Impacts & Collisions

1. Momentum, Impulse, and Collision Modelling

A collision occurs over a brief time interval $\Delta t$ during which interacting bodies exert high mutual contact forces. Linear momentum is the vector quantity defined as the product of mass and velocity:

Momentum is measured in $\text{kg}\cdot\text{m}\cdot\text{s}^{-1}$ or newton-seconds ($\text{N}\cdot\text{s}$). The impulse ($\mathbf{I}$) of a resultant force measures the overall effect of that force acting over time and equals the change in momentum:

On a force–time graph, impulse equals the area under the curve. For example, a triangular force pulse lasting $0.010\text{ s}$ with a peak force of $800\text{ N}$ delivers an impulse of $\frac{1}{2}(800)(0.010) = 4\text{ N}\cdot\text{s}$.

Modelling Assumptions

When modelling impacts in Strand 3, we make several standard assumptions:

• We model the bodies as smooth spheres whose whole mass acts at the centre. Because they are smooth there is no friction at the point of contact, so the contact force acts along the line joining their centres, and we ignore any spin.
• The impact occurs over such a short duration that displacements during contact are negligible, and external impulses from forces such as gravity or friction are negligible compared with the normal contact impulse.
• By Newton's third law, the contact force exerted by sphere A on sphere B is equal in magnitude and opposite in direction to that exerted by B on A. Internal impulses sum to zero: $\mathbf{I}_B = -\mathbf{I}_A$.

Commenting on the Model

If asked to evaluate the mathematical model in an authentic context (such as snooker balls or vehicle impacts), state the main simplifications: real objects deform, air resistance and surface friction operate, and impact duration is non-zero. For example, a snooker ball develops topspin or backspin through cloth friction, so post-collision angles will deviate slightly from the ideal frictionless model.

2. The Principle of Conservation of Momentum (PCM)

Because internal contact impulses are equal and opposite, the total linear momentum of an isolated system remains constant in any direction free from external impulses. This is the Principle of Conservation of Momentum:

For two free spheres colliding on a smooth horizontal table, momentum is conserved in all planar directions. However, momentum is not conserved for a single particle colliding with a fixed barrier (such as a floor or wall). The barrier provides an external impulse to the particle, so PCM cannot be applied to the particle alone.

Completely Inelastic Collisions

In a completely inelastic collision ($e = 0$), colliding particles coalesce or move with a common normal velocity after impact. For two particles moving along a straight line that coalesce upon impact:

For example, a $2\text{ kg}$ trolley travelling at $6\text{ m}\cdot\text{s}^{-1}$ that collides with and couples to a stationary $1\text{ kg}$ trolley moves forward at $V = \frac{2(6) + 1(0)}{2 + 1} = 4\text{ m}\cdot\text{s}^{-1}$.

3. Newton's Experimental Law (NEL) and Restitution

Conservation of momentum gives only one equation for the two unknown post-collision velocities. Newton's Experimental Law of Restitution supplies the second equation. It states that the relative speed of separation along the line of impact equals $e$ times the relative speed of approach:

Using signed velocities along the line of impact, NEL is written as:

The coefficient of restitution $e$ is a dimensionless scalar governed by the physical materials, satisfying $0 \le e \le 1$:

Kinetic energy can never increase in a collision. The best case is that it stays the same, and that only happens when $e = 1$.

4. Direct Collisions: Step-by-Step Method and Inequalities

In a direct collision, all motion occurs along the line of centres. Always draw a clear diagram showing before and after states with labelled arrows, declare a positive direction, and set up a Before-Mass-After table:

Keep $u_1$ and $u_2$ as signed velocities throughout, so the approach speed is always $u_1 - u_2$. If B moves left while positive is to the right, enter B's velocity as $-u$, giving an approach speed of $3u - (-u) = 4u$.

Set up and solve the two governing simultaneous equations:

1. PCM: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
2. NEL: $v_1 - v_2 = -e(u_1 - u_2)$

Presentation of Answers

Examination questions usually ask for the speed of each sphere. If an algebraic velocity expression evaluates to a negative quantity, state the final speed as a positive magnitude together with its direction (for example, 'sphere A reverses direction and moves with speed $\frac{u(8e - 1)}{3}$').

