Further Mechanics: Elastic Collisions and Restitution

Understanding collisions is fundamental to physics and engineering, from designing safer vehicles to analyzing the motion of celestial bodies. This study moves beyond simple momentum conservation to model the realism of impacts—how bouncy or inelastic they are. Mastering the concepts here allows you to predict the aftermath of any collision with precision, a core skill in Further Mechanics.

Newton's Experimental Law and the Coefficient of Restitution

When two bodies collide, their relative speed after impact is related to their relative speed before impact. This relationship is described by Newton's Experimental Law (NEL). It introduces a crucial property: the coefficient of restitution, denoted by $e$. The law is stated mathematically as:

$$\text{Speed of separation}=e\times \text{Speed of approach}$$

For two particles, A and B, colliding directly, this translates to: $$v_B-v_A=e(u_A-u_B)$$ where $u$ and $v$ represent velocities before and after collision, respectively, with direction sign carefully maintained.

The coefficient $e$ is a dimensionless number between 0 and 1. A collision where $e=1$ is perfectly elastic, meaning no kinetic energy is lost. A collision where $e=0$ is perfectly inelastic (or plastic); the bodies coalesce and move together with maximum kinetic energy loss. Any value $0<e<1$ defines a partially elastic or inelastic collision, which is the most common real-world scenario. You can calculate $e$ from experimental data by directly applying NEL to measured approach and separation speeds.

Analysing One-Dimensional Collisions

Solving collision problems in one dimension requires the simultaneous application of two principles: the conservation of linear momentum and Newton's Experimental Law. Momentum is always conserved in a collision if no external forces act, regardless of the value of $e$.

The standard solution strategy is:

1. Define a positive direction.
2. Apply the Principle of Conservation of Linear Momentum:

$$m_A{u}_A+m_B{u}_B=m_A{v}_A+m_B{v}_B$$

1. Apply Newton's Experimental Law:

$$v_B-v_A=e(u_A-u_B)$$

1. Solve the two simultaneous equations for the unknown velocities ($v_A$, $v_B$) or the unknown coefficient $e$.

Worked Example: A particle of mass $2\text{ kg}$ moving at $4{\text{ ms}}^{-1}$ collides directly with a stationary particle of mass $1\text{ kg}$. Given the coefficient of restitution is $0.5$, find their velocities after the collision. Let direction of initial motion be positive.

1. Conservation of Momentum: $(2)(4)+(1)(0)=2v_A+1v_B$ → $8=2v_A+v_B$.
2. NEL: $v_B-v_A=0.5(4-0)$ → $v_B-v_A=2$.
3. Solving simultaneously: Adding the equations $(2v_A+v_B)+(v_B-v_A)=8+2$ gives $v_A+2v_B=10$. Substitute back to find $v_A=2{\text{ ms}}^{-1}$ and $v_B=4{\text{ ms}}^{-1}$ (both positive, so both move in the original direction).

Oblique Collisions in Two Dimensions

Many collisions are not head-on. In an oblique collision with a smooth surface (like a ball striking a smooth wall), or between two smooth bodies, we resolve the velocity vectors into components perpendicular and parallel to the line of impact (the line joining the centres at the moment of collision).

The key rules are:

• Perpendicular to the line of impact: No impulse acts in this direction. Therefore, the velocity components of each particle parallel to the plane of contact remain unchanged.
• Along the line of impact: The collision dynamics apply only to this component. We treat this as a one-dimensional collision problem, applying conservation of momentum and Newton's Experimental Law to these components only.

Worked Scenario: A smooth sphere strikes a smooth, fixed vertical wall obliquely. Its initial velocity makes an angle $\alpha$ with the wall. Since the wall is fixed and smooth, its velocity is always zero. The component of the ball's velocity parallel to the wall is unaffected. The perpendicular component obeys NEL with the wall: $v_{\perp }=-e{u}_{\perp }$, where the negative sign indicates reversal of direction. The ball's rebound angle will therefore be different from its approach angle unless $e=1$.

