Momentum and Collisions

Higher tier: This lesson covers Higher-tier content of OCR Gateway Combined Science A. Foundation-tier students are not assessed on momentum, but the ideas here — especially why crumple zones make cars safer — are well worth understanding.

A moving lorry is far harder to stop than a bicycle travelling at the same speed, and a fast cricket ball stings a lot more than a slow one of the same mass. Both facts are captured by a single quantity called momentum, which combines an object's mass and its velocity. Momentum turns out to be one of the great "conserved" quantities of physics: in any collision or explosion, the total momentum before equals the total momentum after. This lesson, part of Topic P2 (Forces) of OCR Gateway Combined Science A, defines momentum, applies the powerful principle of conservation of momentum to collisions and explosions, introduces force as the rate of change of momentum, and uses that idea to explain why the safety features in a modern car save lives.

By the end of this lesson you should be able to calculate momentum using $p = mv$ , apply conservation of momentum to find an unknown velocity in a collision or explosion, use $F = \dfrac{\Delta p}{\Delta t}$ to relate force to the change in momentum, and explain how car safety features reduce the force on the occupants by increasing the stopping time.

This lesson is AO1 for defining momentum and conservation of momentum, strongly AO2 for applying $p = mv$ and $F = \dfrac{\Delta p}{\Delta t}$ to collisions and explosions, and AO3 when you evaluate why safety features such as crumple zones and airbags reduce the force by extending the stopping time.

What Is Momentum?

Momentum is a measure of how much motion an object has, taking account of both how heavy it is and how fast it is moving. It is defined as the product of mass and velocity:

$p = mv$

where $p$ is the momentum, $m$ is the mass in $\text{kg}$ and $v$ is the velocity in $\text{m/s}$ . The unit of momentum is the kilogram metre per second, $\text{kg}\,\text{m/s}$ .

Momentum is a vector quantity, which means it has a direction as well as a size — the same direction as the velocity. This matters greatly in collisions: momentum to the right is treated as positive and momentum to the left as negative, and the two can cancel. A large lorry moving slowly can have the same momentum as a small car moving quickly, because momentum depends on the product of mass and velocity.

The fact that momentum has a direction is not a technicality to be brushed aside — it is the single most important thing to get right when solving momentum problems. Before starting any calculation you should choose a positive direction (for example, "to the right is positive") and keep to it throughout. Any object moving the other way then has a negative velocity and therefore a negative momentum. This bookkeeping is what lets momenta cancel: two identical objects moving towards each other at the same speed have equal and opposite momenta that add to zero, which is why two such objects can collide and both stop dead. If you ignored the directions and simply added the sizes, you would get the wrong answer entirely. Getting into the habit of writing down your chosen positive direction at the start of every momentum question will save you from one of the most common and costly mistakes in this topic.

Worked Example 1 — Calculating momentum

A car of mass $1400 \text{ kg}$ travels at $20 \text{ m/s}$ . Calculate its momentum.

Step 1 — write the equation: $p = mv$ .

Step 2 — substitute: $p = 1400 \times 20$ .

Step 3 — calculate: $p = 28\,000 \text{ kg}\,\text{m/s}$ .

Answer: the car's momentum is $28\,000 \text{ kg}\,\text{m/s}$ .

Exam Tip: A common misconception is that momentum and kinetic energy are the same. Momentum is $p = mv$ with unit $\text{kg}\,\text{m/s}$ ; kinetic energy is $\dfrac{1}{2}mv^2$ , measured in joules. Momentum uses $v$ to the first power; kinetic energy uses $v^2$ .

Conservation of Momentum

The principle of conservation of momentum states that, in a closed system (one with no external forces acting), the total momentum before an event equals the total momentum after it:

$\text{total momentum before} = \text{total momentum after}$

This applies to collisions (objects hitting each other) and to explosions (objects pushing apart). Because momentum is a vector, you must be careful with directions — objects moving in opposite directions have momenta of opposite sign.