Sanity-Check Tool for Collisions with a Stationary Target

When mass $m_1$ at speed $u$ strikes mass $m_2$ at rest:

Notice that sphere 1 is brought to rest ($v_1 = 0$) exactly when $m_1 = e m_2$. Use this relationship to check your working.

Direction Reversal and Subsequent Collision Conditions

• Direction Reversal: A sphere moving in the positive direction reverses its motion if and only if $v_1 < 0$.
• Trailing-Sphere Test: Two spheres that have just collided can never collide again on their own, because NEL forces the rear sphere's velocity to be less than or equal to the front sphere's velocity ($v_{\text{rear}} \le v_{\text{front}}$). A subsequent collision requires an external intervention, such as a barrier or a third sphere.
• Three Spheres in a Line: If A and B collide, and B moves towards a third sphere C moving with velocity $u_C$, B catches C if and only if $v_B > u_C$. If B rebounds from C with new velocity $v_B'$, B collides with A again only if $v_B'$ overtakes $v_A$ ($v_A < v_B'$ in the shared direction of travel). Always intersect any derived bounds on $e$ with $0 \le e \le 1$.

5. Impulse and Kinetic Energy Loss in Direct Collisions

Impulse in a Two-Sphere Collision

The impulse imparted to one sphere is that sphere's own change in momentum along the line of centres:

By Newton's third law, the impulse exerted on B by A is equal and opposite to that exerted on A by B ($I_B = -I_A$). Calculate the impulse using whichever sphere has simpler algebra, then report its magnitude and direction.

Because expressions for impulse contain the factor $(1 + e)$, substituting the physical limits $e = 0$ and $e = 1$ establishes minimum and maximum bounds on the impulse.

Kinetic Energy Lost in a Direct Collision

The loss of kinetic energy is:

Because speeds are squared, directional signs do not affect the kinetic energy evaluation. However, you must substitute the algebraic velocities determined by PCM and NEL. The resulting expression regularly simplifies to the standard form:

The factor $(1 - e^2)$ confirms that kinetic energy loss is zero when $e = 1$ (perfectly elastic) and reaches its maximum when $e = 0$.

Energy-Transfer Problems

When a problem states that 'a given percentage $x\%$ of the kinetic energy lost by A is transferred to B', formulate the energy relationship as an independent equation:

Combine this energy equation with PCM to find the unknown speeds before applying NEL to find $e$.

6. Oblique Collisions Between Smooth Spheres

In an oblique collision between two smooth spheres, each velocity is split into two perpendicular parts: one along the line joining their centres at the moment of impact (the line of centres, chosen as the $\mathbf{i}$-axis), and one along the common tangent plane at right angles to it (the $\mathbf{j}$-axis).

Always write this core justification in your exam answer: Since the spheres are smooth, there is no force along the common tangent, so the $\mathbf{j}$-components of velocity are unchanged.

The Four-Relationship Framework

1. Tangential velocity of sphere 1 is unaltered: $v_{1y} = u_{1y}$
2. Tangential velocity of sphere 2 is unaltered: $v_{2y} = u_{2y}$
3. PCM along the line of centres ($\mathbf{i}$-axis): $m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}$
4. NEL along the line of centres ($\mathbf{i}$-axis): $v_{1x} - v_{2x} = -e(u_{1x} - u_{2x})$

Once $v_{1x}$ and $v_{2x}$ are found, reassemble the vectors: $\mathbf{v}_1 = v_{1x}\mathbf{i} + u_{1y}\mathbf{j}$ and $\mathbf{v}_2 = v_{2x}\mathbf{i} + u_{2y}\mathbf{j}$.

Geometric Conditions and Deflection

• Perpendicular Paths: Two final velocity vectors are perpendicular if their scalar dot product is zero: $\mathbf{v}_1 \cdot \mathbf{v}_2 = v_{1x}v_{2x} + v_{1y}v_{2y} = 0$. Alternatively, their direction slopes satisfy $s_1 s_2 = -1$ where $s = v_y / v_x$. The dot product is safer because it remains valid if a sphere moves purely along a coordinate axis.
• Deflection Angle: The angle $\theta$ through which an individual sphere is deflected is found from $\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$, or by comparing trajectory angles before and after impact: $\theta = |\alpha - \beta|$, where $\tan\alpha = \frac{u_y}{u_x}$ and $\tan\beta = \frac{v_y}{v_x}$.

