Mastering the Coefficient of Restitution in A-Level Mechanics

In A-Level mechanics, understanding how objects behave during a collision is essential for solving complex dynamics problems. Whether you are analysing the bounce of a ball or the impact of two vehicles, the coefficient of restitution provides the mathematical bridge between initial approach and final separation.

This article explores the definition of the coefficient of restitution, its role in Newton's law of restitution, and how to apply these concepts to solve exam-style problems. Mastering this topic will allow you to confidently distinguish between elastic and inelastic collisions and calculate post-impact velocities with precision.

Defining the Coefficient of Restitution

The coefficient of restitution, denoted by $e$, is a dimensionless quantity that measures the 'springiness' of a collision. It quantifies the ratio of the relative speed of separation to the relative speed of approach along the line of impact. Mathematically, for two bodies with velocities $u_1$ and $u_2$ before impact, and $v_1$ and $v_2$ after impact, Newton's law of restitution is defined as:

$$e = \frac{v_2 - v_1}{u_1 - u_2}$$

This value always lies in the range $0 \le e \le 1$. A value of $e = 1$ represents a perfectly elastic collision where kinetic energy is conserved, while $e = 0$ represents a perfectly inelastic collision where the objects coalesce and move as a single body.

Elastic vs Inelastic Collisions

Collisions are categorised by how they affect the total kinetic energy of the system. In an elastic collision ($e = 1$), both linear momentum and total kinetic energy are conserved. These are idealised scenarios often found in textbook problems involving billiard balls or subatomic particles.

In contrast, most real-world collisions are inelastic ($0 < e < 1$). During these impacts, some kinetic energy is dissipated as heat, sound, or through permanent deformation of the materials. A perfectly inelastic collision ($e = 0$) is the extreme case where the maximum possible kinetic energy is lost, and the objects stick together, sharing a common final velocity.

Worked Example 1: Direct Impact

Consider a sphere of mass $2\text{ kg}$ moving at $5\text{ ms}^{-1}$ towards a stationary sphere of mass $3\text{ kg}$. If the coefficient of restitution is $0.5$, find the velocities of both spheres after the collision.

Step 1: Conservation of Momentum $$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$(2 \times 5) + (3 \times 0) = 2v_1 + 3v_2$$ $$10 = 2v_1 + 3v_2 \quad \text{(Equation 1)}$$

Step 2: Newton's Law of Restitution $$e = \frac{v_2 - v_1}{u_1 - u_2}$$ $$0.5 = \frac{v_2 - v_1}{5 - 0}$$ $$2.5 = v_2 - v_1 \implies v_2 = v_1 + 2.5 \quad \text{(Equation 2)}$$

Step 3: Solve the simultaneous equations Substitute (2) into (1): $$10 = 2v_1 + 3(v_1 + 2.5)$$ $$10 = 2v_1 + 3v_1 + 7.5$$ $$2.5 = 5v_1 \implies v_1 = 0.5\text{ ms}^{-1}$$ $$v_2 = 0.5 + 2.5 = 3.0\text{ ms}^{-1}$$

Both spheres move in the direction of the initial velocity.

Worked Example 2: Rebound from a Surface

A ball is dropped from a height of $4\text{ m}$ onto a horizontal floor. If the coefficient of restitution between the ball and the floor is $0.6$, calculate the height to which the ball rebounds.

Step 1: Velocity before impact Using $v^2 = u^2 + 2as$ (where $u=0, a=9.8, s=4$): $$v^2 = 0 + 2(9.8)(4) = 78.4$$ $$v = \sqrt{78.4} \approx 8.85\text{ ms}^{-1}$$

Step 2: Velocity after impact Since the floor is stationary, $e = \frac{v_{\text{rebound}}}{v_{\text{impact}}}$. $$0.6 = \frac{v_{\text{rebound}}}{8.85}$$ $$v_{\text{rebound}} = 5.31\text{ ms}^{-1}$$

Step 3: Rebound height Using $v^2 = u^2 + 2as$ (where $v=0, u=5.31, a=-9.8$): $$0 = (5.31)^2 + 2(-9.8)h$$ $$19.6h = 28.2$$ $$h \approx 1.44\text{ m}$$

Common Mistakes

1. Sign Convention Errors: Always define a positive direction before starting. If an object moves in the opposite direction after impact, its velocity must be negative.
2. Confusing $e$ with Energy: Remember that $e$ is a ratio of velocities, not a ratio of kinetic energies. Kinetic energy loss is proportional to $1 - e^2$.
3. Ignoring the Line of Impact: Newton's law only applies along the line of impact. For oblique collisions, you must resolve velocities into components parallel and perpendicular to the surface.

Frequently Asked Questions

What does $e=0$ mean in physics? It represents a perfectly inelastic collision where the objects stick together after impact, resulting in the maximum possible loss of kinetic energy.

Is the coefficient of restitution a constant? It is generally treated as a constant for a specific pair of materials, but in reality, it can vary depending on the impact velocity and the geometry of the objects.

How do I calculate kinetic energy loss? Calculate the total kinetic energy before and after the collision using $\frac{1}{2}mv^2$. The difference is the energy lost to heat, sound, or deformation.

Conclusion

Understanding the coefficient of restitution is fundamental to mastering A-Level mechanics. By combining the conservation of momentum with Newton's law of restitution, you can solve almost any collision problem. To see these concepts in action with visual, step-by-step animations, head over to MathInstructor AI and generate your free animated lesson on collisions today.