Momentum

Welcome to the topic of Momentum! You already know that a heavy truck is much harder to stop than a bicycle moving at the same speed. Momentum is the physics concept that puts a number to that idea.

What you'll learn:

How to calculate momentum and use it to solve 1D collision and explosion problems.
The difference between elastic and inelastic collisions.
How force relates to the rate of change of momentum (Newton's Second Law in disguise!).
How to use impulse and force–time graphs to understand car safety features like crumple zones.

What is Momentum?

Momentum is a property of any moving object. It depends on two things: how much "stuff" is moving (mass) and how fast it is going (velocity).

Definition

Momentum

Linear momentum ( $p$ ) of an object is defined as the product of its mass ( $m$ ) and its velocity ( $v$ ).

p=m×v\begin{aligned} p &= m \times v \end{aligned}

Where:

$p$ is momentum in kilogram metres per second ( $s−1\text{kg m s}^{-1}$ )
$m$ is mass in kilograms ( $kg\text{kg}$ )
$v$ is velocity in metres per second ( $s−1\text{m s}^{-1}$ )

Common Mistake

Forgetting the vector nature

Velocity is a vector (it has a direction), which means momentum is also a vector. Always assign a positive direction to your problem (e.g., "right is positive"). If an object is moving to the left, its velocity—and therefore its momentum—must be negative.

Conservation of Linear Momentum

When two objects crash into each other, their individual speeds and directions change, but the total momentum of the system remains exactly the same, provided no external forces get involved.

Key Idea

The Principle of Conservation of Momentum

In a closed system (where no external forces act), the total linear momentum before an event is equal to the total linear momentum after the event.

1D collision diagram showing masses before and after

If we have two objects with masses $m_1$ and $m_2$ , initial velocities $u_1$ and $u_2$ , and final velocities $v_1$ and $v_2$ , we can write this principle as:

m1u1+m2u2=m1v1+m2v2\begin{aligned} m_1 u_1 + m_2 u_2 &= m_1 v_1 + m_2 v_2 \end{aligned}

Let's see how this works in a classic AQA-style question where two objects collide and stick together.

Example

Collision where objects stick together

A railway truck of mass $kg2500\text{ kg}$ is travelling at $s−13.0\text{ m s}^{-1}$ . It collides with a stationary truck of mass $kg1500\text{ kg}$ . They couple together on impact. Calculate the velocity of the trucks immediately after the collision.

Define the system and assign a positive direction. Let the original direction of motion be positive. Let $kgm_1 = 2500\text{ kg}$ , $s−1u_1 = 3.0\text{ m s}^{-1}$ , $kgm_2 = 1500\text{ kg}$ , and $s−1u_2 = 0\text{ m s}^{-1}$ .
Calculate the total initial momentum of the system: $s−1\begin{aligned} p_{\text{initial}} &= m_1 u_1 + m_2 u_2 \\ p_{\text{initial}} &= (2500 \times 3.0) + (1500 \times 0) \\ p_{\text{initial}} &= 7500\text{ kg m s}^{-1} \end{aligned}$
State the total final momentum. Because the trucks couple together, they form a single combined mass $m_1 + m_2)$ moving at a shared final velocity $v$ . $pfinal=(2500+1500)×vpfinal=4000v\begin{aligned} p_{\text{final}} &= (2500 + 1500) \times v \\ p_{\text{final}} &= 4000v \end{aligned}$
Apply the conservation of momentum ( $pinitial=pfinalp_{\text{initial}} = p_{\text{final}}$ ) and solve for $v$ : $s−1\begin{aligned} 7500 &= 4000v \\ v &= \frac{7500}{4000} \\ v &= 1.875\text{ m s}^{-1} \end{aligned}$
Round to appropriate significant figures (usually matching the given data, so $s−11.9\text{ m s}^{-1}$ in the same direction).

Explosions

Conservation of momentum doesn't just apply to things crashing together; it also applies to things flying apart! In an explosion (like a gun firing a bullet, or a stationary astronaut throwing a spanner), the initial momentum is usually zero.

Because total momentum must be conserved, the pieces must fly off in opposite directions so their positive and negative momenta cancel out to zero.

Elastic and Inelastic Collisions

While momentum is always conserved in a closed system, kinetic energy behaves differently. Depending on what happens to the kinetic energy, we classify collisions into two types.

Definition

Elastic and Inelastic

Elastic collision: Kinetic energy is conserved. The total kinetic energy before the collision exactly equals the total kinetic energy after. (This is rare in real life, mostly happening at the atomic level).
Inelastic collision: Kinetic energy is NOT conserved. Some kinetic energy is transferred to other stores, such as thermal energy or sound, or used to permanently deform the objects.

Example

Proving a collision is inelastic

Using the railway trucks from the previous example (initial: $kg2500\text{ kg}$ at $s−13.0\text{ m s}^{-1}$ , final: $kg4000\text{ kg}$ combined at $s−11.875\text{ m s}^{-1}$ ), show whether the collision was elastic or inelastic.

