fast dynamics and high forces during collisions, that are hard to measure. During such a short collision, other forces can usually be neglected, because they are much smaller than the very high collision forces that provide the high accelerations which change the velocity vectors in a short time. As a consequence of the absence of substantial external forces, the law of conservation of momentum Equation 8.12 holds for the system containing both balls. We will now discuss step by step how the dynamics of the collision of two balls can be analysed using momentum conservation and the CoM-frame.
8.5.1 Plane of contact and line of impact
Let us consider two smooth spherical balls \(A, B\) with identical radius, with masses \(m_{A}, m_{B}\) and velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A}\) and \(\overrightarrow{\boldsymbol{v}}_{B}\) that move in the \(x y\)-plane and collide, as shown in Figure 8.3. The motion of the balls consists of 3 phases: 1. before the collision at time \(t_{0}, 2\). during the collision at time \(t_{1}\) and 3 . after the collision at time \(t_{2}\).
We can then choose our Cartesian coordinate system such that the origin \(O\) is at the point of contact. The \(y\)-axis is chosen to be tangential to the surface of the circular disks at the point where they make contact, this surface is called the plane of contact. The \(x\)-axis is chosen perpendicular to the plane of contact and is called the line of impact. The balls \(A\) and \(B\) and velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A}\), \(\overrightarrow{\boldsymbol{v}}_{B}\) can be drawn as shown in Figure 8.3).
The analysis of collisions becomes easier if it is performed in a CoM-frame.
A CoM-frame is a reference frame that is chosen such that the velocity and acceleration of the CoM of the objects is zero.
The analysis process then proceeds along the following steps:
- Determining the velocity vector \(\overrightarrow{\boldsymbol{v}}_{G}\) of the CoM.
- Transforming to the CoM-frame by subtracting the velocity of the centre of mass \(\overrightarrow{\boldsymbol{v}}_{G}\) from all velocity vectors.
- Analysing the system in the centre of mass system.
- Transforming back to the original system by adding \(\overrightarrow{\boldsymbol{v}}_{G}\) to all velocity vectors.
8.5.2 Transformation to the CoM-frame
Since there are no external forces acting on the balls (we neglect gravity), it is a closed system in which momentum conservation holds, such that the velocity \(\overrightarrow{\boldsymbol{v}}_{G}\) of the centre of mass \((\mathrm{CoM})\) is constant. The analysis can be significantly simplified by choosing a coordinate system \(x^{\prime}, y^{\prime}, z^{\prime}\) that moves along with the CoM-frame, such that, measured in these coordinates, the velocity of the CoM is zero \(\overrightarrow{\boldsymbol{v}}_{G}^{\prime}=\overrightarrow{\mathbf{0}}\). A transformation to a different coordinate system like shown in Figure 8.5 was already discussed in Equation 6.24 and is often useful for simplifying the analysis of problems, and therefore an important technique to become familiar with (see Sec. 6.12). The position and velocity vectors in the CoM-frame are found by first determining \(\overrightarrow{\boldsymbol{v}}_{G}\) (using Equation 7.41) and then calculating the velocities \(\overrightarrow{\boldsymbol{v}}^{\prime}\) in the CoM-frame by subtracting \(\overrightarrow{\boldsymbol{v}}_{G}\) :
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{G} & =\frac{m_{A} \overrightarrow{\boldsymbol{v}}_{A}+m_{B} \overrightarrow{\boldsymbol{v}}_{B}}{m_{A}+m_{B}} \tag{8.16} \label{8.16}\\[4pt] \overrightarrow{\boldsymbol{v}}_{A 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{A 0}-\overrightarrow{\boldsymbol{v}}_{G} \tag{8.17} \label{8.17}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{B 0}-\overrightarrow{\boldsymbol{v}}_{G} \tag{8.18} \label{8.18}\\[4pt] \overrightarrow{\boldsymbol{v}}_{G}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{G}-\overrightarrow{\boldsymbol{v}}_{G}=\overrightarrow{\mathbf{0}} \tag{8.19} \label{8.19}\end{align}\]
The result of this transformation to the CoM-frame can be seen in Figure 8.4.
8.5.3 Analysing the collision in the CoM-frame
First we note that in the CoM-frame the velocity vectors of both balls are always parallel, pointing in opposite directions, as can be derived from \(\overrightarrow{\boldsymbol{v}}_{G}^{\prime}=\overrightarrow{\mathbf{0}}\) :
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{G}^{\prime}=\frac{m_{A} \overrightarrow{\boldsymbol{v}}_{A}^{\prime}+m_{B} \overrightarrow{\boldsymbol{v}}_{B}^{\prime}}{m_{A}+m_{B}} & =\overrightarrow{\mathbf{0}} \tag{8.20} \label{8.20}\\[4pt] m_{A} \overrightarrow{\boldsymbol{v}}_{A}^{\prime}+m_{B} \overrightarrow{\boldsymbol{v}}_{B}^{\prime} & =\overrightarrow{\mathbf{0}} \tag{8.21} \label{8.21}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B}^{\prime} & =-\frac{m_{A}}{m_{B}} \overrightarrow{\boldsymbol{v}}_{A}^{\prime} \tag{8.22} \label{8.22}\end{align}\]
Now we can analyse the collision in the CoM-frame in three steps that are illustrated in Figs. 8.4):
- Before the impact, at time \(t=t_{0}\) the velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A 0}^{\prime}, \overrightarrow{\boldsymbol{v}}_{B 0}^{\prime}\) point toward each other.