Energy Transfers and Kinetic Energy Loss

While momentum is conserved, kinetic energy often is not. The coefficient of restitution $e$ directly governs the proportion of kinetic energy lost during the impact. For a direct collision between two particles, the kinetic energy lost, $\Delta KE$, can be derived as:

$$\Delta KE=\frac{1}{2}\frac{m_A{m}_B}{m_A+m_B}(1-e^2)(u_A-u_B{)}^2$$

This formula reveals critical insights:

• The energy loss is proportional to $(1-e^2)$. For a perfectly elastic collision ($e=1$), $\Delta KE=0$.
• The loss depends on the square of the relative speed of approach, $(u_A-u_B{)}^2$.
• It involves the reduced mass $\frac{m_A{m}_B}{m_A+m_B}$ of the system.

You can calculate the proportion of initial kinetic energy lost as: $$\text{Proportion lost}=\frac{\Delta KE}{K{E}_{\text{initial}}}=1-\frac{K{E}_{\text{final}}}{K{E}_{\text{initial}}}$$ In a perfectly inelastic collision ($e=0$) with one body initially stationary, this proportion simplifies to $\frac{m_B}{m_A+m_B}$, showing maximum energy loss is not 100% unless the target mass is infinitely large.

Modelling Multi-Impact Problems: The Bouncing Ball

A classic application is modelling a ball dropped from a height onto a fixed horizontal surface. This uses the restitution coefficient iteratively to find rebound heights and times.

If a ball is dropped from an initial height $H_0$, its speed at impact is $u=\sqrt{2g{H}_0}$. Applying NEL with the fixed surface ($v=-eu$), the rebound speed is $eu$. Using $v^2=u^2+2as$ upwards, the next rebound height is: $$H_1=\frac{(eu{)}^2}{2g}=e^2{H}_0$$

This creates a geometric sequence for rebound heights: $H_0,e^2{H}_0,e^4{H}_0,...$ The total vertical distance travelled until the ball comes to rest is the sum of the infinite series of up-and-down journeys: $D=H_0+2e^2{H}_0+2e^4{H}_0+...=H_0+\frac{2e^2{H}_0}{1-e^2}$.

Common Pitfalls

1. Misapplying NEL in Two Dimensions: The most frequent error is applying the restitution law to the full velocity vectors instead of only the components along the line of impact. Always resolve velocities first. The parallel components are unchanged for smooth surfaces/bodies.

1. Sign Errors in One Dimension: Failing to consistently define and adhere to a positive direction when applying both momentum conservation and NEL will lead to incorrect signs (and thus directions) for final velocities. Write "Let → be positive" clearly at the start.

1. Assuming Energy Conservation: Unless explicitly told $e=1$, you cannot assume kinetic energy is conserved. Kinetic energy loss is the norm. Use the $(1-e^2)$ factor to check your energy calculations post-collision.

1. Confusing 'Inelastic' Definitions: Remember, a perfectly inelastic collision ($e=0$) means the bodies stick together, but they do not necessarily stop. They move with a common velocity dictated by momentum conservation. The kinetic energy lost is at its maximum for that particular collision scenario, but not necessarily 100% of the initial energy.

Summary

• Newton's Experimental Law, $v_B-v_A=e(u_A-u_B)$, defines the coefficient of restitution $e$, which quantifies the "bounciness" of a collision on a scale from 0 (perfectly inelastic) to 1 (perfectly elastic).
• All collision analyses combine Conservation of Linear Momentum with NEL. In two-dimensional oblique collisions, these principles apply only to the velocity components along the line of impact.
• The proportion of kinetic energy lost in a direct collision is proportional to $(1-e^2)$. Energy is conserved only when $e=1$.
• Multi-impact scenarios, like a bouncing ball, generate geometric sequences in rebound height and time, with the common ratio $e^2$.
• Consistent sign convention and correct vector resolution are essential to avoid errors in problem-solving.