Kinetic Energy Loss in Oblique Collisions

Because tangential components are unchanged, they subtract to zero when calculating the absolute energy lost:

However, when computing the fraction or percentage of kinetic energy lost, the denominator must include the entire initial kinetic energy of the system across both $\mathbf{i}$ and $\mathbf{j}$ components:

7. Impacts with a Fixed Plane: Floors, Walls, and Bouncing

When a smooth particle collides with a fixed surface, set up coordinate axes parallel and perpendicular to the surface. Because the barrier is fixed and smooth:

• Velocity parallel to the plane is unchanged: $v_\parallel = u_\parallel$
• Velocity perpendicular to the plane reverses and scales by $e$: $v_\perp = -e u_\perp$

Impulse on the Plane

The velocity change is entirely perpendicular to the surface: $\Delta \mathbf{v} = v_\perp - u_\perp = -e u_\perp - u_\perp = -(1 + e)u_\perp$. The magnitude of the impulse on the particle is:

By Newton's third law, the particle delivers an equal and opposite impulse of magnitude $m(1 + e)|u_\perp|$ directly into the plane.

Successive Vertical Bounces as a Geometric Sequence

A particle dropped vertically from height $h$ strikes a horizontal floor with impact speed $u_0 = \sqrt{2gh}$.

• Rebound speed after bounce $n$: $v_n = e^n \sqrt{2gh}$
• Height reached after bounce $n$: $h_n = \frac{v_n^2}{2g} = e^{2n} h$
• Time of flight between bounce $n$ and $n+1$: $t_n = \frac{2v_n}{g} = \frac{2e^n \sqrt{2gh}}{g}$

For $e < 1$, the total vertical distance travelled before the ball comes to rest forms an infinite geometric series:

Successive Hops Down an Inclined Plane

For a particle hopping down a plane inclined at angle $\beta$, resolve gravity into $g\sin\beta$ down the plane and $g\cos\beta$ into the plane. For each hop:

• Time of flight: $T = \frac{2u_\perp}{g\cos\beta}$
• Range along the plane: $R = u_\parallel T + \frac{1}{2}(g\sin\beta)T^2$

The impact itself leaves u∥ unchanged and multiplies u⊥ by e, so the next time of flight is T′ = eT. But during a hop gravity's component g sin β along the plane speeds the particle up, so the parallel velocity entering the next hop is u∥′ = u∥ + (g sin β)T, not u∥. Recompute the next range from R′ = u∥′T′ + ½(g sin β)T′² using these new values — the range does not simply scale by e or e², which is why questions ask you to find the ratio k rather than quoting it.

Key terms

Linear Momentum

The vector quantity p = mv, measuring the quantity of motion of a body in kg m s⁻¹ or N s.

Impulse

The time integral of resultant force, equivalent to the change in momentum (I = Δp = mv - mu) imparted to a body during a collision.

Principle of Conservation of Momentum (PCM)

The fundamental law stating that the total linear momentum of an isolated system remains constant along any direction free from external impulses.

Newton's Experimental Law (NEL)

The empirical collision law stating that relative speed of separation along the line of centres equals e times the relative speed of approach: v₁ - v₂ = -e(u₁ - u₂).

Coefficient of Restitution (e)

A dimensionless measure of elasticity (0 ≤ e ≤ 1) governing the ratio of separation speed to approach speed along the line of impact.

Line of Centres

The normal axis passing through the geometric centres of two colliding spheres at the point of contact, along which all contact impulses act.

Common Tangent

The interface direction perpendicular to the line of centres at the point of impact, along which smooth bodies exert zero contact force.

Direct Collision

An impact where particle velocities lie entirely along the line of centres before and after the collision.

Oblique Collision

An impact where initial velocities are inclined at an angle to the line of centres, requiring resolution into normal and tangential components.

Perfectly Elastic

An ideal collision with e = 1 in which no kinetic energy is lost.

Completely Inelastic

A collision with e = 0 where colliding bodies have zero relative separation velocity along the line of impact, dissipating maximum kinetic energy.

Deflection Angle

The angle through which a particle's trajectory turns as a result of an impact, calculated using the scalar dot product cos θ = (u · v) / (|u||v|).