Welcome to Restitution: The Physics of Collisions
Have you ever wondered why a basketball bounces right back up to your hands, while a lump of wet clay hits the floor with a dull thud and stays completely flat? Or why a cue ball in snooker transfers its motion so cleanly to another ball?
The answer lies in a property called restitution. In Mechanics, restitution describes how "bouncy" or elastic a collision between two objects is. In this chapter, you will learn how to model collisions mathematically using two core tools: Conservation of Linear Momentum (CLM) and Newton’s Experimental Law of Restitution (NLR).
Don't worry if this seems tricky at first! Once you master a standard step-by-step method for setting up your equations, solving collision problems becomes very systematic and rewarding.
1. Newton's Experimental Law (NLR) and the Coefficient of Restitution
Sir Isaac Newton discovered experimentally that when two objects collide directly along their line of motion, their relative speed of separation is directly proportional to their relative speed of approach.
The Fundamental Definition
The constant of proportionality is called the coefficient of restitution, denoted by the letter \(e\):
\(e = \frac{\text{Speed of Separation}}{\text{Speed of Approach}}\)
Alternatively, we can write this in a rearranged form that is often easier to use in calculations:
Speed of Separation = \(e \times\) Speed of Approach
The Value of \(e\)
The value of \(e\) is a dimensionless number (it has no units) that always lies in the range:
\(0 \le e \le 1\)
• Perfectly Elastic Collision (\(e = 1\)): No kinetic energy is lost. Objects bounce apart with the exact same relative speed with which they approached each other (like idealised pool balls or gas molecules).
• Inelastic / Semi-Elastic Collision (\(0 < e < 1\)): Real-world collisions (like tennis balls or cars). Some kinetic energy is converted into heat and sound.
• Perfectly Inelastic / Coalescing Collision (\(e = 0\)): The objects do not bounce apart at all; they stick together (coalesce) and move with a single common velocity.
Did You Know?
In professional tennis, the International Tennis Federation requires tournament balls to have a coefficient of restitution of roughly \(e \approx 0.73\) to \(0.76\) when dropped onto concrete from a height of 2.54 metres!
Direct Impact with a Fixed, Smooth Surface
When a particle hits a solid, fixed barrier (such as a rigid wall or the floor) perpendicular to the surface with speed \(u\), the barrier cannot move (speed of wall = \(0\)). Therefore:
• Speed of approach = \(u\)
• Speed of separation = \(v\)
Applying Newton's Law gives:
\(v = e u\)
The particle simply rebounds in the opposite direction with speed scaled down by \(e\).
Key Takeaway for Section 1
The coefficient of restitution \(e\) measures elasticity. Always remember: Separation = \(e \times\) Approach, with \(0 \le e \le 1\).
2. Direct Impact Between Two Moving Particles
A direct collision occurs when two particles or smooth spheres are moving along the same straight line joining their centres before they collide.
Setting Up the Problem: The Standard Diagram
Setting up your diagram with consistent positive directions is the secret to getting every collision question right.
Let two particles \(A\) and \(B\) have masses \(m_1\) and \(m_2\):
• Before collision: \(A\) moves with velocity \(u_1\), \(B\) moves with velocity \(u_2\).
• After collision: \(A\) moves with velocity \(v_1\), \(B\) moves with velocity \(v_2\).
Memory Aid: Always choose one direction (usually from left to right, \(\rightarrow\)) as positive for all velocities.
The Two Golden Equations
Every standard direct collision problem requires you to set up and solve a pair of simultaneous equations:
1. Conservation of Linear Momentum (CLM):
Since no external impulse acts on the two-particle system along the line of impact:
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
2. Newton's Experimental Law (NLR):
If both particles are moving to the right and \(A\) catches up with \(B\) (so \(u_1 > u_2\)), then after the bounce \(B\) must move faster to the right than \(A\) (so \(v_2 > v_1\)):
\(\text{Speed of Separation} = v_2 - v_1\)
\(\text{Speed of Approach} = u_1 - u_2\)
\(v_2 - v_1 = e(u_1 - u_2)\)
Step-by-Step Worked Example
Problem: Sphere \(A\) of mass \(2\text{ kg}\) moves with speed \(6\text{ m s}^{-1}\) towards sphere \(B\) of mass \(3\text{ kg}\) which is moving in the same direction with speed \(1\text{ m s}^{-1}\). The coefficient of restitution is \(e = 0.5\). Find the velocities of both spheres after the impact.
