Momentum and Collisions

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

A moving lorry is far harder to stop than a moving bicycle travelling at the same speed, and a fast bullet does more damage 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 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 — crumple zones, airbags and seat belts — save lives.

By the end of this lesson you should be able to calculate momentum using $p = mv$ , apply conservation of momentum to work out 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 time over which they stop.

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 a great deal in collisions: momentum to the right is treated as positive and momentum to the left as negative, and they can cancel out. 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 stick to it throughout. Any object moving the other way then has a negative velocity and therefore a negative momentum. This bookkeeping is what allows momenta to 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: Momentum is $p = mv$ and its unit is $\text{kg}\,\text{m/s}$ — do not confuse it with kinetic energy $\dfrac{1}{2}mv^2$ , 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 is equal to 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 rifle of mass $4 \text{ kg}$ fires a bullet of mass $0.02 \text{ kg}$ at $200 \text{ m/s}$ . Calculate the recoil velocity of the rifle.

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

Step 2 — by conservation, the total momentum after must also be $0$ . Taking the bullet's direction as positive: $(0.02 \times 200) + (4 \times v) = 0$ .

Step 3 — evaluate the bullet's momentum: $4 + 4v = 0$ .

Step 4 — solve for the rifle's velocity: $4v = -4$ , so $v = -1 \text{ m/s}$ .

Answer: the rifle recoils at $1 \text{ m/s}$ in the opposite direction to the bullet (the minus sign shows the direction).

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.

Force as the Rate of Change of Momentum (Higher)

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

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

where $\Delta p$ is the change in momentum, $\Delta t$ is the time taken for the change, $m$ is the mass and $\Delta v$ is the change in velocity. Reading this equation carefully reveals the key safety idea: for a given change in momentum $\Delta p$ , making the time $\Delta t$ longer makes the force $F$ smaller. The same change in momentum spread over a longer time produces a gentler force.

Worked Example 4 — Force from change in momentum

A $0.15 \text{ kg}$ ball travelling at $12 \text{ m/s}$ is brought to rest by a wall in $0.02 \text{ s}$ . Calculate the average force on the ball.

Step 1 — find the change in velocity: $\Delta v = 12 - 0 = 12 \text{ m/s}$ .

Step 2 — find the change in momentum: $\Delta p = m\Delta v = 0.15 \times 12 = 1.8 \text{ kg}\,\text{m/s}$ .

Step 3 — use $F = \dfrac{\Delta p}{\Delta t} = \dfrac{1.8}{0.02}$ .

Step 4 — calculate: $F = 90 \text{ N}$ .

Answer: the average force on the ball is $90 \text{ N}$ .

Momentum and Collisions (Higher)