Calculate the total initial kinetic energy ( $Ek=12mv2E_k = \frac{1}{2}mv^2$ ): $J\begin{aligned} E_{k(\text{initial})} &= \frac{1}{2} \times 2500 \times (3.0)^2 + 0 \\ E_{k(\text{initial})} &= 11250\text{ J} \end{aligned}$
Calculate the total final kinetic energy: $J\begin{aligned} E_{k(\text{final})} &= \frac{1}{2} \times 4000 \times (1.875)^2 \\ E_{k(\text{final})} &= 7031.25\text{ J} \end{aligned}$
Compare the two values and state your conclusion. Since $J11250\text{ J} \neq 7031.25\text{ J}$ , kinetic energy is lost to the surroundings. Therefore, the collision is inelastic.

Force and Rate of Change of Momentum

When you learned Newton's Second Law at GCSE, you were taught $F = ma$ . But Isaac Newton actually wrote it differently. He said that force is the rate of change of momentum.

If an object's momentum changes over a period of time, a force must be acting on it. The faster the momentum changes, the larger the force.

F=Δ(mv)Δt\begin{aligned} F &= \frac{\Delta (mv)}{\Delta t} \end{aligned}

Where:

$F$ is the average force acting ( $N\text{N}$ )
$Δ(mv)\Delta(mv)$ is the change in momentum ( $s−1\text{kg m s}^{-1}$ )
$Δt\Delta t$ is the time taken for the change ( $s\text{s}$ )

Tip

Where does F=ma come from?

If mass is constant, we can pull $m$ out of the change brackets: $Δ(mv)=mΔv\Delta(mv) = m\Delta v$ . So the equation becomes $\frac{\Delta v}{\Delta t}$ . Since $ΔvΔt\frac{\Delta v}{\Delta t}$ is acceleration ( $a$ ), we get back to $F = ma$ ! The momentum version is just more powerful because it can also handle situations where mass changes (like a rocket burning fuel).

Impulse

If we rearrange the formula above by multiplying both sides by $Δt\Delta t$ , we get:

FΔt=Δ(mv)\begin{aligned} F \Delta t &= \Delta (mv) \end{aligned}

The product of force and time ( $\Delta t$ ) is called impulse. Therefore, Impulse = Change in momentum.

Because of this, you'll sometimes see momentum given in units of Newton seconds ( $s\text{N s}$ ). This is exactly equivalent to $s−1\text{kg m s}^{-1}$ .

Force–Time Graphs

In real life, forces during impacts (like a football being kicked) are rarely constant. They start at zero, peak quickly as the object compresses, and fall back to zero as the object flies away.

Force-time graph showing area as impulse

Key Idea

Area under a Force-Time Graph

The area under a force–time graph represents the impulse (and therefore the change in momentum).

Example

Estimating force from a graph

A tennis ball of mass $kg0.060\text{ kg}$ is struck by a racket. A force-time graph shows the impact forms a rough triangle with a base (time) of $s0.008\text{ s}$ and a peak force of $N450\text{ N}$ . The ball starts from rest. Calculate its final velocity.

Calculate the area under the force-time graph (area of a triangle = $12×base×height\frac{1}{2} \times \text{base} \times \text{height}$ ) to find the impulse. $s\begin{aligned} \text{Impulse} &= \frac{1}{2} \times 0.008 \times 450 \\ \text{Impulse} &= 1.8\text{ N s} \end{aligned}$
Equate the impulse to the change in momentum ( $Δp=mv−mu\Delta p = mv - mu$ ). Since the ball started from rest, $u = 0$ , so $Δp=mv\Delta p = mv$ . $1.8=0.060×v\begin{aligned} 1.8 &= 0.060 \times v \end{aligned}$
Solve for the final velocity $v$ : $s−1\begin{aligned} v &= \frac{1.8}{0.060} \\ v &= 30\text{ m s}^{-1} \end{aligned}$

Real-World Applications: Contact Times and Safety

The equation $\frac{\Delta(mv)}{\Delta t}$ is incredibly important for engineers, particularly when designing transport systems and packaging.

If a car crashes at $mph30\text{ mph}$ , the driver's change in momentum ( $Δ(mv)\Delta(mv)$ ) to reach $mph0\text{ mph}$ is fixed. You cannot change the initial mass or the velocity. However, you can change the impact force ( $F$ ) by increasing the contact time ( $Δt\Delta t$ ).

If you make the collision take ten times longer, the average force experienced by the occupants is ten times smaller. This appreciation of momentum conservation is a key issue in ethical transport design.

Analogy

Catching a water balloon

If someone throws a water balloon at you, you naturally move your hands backward as you catch it. You are increasing $Δt\Delta t$ (the time it takes to stop the balloon). Because $\frac{\Delta p}{\Delta t}$ and $Δp\Delta p$ is fixed, increasing $Δt\Delta t$ decreases the force $F$ , meaning the balloon doesn't burst in your hands!