Step 1: Identify values and define positive direction (\(\rightarrow\))
\(m_1 = 2\text{ kg}\), \(u_1 = 6\text{ m s}^{-1}\)
\(m_2 = 3\text{ kg}\), \(u_2 = 1\text{ m s}^{-1}\)
\(e = 0.5\)
Step 2: Apply Conservation of Linear Momentum (CLM)
\((2)(6) + (3)(1) = 2 v_1 + 3 v_2\)
\(12 + 3 = 2 v_1 + 3 v_2\)
\(15 = 2 v_1 + 3 v_2\) --- [Equation 1]
Step 3: Apply Newton's Law of Restitution (NLR)
\(v_2 - v_1 = e(u_1 - u_2)\)
\(v_2 - v_1 = 0.5(6 - 1)\)
\(v_2 - v_1 = 2.5 \implies v_2 = v_1 + 2.5\) --- [Equation 2]
Step 4: Solve simultaneously
Substitute Equation 2 into Equation 1:
\(15 = 2 v_1 + 3(v_1 + 2.5)\)
\(15 = 2 v_1 + 3 v_1 + 7.5\)
\(15 - 7.5 = 5 v_1\)
\(7.5 = 5 v_1 \implies v_1 = 1.5\text{ m s}^{-1}\)
Now calculate \(v_2\):
\(v_2 = 1.5 + 2.5 = 4\text{ m s}^{-1}\)
Result: Sphere \(A\) moves at \(1.5\text{ m s}^{-1}\) and sphere \(B\) moves at \(4\text{ m s}^{-1}\), both in the original direction of motion.
Common Pitfall to Avoid
If particles are moving towards each other before collision, remember that their initial velocities have opposite signs! If \(A\) moves right at \(4\text{ m s}^{-1}\) and \(B\) moves left at \(3\text{ m s}^{-1}\), then \(u_1 = 4\) and \(u_2 = -3\). The speed of approach is \(4 - (-3) = 7\text{ m s}^{-1}\).
Key Takeaway for Section 2
Always solve direct collisions by combining CLM (\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)) with NLR (\(v_2 - v_1 = e(u_1 - u_2)\)). Take careful note of positive and negative signs for directions.
3. Loss of Kinetic Energy
In any collision where \(e < 1\), some mechanical energy is lost as thermal energy, acoustic energy, or internal deformation work.
Calculating the Loss of Kinetic Energy
The total initial kinetic energy of the system is:
\(E_{k,\text{initial}} = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2\)
The total final kinetic energy of the system is:
\(E_{k,\text{final}} = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2\)
The loss of kinetic energy, \(\Delta E_k\), is simply:
\(\Delta E_k = E_{k,\text{initial}} - E_{k,\text{final}}\)
Useful Algebraic Result
By combining CLM and NLR, the kinetic energy lost during a direct collision between two particles can also be expressed in the compact form:
\(\Delta E_k = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2 (1 - e^2)\)
Why this formula makes intuitive sense:
• If \(e = 1\), then \((1 - e^2) = 0\), so \(\Delta E_k = 0\) (no kinetic energy is lost).
• If \(e = 0\), \((1 - e^2) = 1\), producing the maximum possible loss of kinetic energy.
Key Takeaway for Section 3
Loss in \(E_k\) is always \(\text{Initial } E_k - \text{Final } E_k\). A positive value means energy was converted into heat/sound.
4. Successive Collisions
In some examination problems, a particle might bounce between multiple objects or hit a barrier and rebound to collide with another particle a second time.
Condition for a Second Collision
Consider three particles, \(A\), \(B\), and \(C\), lined up in order from left to right along a straight line.
If \(A\) strikes \(B\), then \(B\) moves off and strikes \(C\):
• After the first impact, \(B\) gets a velocity \(v_B\) and moves towards \(C\).
