Further Mathematics | Collisions

Unit M2: Mechanics 2 — Collisions

Welcome to the World of Bouncing!

Hello future Further Mathematician! The topic of Collisions is where we bring together everything you learned about momentum and apply it to real-world interactions—from snooker balls hitting each other to cars crashing.

While it might seem challenging at first because we deal with two simultaneous equations, the physics behind it is beautiful and logical. By the end of these notes, you will be able to predict the exact motion of objects after they hit each other, simply by knowing how fast they were going before the impact and how "bouncy" they are.

Let’s dive into predicting the consequences of impact!

1. The Foundation: Momentum and Impulse

Before we look at the collision itself, we need to quickly recap the fundamental concept that governs all impacts: Linear Momentum.

1.1 Momentum and Its Conservation

Momentum (\(p\)) is the measure of the mass and velocity of an object. It is a vector quantity, meaning direction matters!

• Definition: Momentum \(p\) is given by \(p = mv\), where \(m\) is mass (kg) and \(v\) is velocity (m/s).
• Impulse: When two objects collide, they exert massive forces on each other for a short time. The effect of this force is called Impulse, which is equal to the change in momentum of a single particle.

1.2 Setting up the Conservation of Momentum Equation

Imagine two particles, $P_1$ (mass $m_1$) and $P_2$ (mass $m_2$), moving along the same straight line (a direct impact).

• $u_1, u_2$: Velocities before collision.
• $v_1, v_2$: Velocities after collision.

The Conservation of Momentum equation is:
$$\text{(Total Momentum Before)} = \text{(Total Momentum After)}$$ $$\mathbf{m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2}$$

Crucial Rule: Sign Convention

Since velocity is a vector, you must immediately choose a positive direction (e.g., right). Any velocity moving in the opposite direction must be substituted into the equation as a negative value. This is the source of many common errors!

Key Takeaway (CLM)

The Conservation of Linear Momentum always provides us with our first simultaneous equation when solving collision problems.

2. The Coefficient of Restitution (The "Bounciness" Factor)

The momentum equation alone usually isn't enough to solve for the two unknown final velocities ($v_1$ and $v_2$). We need a second equation that tells us how the energy was transferred—in other words, how bouncy the collision was.

2.1 Introducing Newton’s Experimental Law (NEL)

The Coefficient of Restitution, denoted by the letter \(e\), is a value between 0 and 1 that describes the elasticity of the collision. This concept forms the basis of Newton's Experimental Law (NEL).

The NEL relates the relative speed of approach before the collision to the relative speed of separation after the collision.

$$\mathbf{v_2 - v_1 = e(u_1 - u_2)}$$ $$(\text{Relative Speed of Separation}) = e \times (\text{Relative Speed of Approach})$$

Important Note: This equation is sometimes rearranged to include a minus sign, but it is easiest to remember it as: Relative separation speed (which must be positive) is equal to $e$ times the relative approach speed (which must also be positive).

• The term \((v_2 - v_1)\) calculates the speed at which the particles move apart.
• The term \((u_1 - u_2)\) calculates the speed at which the particles move together (we assume $u_1 > u_2$ for them to collide).

2.2 The Range of \(e\)

The value of \(e\) dictates the nature of the collision:

1. $e = 1$: Perfectly Elastic Collision

   This is a perfectly bouncy collision. There is no loss of kinetic energy during the impact. (Example: Idealised gas particle collisions.)

2. $e = 0$: Perfectly Inelastic Collision

   The particles have maximum kinetic energy loss and stick together after impact. This means $v_1 = v_2 = V$. In this case, the NEL is not needed, as the single velocity $V$ can be found directly using CLM.

3. $0 < e < 1$: Inelastic Collision

   This is the most common scenario. Some kinetic energy is lost (usually converted to sound and heat). We must use both the CLM and NEL equations to solve the problem.

Key Takeaway (NEL)

The Coefficient of Restitution (\(e\)) gives us the second simultaneous equation required to solve for the two final velocities, $v_1$ and $v_2$.

3. Step-by-Step Guide to Solving Direct Collisions

Don’t worry if this seems like a lot of equations! Every direct collision problem follows the exact same procedure, leading to two equations in two unknowns.

Problem Example Setup: Particle $P_1$ (2 kg) moves right at 5 m/s, Particle $P_2$ (3 kg) moves left at 1 m/s. They collide with $e = 0.5$. Find $v_1$ and $v_2$.

Step 1: Define Direction and Substitute Initial Velocities

Choose RIGHT as POSITIVE.