Worked Example 2 — A collision where objects stick together

A $2 \text{ kg}$ trolley moving at $3 \text{ m/s}$ collides with a stationary $1 \text{ kg}$ trolley. They stick together and move off as one. Calculate their common velocity after the collision.

Step 1 — momentum before: only the first trolley is moving, so $p_{\text{before}} = (2 \times 3) + (1 \times 0) = 6 \text{ kg}\,\text{m/s}$ .

Step 2 — after the collision the combined mass is $2 + 1 = 3 \text{ kg}$ , moving at velocity $v$ . So $p_{\text{after}} = 3v$ .

Step 3 — apply conservation: $p_{\text{before}} = p_{\text{after}}$ , so $6 = 3v$ .

Step 4 — solve: $v = \dfrac{6}{3} = 2 \text{ m/s}$ .

Answer: the two trolleys move off together at $2 \text{ m/s}$ in the original direction.

Worked Example 3 — An explosion (recoil)

A stationary trolley system holds a compressed spring. When released, it pushes apart a $4 \text{ kg}$ trolley and a $1 \text{ kg}$ trolley. The $1 \text{ kg}$ trolley moves off at $8 \text{ m/s}$ . Calculate the velocity of the $4 \text{ kg}$ trolley.

Step 1 — before release, everything is at rest, so the total momentum before $= 0$ .

Step 2 — by conservation, the total momentum after must also be $0$ . Taking the $1 \text{ kg}$ trolley's direction as positive: $(1 \times 8) + (4 \times v) = 0$ .

Step 3 — evaluate the first momentum: $8 + 4v = 0$ .

Step 4 — solve for the velocity: $4v = -8$ , so $v = -2 \text{ m/s}$ .

Answer: the $4 \text{ kg}$ trolley moves at $2 \text{ m/s}$ in the opposite direction (the minus sign shows the direction).

Worked Example 3b — A collision where one object rebounds

A $2 \text{ kg}$ ball moving to the right at $5 \text{ m/s}$ collides head-on with a $3 \text{ kg}$ ball moving to the left at $2 \text{ m/s}$ . After the collision the $2 \text{ kg}$ ball rebounds and moves to the left at $1 \text{ m/s}$ . Calculate the velocity of the $3 \text{ kg}$ ball after the collision.

Step 1 — choose a positive direction. Take right as positive, so leftward velocities are negative.

Step 2 — total momentum before: $p_{\text{before}} = (2 \times 5) + (3 \times (-2)) = 10 - 6 = 4 \text{ kg}\,\text{m/s}$ .

Step 3 — total momentum after, with the $2 \text{ kg}$ ball now moving left at $1 \text{ m/s}$ (so $-1 \text{ m/s}$ ) and the $3 \text{ kg}$ ball at unknown velocity $v$ : $p_{\text{after}} = (2 \times (-1)) + (3 \times v) = -2 + 3v$ .

Step 4 — apply conservation, $p_{\text{before}} = p_{\text{after}}$ : $4 = -2 + 3v$ .

Step 5 — solve: $3v = 4 + 2 = 6$ , so $v = \dfrac{6}{3} = +2 \text{ m/s}$ .

Answer: the $3 \text{ kg}$ ball moves at $2 \text{ m/s}$ to the right (positive, so in the original positive direction). The whole calculation hinges on the signs: because the two balls started moving in opposite directions, one momentum had to be entered as negative, and the same care was needed for the rebounding ball afterwards. Get a sign wrong and the final answer is wrong — which is why fixing a positive direction at the very start is not optional.

Exam Tip: In an explosion, the total momentum is zero before and after if everything started at rest. A negative answer is not a mistake — it simply tells you the object moves in the opposite direction. Always state the direction in your answer. In a head-on collision where objects approach from opposite sides, one of the "before" velocities must be negative — the single most common momentum error is adding both as if they were positive.

Force as the Rate of Change of Momentum

When a force acts on an object it changes the object's momentum. In fact, force equals the rate of change of momentum:

$F = \dfrac{\Delta p}{\Delta t} = \dfrac{m\Delta v}{\Delta t}$

Momentum and Collisions (Higher)