18 Collisions
18.1 Collision of Two Bodies
Question: What happens in reality when two cars collide?
In reality, colliding bodies deform — possibly permanently — and their rotational inertia may change.
Ignoring rotational motion, the simplest model uses a mass particle for each body. Since particles do not deform, we introduce the coefficient of restitution \(e\).
18.2 Frictionless, Oblique, Central Impact of Two Bodies
18.2.1 Assumptions
- Model each body as a mass particle.
- We ignore the rotational inertias of bodies (purely translational motion).
- All points on a body have the same velocity.
- No friction between particles during collision — all impact forces are along \({\bf n}\).
- The duration of compression (\(t_1-t_0\)) and restitution (\(t_2-t_1\)) are very small.
- At time \(t_1\), both particles have the same velocity along \({\bf n}\): \(v_{II}\).
- \({\bf F}_{1C} = -{\bf F}_{2C}\), \({\bf F}_{1R} = -{\bf F}_{2R}\).
18.2.2 Impact Stages
Let \({\bf n}\) be the common unit normal between the bodies at contact. Define \(\{{\bf n},{\bf t}_1,{\bf t}_2\}\) as an orthonormal set. These vectors are assumed constant for the duration of impact.
18.2.3 Linear Impulses During Impact
Question: Draw the FBDs of \(m_1\) and \(m_2\) separately during compression and restitution.
- \({\bf F}_{1_d} = F_{1_d}{\bf n}\): force exerted by body 2 on body 1 during compression.
- \({\bf F}_{1_r} = F_{1_r}{\bf n}\): force exerted by body 2 on body 1 during restitution.
- \({\bf F}_{2_d}\), \({\bf F}_{2_r}\): reaction forces by body 1 on body 2.
- All other forces have resultants \({\bf R}_1\) and \({\bf R}_2\).
Since the collision duration is very small, the impulses of \({\bf R}_1\) and \({\bf R}_2\) vanish. Approximating: \[\begin{align} \int_{t_0}^{t_1}\lp{\bf F}_{1_d}+{\bf R}_1\rp d\tau &\approx \int_{t_0}^{t_1}{\bf F}_{1_d}\,d\tau,\\ \int_{t_1}^{t_2}\lp{\bf F}_{1_r}+{\bf R}_1\rp d\tau &\approx \int_{t_1}^{t_2}{\bf F}_{1_r}\,d\tau, \end{align}\] and similarly for body 2.
With equal and opposite collisional forces (\({\bf F}_{1_d}=-{\bf F}_{2_d}\), \({\bf F}_{1_r}=-{\bf F}_{2_r}\)), we define the coefficient of restitution: \[\begin{align} e = \frac{\int_{t_1}^{t_2}{\bf F}_{1_r}\cdot{\bf n}\,d\tau}{\int_{t_0}^{t_1}{\bf F}_{1_d}\cdot{\bf n}\,d\tau}. \end{align}\]
Question: What does \(e=0\) and \(e=1\) mean?
- \(e=1\): perfectly elastic collision — compression impulse equals restitution impulse.
- \(e=0\): perfectly plastic collision — no restitution impulse.
- In general \(0\leq e\leq 1\), determined experimentally.
18.2.4 The Coefficient of Restitution in Terms of Velocities
From the linear impulse-momentum equations for each particle and stage: \[\begin{align} \begin{split} m_1 v_{II} - m_1{\bf v}_1\cdot{\bf n} &= \int_{t_0}^{t_1}{\bf F}_{1_d}\cdot{\bf n}\,d\tau,\\ m_2 v_{II} - m_2{\bf v}_2\cdot{\bf n} &= \int_{t_0}^{t_1}{\bf F}_{2_d}\cdot{\bf n}\,d\tau,\\ m_1{\bf v}_1'\cdot{\bf n}-m_1 v_{II} &= e\int_{t_0}^{t_1}{\bf F}_{1_d}\cdot{\bf n}\,d\tau,\\ m_2{\bf v}_2'\cdot{\bf n}-m_2 v_{II} &= e\int_{t_0}^{t_1}{\bf F}_{2_d}\cdot{\bf n}\,d\tau. \end{split} \end{align}\]
Eliminating \(v_{II}\) yields the familiar expression: \[\begin{align} e = \frac{{\bf v}_2'\cdot{\bf n}-{\bf v}_1'\cdot{\bf n}}{{\bf v}_1\cdot{\bf n}-{\bf v}_2\cdot{\bf n}}. \end{align}\]
When one object impacts a massive constrained object (e.g. a wall), the velocity of the massive object is unaffected by the collision.
18.3 Example: The Most General Planar Impact Problem
Question: Given particles \(m_1\) and \(m_2\) with pre-impact velocities \({\bf v}_1\) and \({\bf v}_2\), find the post-impact velocities. The coefficient of restitution \(e\) is given.
There are 4 unknowns (components of \({\bf v}_1'\) and \({\bf v}_2'\) along \({\bf n}\), \({\bf t}_1\), \({\bf t}_2\)). Along the tangent directions: velocities are unchanged. Along \({\bf n}\): we have the conservation of linear momentum and the restitution equation.
18.4 Energy Loss During Impact
Pre-impact and post-impact kinetic energies: \[\begin{align} T &= \frac{1}{2}m_1{\bf v}_1\cdot{\bf v}_1+\frac{1}{2}m_2{\bf v}_2\cdot{\bf v}_2,\\ T' &= \frac{1}{2}m_1{\bf v}_1'\cdot{\bf v}_1'+\frac{1}{2}m_2{\bf v}_2'\cdot{\bf v}_2'. \end{align}\]
The energy lost: \[\begin{align} T-T' = \frac{m_1 m_2}{2(m_1+m_2)}\lp{\bf v}_1\cdot{\bf n}-{\bf v}_2\cdot{\bf n}\rp^2(1-e^2). \end{align}\]
Hence, for \(0\leq e\leq 1\), kinetic energy cannot increase as a result of impact.
18.5 Negative Values of the Coefficient of Restitution
Question: What does a negative \(e\) mean for a ball impacting a wall?
When \(e < 0\), \({\bf v}_1'\cdot{\bf n}-{\bf v}_2'\cdot{\bf n}\) has the same sign as \({\bf v}_1\cdot{\bf n}-{\bf v}_2\cdot{\bf n}\), meaning the bodies pass through each other during impact. To prevent this, we generally require \(e \geq 0\).