• $m_1 = 2$, $u_1 = +5$
• $m_2 = 3$, $u_2 = -1$ (Negative because it moves left)
• $e = 0.5$

Step 2: Apply Conservation of Linear Momentum (CLM)

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$2(5) + 3(-1) = 2v_1 + 3v_2$$ $$10 - 3 = 2v_1 + 3v_2$$ $$\mathbf{7 = 2v_1 + 3v_2} \quad \text{ (Equation 1)}$$

Step 3: Apply Newton’s Experimental Law (NEL)

Remember the relative speeds: $v_2 - v_1 = e(u_1 - u_2)$.
$$v_2 - v_1 = 0.5(5 - (-1))$$ $$v_2 - v_1 = 0.5(6)$$ $$v_2 - v_1 = 3$$ $$\mathbf{v_2 = v_1 + 3} \quad \text{ (Equation 2)}$$

Step 4: Solve Simultaneously

Substitute Equation 2 into Equation 1:
$$7 = 2v_1 + 3(v_1 + 3)$$ $$7 = 2v_1 + 3v_1 + 9$$ $$7 - 9 = 5v_1$$ $$-2 = 5v_1 \implies v_1 = -0.4 \text{ m/s}$$
Substitute $v_1$ back into Equation 2:
$$v_2 = -0.4 + 3 \implies v_2 = 2.6 \text{ m/s}$$

Step 5: Interpret the Results

• $v_1 = -0.4$ m/s. Particle $P_1$ moves 0.4 m/s to the left (it rebounded).
• $v_2 = +2.6$ m/s. Particle $P_2$ moves 2.6 m/s to the right.

4. Energy Considerations in Collisions

In most real-world collisions ($e < 1$), mechanical energy is lost, primarily converted into heat, sound, or permanent deformation (like crushing metal). We often need to calculate this energy loss.

4.1 Kinetic Energy (KE)

The kinetic energy of a single particle is given by: $$\mathbf{KE = \frac{1}{2} m v^2}$$ (Note: Since $v^2$ is always positive, KE is a scalar quantity and is always positive, regardless of the direction of motion.)

4.2 Calculating Loss of Kinetic Energy

The total KE of the system is the sum of the individual KE of the particles.

$$\text{Total KE Before} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2$$ $$\text{Total KE After} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$

The loss in kinetic energy (\(L\)) is: $$\mathbf{L = (\text{Total KE Before}) - (\text{Total KE After})}$$

Did you know? If $e=1$ (a perfectly elastic collision), the loss of KE is zero. This happens because the forces involved in the collision are conservative, meaning no energy is dissipated as heat or sound.

Check: For any collision where $e < 1$, the loss $L$ must be a positive value. If your calculation yields a negative loss (meaning KE increased), you have made a mistake in calculating one of the velocities!

Example Using Previous Results

$m_1=2, u_1=5, v_1=-0.4$. $m_2=3, u_2=-1, v_2=2.6$.

KE Before: $$\frac{1}{2}(2)(5^2) + \frac{1}{2}(3)((-1)^2) = 25 + 1.5 = 26.5 \text{ J}$$

KE After: $$\frac{1}{2}(2)((-0.4)^2) + \frac{1}{2}(3)((2.6)^2)$$ $$= (0.16) + (1.5 \times 6.76)$$ $$= 0.16 + 10.14 = 10.3 \text{ J}$$

Loss in KE: $$L = 26.5 - 10.3 = 16.2 \text{ J}$$

5. Collision with a Fixed Surface (Wall)

Sometimes, a particle strikes a smooth fixed surface, like a wall or the ground. Since the surface is fixed, it is considered to have infinite mass and zero velocity both before and after the collision.

Because the mass of the wall is infinite, the Conservation of Linear Momentum equation is not useful for finding the final velocity of the particle, as the wall absorbs all momentum change.

Therefore, when dealing with a fixed surface, we rely only on the Coefficient of Restitution (\(e\)), which simplifies significantly.

If a particle of mass $m$ hits a wall with speed $u$ and rebounds with speed $v$:
$$\text{(Speed of separation)} = e \times \text{(Speed of approach)}$$ $$\mathbf{v = eu}$$

This is a very common scenario, often used in problems involving balls bouncing off the floor. Note that $v$ and $u$ here are the magnitudes (speeds) of the velocities perpendicular to the wall.

Example: Ball Bouncing

A ball drops and hits the ground at 6 m/s. If $e=0.7$, what is its speed immediately after impact?
$$v = eu = 0.7 \times 6 = 4.2 \text{ m/s}$$

Key Takeaway (Fixed Surface)

For collisions against fixed surfaces, ignore CLM. The final speed is simply the initial speed multiplied by $e$: $v = eu$.

6. Collision Checklist and Encouragement

You have now mastered the two fundamental tools needed for collision problems: CLM and NEL. Remember that practice is key to avoiding sign errors!

Your Collision Problem Checklist:

1. Diagram & Direction: Draw a sketch showing all masses and initial velocities. Define your positive direction (e.g., Right is positive).
2. CLM Equation (Eq 1): Write down the Conservation of Momentum equation, being extremely careful with negative signs for initial velocities.
3. NEL Equation (Eq 2): Write down the Coefficient of Restitution equation, ensuring the relative speeds are ordered correctly to maintain consistency.
4. Solve: Use substitution or elimination to solve the two simultaneous equations for $v_1$ and $v_2$.
5. Interpret: If your calculated $v$ is negative, remember the particle is moving in the opposite direction to your chosen positive direction.

You’ve got this! While Further Maths is meant to be challenging, collision problems are highly procedural. Master the two core equations and you will succeed.

Good luck!