- During the impact both balls exert forces \(\overrightarrow{\boldsymbol{F}}_{A B}=-\overrightarrow{\boldsymbol{F}}_{B A}=F_{A B} \hat{\boldsymbol{\imath}}\) on each other along the \(x\)-axis and slightly deform at the contact point. At the time of maximal deformation \(t=t_{1}\), the velocity \(x\)-components of both balls are zero \({ }^{2} v_{A 1, x}^{\prime}=v_{B 1, x}^{\prime}=0\).
- After the collision, the energy that was stored in the deformation is converted back to kinetic energy, at \(t=t_{2}\) the velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}, \overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}\) point away from each other.
The subscripts \(0,1,2\) are used to designate the velocities of the balls at times \(t_{0}, t_{1}\) and \(t_{2}\). To analyse the change in momentum during the impact, we apply the principle of impulse and momentum Equation 8.9 along the \(y\) and \(x\) axes. Because the balls are smooth and frictionless, there are no forces in the \(y\)-direction tangential to the plane of contact, such that all forces act along the line of impact and are parallel to the \(x\)-axis. Since the forces and impulses along the \(y\)-axis are zero, the momentum and velocity components of both balls are conserved \(\Delta p_{y}^{\prime}=m \Delta v_{y}^{\prime}=0\) along that axis:
\[\begin{align} v_{A 2, y}^{\prime} & =v_{A 0, y}^{\prime} \tag{8.23} \label{8.23}\\[4pt] v_{B 2, y}^{\prime} & =v_{B 0, y}^{\prime} \tag{8.24} \label{8.24}\end{align}\]
Along the \(x\)-axis it is more difficult to analyse the collision, since the contact forces along this axis are unknown, and can depend on the material properties of the balls and other conditions that affect the collision. These properties are conveniently captured by the coefficient of restitution \(e\), that relates the velocity of the ball in the CoM at time \(t_{2}\) to that at \(t_{0}\) :
\[\begin{align} v_{A 2, x}^{\prime} & =-e v_{A 0, x}^{\prime} \tag{8.25} \label{8.25}\\[4pt] v_{B 2, x}^{\prime} & =-e v_{B 0, x}^{\prime} \tag{8.26} \label{8.26}\end{align}\]
Note that these equations follow from momentum conservation, ensuring that Equation 8.21 still holds after the collision. The coefficient \(e\), that can have any value between 0 and 1 , is closely related to the energy loss during the collision as we will discuss later. Combining Eqs. (8.23)-(8.26) we have:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2}^{\prime} & =-e v_{A 0, x}^{\prime} \hat{\boldsymbol{\imath}}+v_{A 0, y}^{\prime} \hat{\boldsymbol{\jmath}} \tag{8.27} \label{8.27}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2}^{\prime} & =-e v_{B 0, x}^{\prime} \hat{\boldsymbol{\imath}}+v_{B 0, y}^{\prime} \hat{\boldsymbol{\jmath}} \tag{8.28} \label{8.28}\end{align}\]
Finally, we transform back to the original coordinate system (Figure 8.3 at \(t_{2}\) ) by adding the CoM velocity \(\overrightarrow{\boldsymbol{v}}_{G}\) to all velocity vectors:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2} & =\overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G} \tag{8.29} \label{8.29}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2} & =\overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G} \tag{8.30} \label{8.30}\end{align}\]
The whole outlined procedure can also be captured in a single set of equations:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2} & =\left[-e\left(v_{A 0, x}-v_{G, x}\right)+v_{G, x}\right] \hat{\boldsymbol{\imath}}+v_{A 0, y} \hat{\boldsymbol{\jmath}} \tag{8.31} \label{8.31}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2} & =\left[-e\left(v_{B 0, x}-v_{G, x}\right)+v_{G, x}\right] \hat{\boldsymbol{\imath}}+v_{B 0, y} \hat{\boldsymbol{\jmath}} \tag{8.32} \label{8.32}\end{align}\]
Thus we have analysed the dynamics of collisions with oblique impact (impact at an angle) by only using the conservation of momentum principle, without knowing any details about the forces, except the restitution coefficient \(e\). In the special case that the balls both move parallel to the line of impact ( \(x\)-axis) they experience central impact, where all previous equations still hold with all velocity components along the \(y\)-axis being zero \(\left(v_{A 0, y}=v_{B 0, y}=0\right)\). Note that since we are dealing with spherical balls, the centres of mass of both objects always coincide with the line of impact. For non-circular objects, this is not always the case, and the impact can become eccentric such that the objects can start rotating after the collision. Also if the surface of the objects is not smooth, friction forces might generate rotations under oblique impact. Impact with rotations will be discussed in chapter Ch. 12.
