Momentum — p = mv

Linear momentum is arguably the single most important concept in A-Level mechanics. It sits at the heart of Newton's Second Law (F = Δp/Δt), underpins the conservation principle that dominates all collision problems, and reappears later in circular motion, orbital mechanics, and even quantum physics. This lesson defines momentum carefully, works through its vector properties, and establishes the calculation habits you will need for the rest of OCR Module 3.5.

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Definition

Linear momentum is the product of an object's mass and its velocity.

p = m v

• p is momentum, a vector quantity.
• m is mass, measured in kilograms (kg).
• v is velocity, measured in metres per second (m s⁻¹).
• Units: kg m s⁻¹ (equivalently, N s — see Lesson 5 on impulse).

Because velocity is a vector, momentum is a vector: it has both magnitude and direction, and the direction of p is the same as the direction of v. A lorry and a bicycle can have the same momentum if the bicycle is moving very fast and the lorry very slowly — but only along the same line.

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Scalars vs Vectors — Why It Matters

A common OCR exam question is: "State what is meant by linear momentum and explain why it is a vector quantity."

Model answer: "Linear momentum is the product of an object's mass and its velocity (p = m v). It is a vector because velocity is a vector, and multiplying a vector by a scalar (mass) gives another vector with the same direction."

If your definition omits "vector" or "direction" you will lose marks.

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Worked Example 1 — Bullet and Lorry

Compare the magnitudes of the momentum of: (a) a 0.010 kg bullet travelling at 400 m s⁻¹, (b) a 20 000 kg lorry moving at 0.50 m s⁻¹.

• (a) p = 0.010 × 400 = 4.0 kg m s⁻¹
• (b) p = 20 000 × 0.50 = 10 000 kg m s⁻¹

The lorry has 2500 times the momentum of the bullet, even though the bullet travels 800 times faster. This is why slow-moving massive objects are dangerous in traffic accidents.

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Worked Example 2 — Car on a Bend

A 1200 kg car travels at 25 m s⁻¹ due East. 10 s later it is travelling at 25 m s⁻¹ due North (same speed, different direction). Find the change in momentum.

Because the speed is the same, beginners often say "no change". But momentum is a vector:

• Initial: p₁ = 1200 × 25 east = 30 000 kg m s⁻¹ east
• Final: p₂ = 1200 × 25 north = 30 000 kg m s⁻¹ north
• Change: Δp = p₂ − p₁

Using vector subtraction (draw p₂ − p₁ as a tip-to-tail diagram), the magnitude is:

|Δp| = √(30 000² + 30 000²) = 30 000 × √2 ≈ 4.24 × 10⁴ kg m s⁻¹

pointing north-west. Because the momentum has changed, Newton's Second Law tells us a resultant force acted on the car — in this case, friction from the tyres during the turn.

This example is critical: a change of direction is a change of momentum, even at constant speed. Centripetal force exists precisely because circular motion demands a constant change of momentum.

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Worked Example 3 — Ball Bouncing Off a Wall

A 0.20 kg ball moving horizontally at 5.0 m s⁻¹ hits a wall and rebounds at 4.0 m s⁻¹ in the opposite direction. Find the change in momentum.

Take the initial direction as positive.

• p_initial = 0.20 × (+5.0) = +1.0 kg m s⁻¹
• p_final = 0.20 × (−4.0) = −0.80 kg m s⁻¹
• Δp = −0.80 − 1.0 = −1.80 kg m s⁻¹

The negative sign tells you the change is in the opposite direction to the initial motion (i.e. away from the wall, towards the source). Notice that the magnitude of the change (1.80 kg m s⁻¹) is greater than either initial or final momentum alone; this is because the direction reverses.

A classic OCR mistake is to compute Δp = 0.80 − 1.0 = −0.20 kg m s⁻¹ by forgetting the negative sign on the final velocity. Always define and stick to a positive direction.

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Momentum and Newton's Second Law Revisited

From Lesson 2:

F = Δp/Δt

So the resultant force on a body is the rate at which its momentum is changing. Equivalently:

Δp = F · Δt (when F is constant)

This product is called the impulse, and is the subject of Lesson 5.