Engineering features that increase $Δt\Delta t$ :

Crumple zones in cars: The front of the car is designed to deform slowly on impact.
Airbags: They slow down the driver's head gradually, rather than it hitting a hard steering wheel instantly.
Seatbelts: They stretch slightly, increasing the time taken for the passenger to stop.
Packaging: Bubble wrap and polystyrene compress to extend the contact time if a parcel is dropped.

Exam technique

In the exam

Draw a quick sketch: Always sketch "before" and "after" diagrams for collisions. Draw arrows indicating the direction of travel and write down the velocities with their correct signs (+ or -).
Watch your signs: If an object rebounds off a wall, its final velocity has the opposite sign to its initial velocity. The change in momentum is $m v - m (- u) = m (v + u)$ . A very common trap!
Count the shapes: If asked to find the area under a curve on a force-time graph, count the grid squares carefully. Figure out what one tiny square represents in terms of impulse (e.g., $s1\text{ N} \times 0.01\text{ s} = 0.01\text{ N s}$ ).
State the obvious for safety questions: If a 2-mark question asks why an airbag is safer, explicitly state: "Increases impact time" (1 mark) "which reduces the rate of change of momentum / reduces the impact force" (1 mark).

Self review

Check yourself

Can you state the definition of linear momentum and its SI unit?
What is the difference between an elastic and an inelastic collision?
How do you calculate impulse from a force-time graph?
Why do modern cars have crumple zones, explained using the formula for the rate of change of momentum?

Welcome to the topic of Momentum! You already know that a heavy truck is much harder to stop than a bicycle moving at the same speed. Momentum is the physics concept that puts a number to that idea.

What you'll learn:

How to calculate momentum and use it to solve 1D collision and explosion problems.
The difference between elastic and inelastic collisions.
How force relates to the rate of change of momentum (Newton's Second Law in disguise!).
How to use impulse and force–time graphs to understand car safety features like crumple zones.

What is Momentum?

Momentum is a property of any moving object. It depends on two things: how much "stuff" is moving (mass) and how fast it is going (velocity).

Definition

Momentum

Linear momentum ( $p$ ) of an object is defined as the product of its mass ( $m$ ) and its velocity ( $v$ ).

p=m×v\begin{aligned} p &= m \times v \end{aligned}

Where:

$p$ is momentum in kilogram metres per second ( $s−1\text{kg m s}^{-1}$ )
$m$ is mass in kilograms ( $kg\text{kg}$ )
$v$ is velocity in metres per second ( $s−1\text{m s}^{-1}$ )

Common Mistake

Forgetting the vector nature

Conservation of Linear Momentum

When two objects crash into each other, their individual speeds and directions change, but the total momentum of the system remains exactly the same, provided no external forces get involved.

Key Idea

The Principle of Conservation of Momentum

In a closed system (where no external forces act), the total linear momentum before an event is equal to the total linear momentum after the event.

1D collision diagram showing masses before and after

If we have two objects with masses $m_1$ and $m_2$ , initial velocities $u_1$ and $u_2$ , and final velocities $v_1$ and $v_2$ , we can write this principle as:

m1u1+m2u2=m1v1+m2v2\begin{aligned} m_1 u_1 + m_2 u_2 &= m_1 v_1 + m_2 v_2 \end{aligned}

Let's see how this works in a classic AQA-style question where two objects collide and stick together.

Example

Collision where objects stick together

Define the system and assign a positive direction. Let the original direction of motion be positive. Let $kgm_1 = 2500\text{ kg}$ , $s−1u_1 = 3.0\text{ m s}^{-1}$ , $kgm_2 = 1500\text{ kg}$ , and $s−1u_2 = 0\text{ m s}^{-1}$ .
Calculate the total initial momentum of the system: $s−1\begin{aligned} p_{\text{initial}} &= m_1 u_1 + m_2 u_2 \\ p_{\text{initial}} &= (2500 \times 3.0) + (1500 \times 0) \\ p_{\text{initial}} &= 7500\text{ kg m s}^{-1} \end{aligned}$
State the total final momentum. Because the trucks couple together, they form a single combined mass $m_1 + m_2)$ moving at a shared final velocity $v$ . $pfinal=(2500+1500)×vpfinal=4000v\begin{aligned} p_{\text{final}} &= (2500 + 1500) \times v \\ p_{\text{final}} &= 4000v \end{aligned}$
Apply the conservation of momentum ( $pinitial=pfinalp_{\text{initial}} = p_{\text{final}}$ ) and solve for $v$ : $s−1\begin{aligned} 7500 &= 4000v \\ v &= \frac{7500}{4000} \\ v &= 1.875\text{ m s}^{-1} \end{aligned}$
Round to appropriate significant figures (usually matching the given data, so $s−11.9\text{ m s}^{-1}$ in the same direction).