• \(B\) hits \(C\), and \(B\)'s velocity changes to \(v_B'\).
• A second collision between \(A\) and \(B\) will occur if and only if \(v_A > v_B'\) (meaning \(A\) is moving faster to the right than \(B\) is after \(B\) rebounds or slows down from \(C\)).
Bouncing Ball Dropped Under Gravity
When a ball is dropped from an initial height \(h_0\) onto a smooth horizontal floor:
• Speed just before the first impact: \(u_1 = \sqrt{2gh_0}\)
• Speed just after the first rebound: \(v_1 = e u_1 = e\sqrt{2gh_0}\)
• Maximum height reached after 1st bounce: \(h_1 = \frac{v_1^2}{2g} = e^2 h_0\)
• By extension, after \(n\) bounces, the rebound height is:
\(h_n = e^{2n} h_0\)
Key Takeaway for Section 4
For successive impacts, treat each collision as a separate problem. The final velocities from the first collision become the initial velocities for subsequent collisions.
5. Oblique Impact (Collisions at an Angle)
An oblique impact occurs when the velocity of a particle is not perpendicular to the surface it strikes, or not along the line of centres of two colliding spheres.
Case A: Oblique Impact with a Smooth Fixed Plane
Imagine a smooth sphere striking a smooth, fixed horizontal or vertical wall at an angle.
Let the line perpendicular (normal) to the wall be the \(y\)-axis, and the line parallel (tangential) to the wall be the \(x\)-axis.
1. Parallel to the Plane (Tangential Direction):
Because the surface is smooth, there is no frictional force along the plane. Therefore, the component of velocity parallel to the plane remains unchanged:
\(v_x = u_x\)
2. Perpendicular to the Plane (Normal Direction):
Newton's Law of Restitution applies directly along the normal to the plane:
\(v_y = e u_y\)
Angle and Speed Relations
If the incident velocity makes an angle \(\alpha\) with the normal to the wall, and the rebound velocity makes an angle \(\theta\) with the normal:
• \(u_x = u \sin \alpha\) and \(u_y = u \cos \alpha\)
• \(v_x = v \sin \theta\) and \(v_y = v \cos \theta\)
Since \(v_x = u_x \implies v \sin \theta = u \sin \alpha\)
And \(v_y = e u_y \implies v \cos \theta = e u \cos \alpha\)
Dividing the two equations gives:
\(\tan \theta = \frac{1}{e}\tan \alpha\)
The speed after impact is given by Pythagoras:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{(u \sin \alpha)^2 + (e u \cos \alpha)^2}\)
Case B: Oblique Impact Between Two Smooth Spheres
When two smooth uniform spheres collide obliquely:
• Along the Line of Centres: The impact forces act entirely along this line. Apply CLM and NLR along the line of centres exactly like a direct collision.
• Perpendicular to the Line of Centres: Since the spheres are smooth, no forces act perpendicular to the line of centres. Each sphere's velocity component in this perpendicular direction remains unchanged.
Key Takeaway for Section 5
In any oblique collision with smooth surfaces: components parallel to the contact plane never change; components perpendicular to the contact plane follow standard 1D restitution rules.
Quick Reference Summary & Exam Tips
Core Formulas at a Glance:
• Restitution Definition: \(\text{Speed of Separation} = e \times \text{Speed of Approach}\)
• CLM: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
• Direct Rebound from Wall: \(v = eu\)
• Oblique Wall Impact: \(v_x = u_x\), \(v_y = eu_y\), \(\tan \theta = \frac{1}{e}\tan \alpha\)
• Loss of KE: \(\Delta E_k = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \left(\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2\right)\)
Top 3 Exam Tips:
1. Always draw a clear diagram with arrows showing assumed directions of velocities after impact. If your calculated velocity turns out negative, it simply means the object moves in the opposite direction to your arrow.
2. Be careful with angles: Double check whether the angle given in the question is measured with the surface/wall or with the normal (perpendicular) to the surface.
3. Check your units: Ensure masses are in \(\text{kg}\) and speeds are in \(\text{m s}^{-1}\) before calculating kinetic energy.