Let us analyse a numerical example of the situation sketched in Figure 8.3 and Figure 8.4 with \(e=0.6, m_{A}=3 \mathrm{~kg}\) and \(m_{B}=5 \mathrm{~kg}\). First we determine the velocity vector \(\overrightarrow{\boldsymbol{v}}_{G}\) of the centre of mass:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 0} & =[3.5 \hat{\boldsymbol{\imath}}-1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.33} \label{8.33}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 0} & =[-0.5 \hat{\imath}+2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.34} \label{8.34}\\[4pt] \overrightarrow{\boldsymbol{v}}_{G} & =\left[\frac{3 \cdot 3.5+5 \cdot-0.5}{3+5} \hat{\boldsymbol{\imath}}+\frac{3 \cdot-1.5+5 \cdot 2.5}{3+5} \hat{\boldsymbol{\jmath}}\right] \mathrm{m} / \mathrm{s} \tag{8.35} \label{8.35}\\[4pt] & =[1 \hat{\imath}+1 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.36} \label{8.36}\end{align}\]
Then we transform to the centre of mass system:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{A 0}-\overrightarrow{\boldsymbol{v}}_{G}=[(3.5-1) \hat{\boldsymbol{\imath}}+(-1.5-1) \hat{\jmath}] \mathrm{m} / \mathrm{s} \tag{8.37} \label{8.37}\\[4pt] & =[2.5 \hat{\boldsymbol{\imath}}-2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.38} \label{8.38}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{B 0}-\overrightarrow{\boldsymbol{v}}_{G}=[(-0.5-1) \hat{\boldsymbol{\imath}}+(2.5-1) \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.39} \label{8.39}\\[4pt] & =[-1.5 \hat{\boldsymbol{\imath}}+1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.40} \label{8.40}\end{align}\]
We multiply the \(x\)-components by \(-e=-0.6\) to obtain the velocities in the CoM-frame after the collision:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}=[-e \cdot 2.5 \hat{\boldsymbol{\imath}}-2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s}=[-1.5 \hat{\boldsymbol{\imath}}-2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.41} \label{8.41}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}=[-e \cdot-1.5 \hat{\boldsymbol{\imath}}+1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s}=[0.9 \hat{\boldsymbol{\imath}}+1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.42} \label{8.42}\end{align}\]
Finally we add again the centre of mass velocity vector to obtain the velocities in the original system after the impact:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2} & =\overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G}=[(-1.5+1) \hat{\boldsymbol{\imath}}+(-2.5+1) \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.43} \label{8.43}\\[4pt] & =[-0.5 \hat{\boldsymbol{\imath}}-1.5 \hat{\jmath}] \mathrm{m} / \mathrm{s} \tag{8.44} \label{8.44}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2} & =\overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G}=[(0.9+1) \hat{\boldsymbol{\imath}}+(1.5+1) \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.45} \label{8.45}\\[4pt] & =[1.9 \hat{\boldsymbol{\imath}}+2.5 \hat{\jmath}] \mathrm{m} / \mathrm{s} \tag{8.46} \label{8.46}\end{align}\]
As an exercise, you can check if there is momentum conservation during this collision.
8.5.4 Coefficient of restitution
The coefficient of restitution \(e\), that was introduced to capture the effect of the forces during impact, can be used as a measure for the kinetic energy that is lost during the collision in the CoM-frame. To analyse this energy loss, we use that the kinetic energy of ball \(A\) is given by:
\[T_{A}^{\prime}=\frac{1}{2} m_{A}\left(v_{A, x}^{\prime}{ }^{2}+v_{A, y}^{\prime}{ }^{2}\right) \tag{8.47} \label{8.47}\]
If we only consider the kinetic energy \(T_{A, x}^{\prime}\) contributed by the \(x\)-component of the velocity vector, we find, using Equation 8.25 that the kinetic energy after the collision is:
\[T_{A 2, x}^{\prime}=\frac{1}{2} m_{A} v_{A 2, x}^{\prime}{ }^{2}=\frac{1}{2} m_{A} e^{2} v_{A 0, x}^{\prime}{ }^{2}=e^{2} T_{A 0, x}^{\prime} \tag{8.48} \label{8.48}\]
And the same equation holds for ball \(B\). So, the \(x\)-axis kinetic energy \(T_{A, x}^{\prime}\)
reduces by a factor \(e^{2}\) during the collision, while the \(y\)-axis contribution remains constant \(T_{A 2, y}^{\prime}=T_{A 0, y}^{\prime}\). The largest energy reduction occurs for \(e=0\), which is called a plastic or perfectly inelastic collision. Kinetic energy is conserved \(\left(T_{A 2, x}^{\prime}=T_{A 0, x}^{\prime}\right)\) for \(e=1\), which is called an elastic collision. The restitution coefficient can have any value between 0 and \(1(0 \leq e \leq 1)\). Thus the coefficient of restitution provides information on energy conservation during collisions.