Linear Momentum: Study Guide (Physics C: Mechanics 4.1)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both open Unit 4 with a topic called Linear Momentum. This guide is the calculus-based one. It finds momentum from position functions by differentiating and treats momentum as a two-dimensional vector throughout. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Unit 3 described motion with energy, a scalar. Unit 4 adds a second quantity that every moving object carries: linear momentum. Unlike kinetic energy, it keeps track of direction. That makes it the right tool for short, violent interactions such as impacts, kicks and things breaking apart.

The definition: p = mv

The linear momentum of an object of mass m moving with velocity v is

p = mv

• p is the momentum in kg·m/s. Because 1 N = 1 kg·m/s², this is the same unit as N·s.
• m is the mass in kg, a positive scalar.
• v is the velocity in m/s, a vector.

“Linear” separates it from angular momentum, which comes later in the course. On its own, “momentum” in this unit means linear momentum.

Multiplying a vector by a positive scalar changes its length but not its direction. So momentum always points the same way as the velocity. In one dimension that means p_x has the same sign as v_x. Take +x to the right: a 3.0 kg trolley moving left at 2.0 m/s has p_x = 3.0 × (−2.0) = −6.0 kg·m/s.

Momentum as a vector in two dimensions

In two dimensions, work component by component:

p_x = m v_x and p_y = m v_y

The size of the momentum is |p| = √(p_x² + p_y²) = m|v|, and its direction is the angle θ with tan θ = p_y/p_x. Always check the quadrant from the signs of the components, because a calculator’s inverse tangent cannot tell (+, +) from (−, −).

Momentum of a system

A system is whatever collection of objects you choose to study (Topic 2.1). The momentum of a system is the vector sum of the momenta of its parts:

p_sys = p₁ + p₂ + p₃ + …

Two features follow at once:

• Momenta in opposite directions cancel. Two identical carts moving apart at equal speeds have a total momentum of zero, even though both are moving.
• Kinetic energies never cancel. K is a scalar and is never negative, so the total kinetic energy of the same two carts is the sum of two positive numbers.

Topic 4.3 shows that, in the absence of a net external force, the system momentum stays constant. For now, the job is to find it correctly.

Worked example 1: the momentum of two pucks

Question. Take +x to the right and +y up across an air table seen from above. Puck A (0.40 kg) moves with velocity (2.5, 1.0) m/s. Puck B (0.25 kg) moves with velocity (−2.0, 3.2) m/s. Find (a) the momentum of each puck, (b) the momentum of the two-puck system, and (c) the total kinetic energy.

1. Puck A: p_A = 0.40 × (2.5, 1.0) = (1.0, 0.40) kg·m/s. Size √(1.0² + 0.40²) ≈ 1.08 kg·m/s.
2. Puck B: p_B = 0.25 × (−2.0, 3.2) = (−0.50, 0.80) kg·m/s. Size ≈ 0.94 kg·m/s.
3. System, by components: p_x = 1.0 + (−0.50) = 0.50 kg·m/s; p_y = 0.40 + 0.80 = 1.20 kg·m/s.
4. Size: √(0.50² + 1.20²) = 1.30 kg·m/s. Direction: tan θ = 1.20 ÷ 0.50, so θ ≈ 67° above +x. Both components are positive, so the angle is in the first quadrant.
5. Kinetic energy: K_A = ½ × 0.40 × (2.5² + 1.0²) = 1.45 J; K_B = ½ × 0.25 × (2.0² + 3.2²) = 1.78 J. Total K = 3.23 J.

Check. Adding the sizes, 1.08 + 0.94 ≈ 2.02 kg·m/s, would ignore direction. The x-components point opposite ways and partly cancel, so the true total is smaller. The energies, being scalars, simply add.

Momentum that changes with time

If an object’s position is known as a function of time, its momentum follows by differentiating:

p_x(t) = m v_x(t) = m dx/dt

For constant mass, the momentum–time graph has exactly the shape of the velocity–time graph, stretched vertically by m. So the features you learned to read in Topic 1.2 carry over:

• p_x = 0 wherever the object is momentarily at rest.
• p_x changes sign wherever the object turns around.
• p_x has a maximum or minimum where its slope dp_x/dt is zero.

The slope dp_x/dt has the unit kg·m/s², which is a newton. Topic 4.2 shows that this slope is the net force on the object.

Worked example 2: momentum from a position function

Question. Take +x along a straight track. A 2.0 kg box moves with x(t) = (1.5 m/s²)t² − (0.25 m/s³)t³ for 0 ≤ t ≤ 6.0 s. Find p_x(t), when the box has zero momentum, its greatest positive momentum, and its momentum and kinetic energy at 6.0 s. Sketch p_x against t.

1. Velocity: v_x = dx/dt = 3.0t − 0.75t² (m/s).
2. Momentum: p_x = m v_x = 2.0(3.0t − 0.75t²) = 6.0t − 1.5t² (kg·m/s).
3. Zero momentum: 1.5t(4.0 − t) = 0, so t = 0 and t = 4.0 s. At 4.0 s the box turns round, at x = 8.0 m.
4. Greatest positive momentum: dp_x/dt = 6.0 − 3.0t = 0 at t = 2.0 s, so p_max = 12 − 6.0 = 6.0 kg·m/s.
5. At t = 6.0 s: p_x = 36 − 54 = −18 kg·m/s (18 kg·m/s backwards along the track).
6. Kinetic energy at 6.0 s: K = p²/(2m) = 18² ÷ (2 × 2.0) = 81 J.

Check. At t = 2.0 s, v_x = 3.0 m/s, so ½mv² = ½ × 2.0 × 9.0 = 9.0 J, and p²/(2m) = 36 ÷ 4.0 = 9.0 J. The two routes to K agree.

Interpretation. Between 4.0 s and 6.0 s the size of the momentum grows from 0 to 18 kg·m/s while the box moves backwards. A growing size of momentum with a negative sign is still “speeding up”.

Momentum and kinetic energy

Both p and K depend on mass and speed, but differently. Substitute v = p/m into K = ½mv²:

K = p²/(2m), or equivalently |p| = √(2mK)

This makes factor-of-change questions quick.

Change	Factor on p	Factor on K
Speed doubled, mass fixed	× 2	× 4
Mass doubled, speed fixed	× 2	× 2
Mass halved, speed × 4	× 2	× 8
Mass tripled, speed ÷ 3	× 1 (unchanged)	÷ 3

The last row shows why the two quantities tell different stories. With the same momentum, the lighter object always has the larger kinetic energy, because K = p²/(2m) falls as m rises.

Collisions and explosions as models

Momentum is most useful for interactions that are over quickly. Two models describe them.

The collision model. An interaction is a collision when the forces the objects exert on each other are much larger than the net external force on them during the interaction. Then, over the short contact time, external forces have almost no effect.

For example, suppose a bat pushes on a 0.15 kg ball with an average force of about 8000 N for 1.2 ms (invented but realistic figures). The ball’s weight is 0.15 × 9.8 = 1.47 N, more than 5000 times smaller. Over the contact, the bat changes the ball’s velocity by about (8000 ÷ 0.15) × 0.0012 = 64 m/s; gravity changes it by only 9.8 × 0.0012 ≈ 0.012 m/s. Ignoring gravity during the hit is safe.

The object model in collisions. In a collision you usually compare only the state just before with the state just after. You do not follow the squashing and twisting during contact. So each colliding body can be treated as a single object, even though it deforms.

The explosion model. An interaction is an explosion when forces internal to the system push its parts apart. A firework shell bursting, a spring released between two carts, a person stepping off a canoe and a gun recoiling as it fires are all explosions in this sense. The word does not require fire or noise.

When neither model applies, for example a box sliding slowly into a soft cushion while friction from the floor is comparable to the cushion’s push, external forces matter during the interaction and must be included.

Common misconceptions

• “Momentum is just another word for kinetic energy.” Momentum is a vector, proportional to v; kinetic energy is a scalar, proportional to v². They scale differently and only momentum can cancel.
• Adding sizes of momenta. Add components with signs (Worked example 1).
• Dropping the sign in one dimension. p_x takes the sign of v_x.
• Forgetting the quadrant. Check the signs of p_x and p_y after using an inverse tangent.
• “Zero total momentum means nothing is moving.” Equal and opposite momenta give zero total while both objects move.
• “Same momentum means same kinetic energy.” For equal momenta, the lighter object has more kinetic energy.
• “An explosion must involve burning.” The model only requires internal forces that push parts of the system apart.
• Applying the collision model to slow interactions. If external forces act for a long time compared with the impact, they cannot be ignored.

Where this leads

Topic 4.2 asks what changes momentum. It shows that the slope of a momentum–time graph is the net force, and that the area under a force–time graph is the change in momentum: start the Change in Momentum and Impulse study guide next. Topics 4.3 and 4.4 then use system momentum to analyse collisions and explosions. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.