P16: Newton's Laws and Momentum
Newton's laws, inertia and momentum
Newton S Laws And Momentum
📋 Key Definitions
Newton's 1st Law: If the resultant force on an object is zero, the object will remain stationary or continue moving at a constant velocity.
Newton's 2nd Law: The acceleration of an object is proportional to the resultant force acting on it and inversely proportional to its mass. F = ma.
Newton's 3rd Law: When two objects interact, they exert equal and opposite forces on each other.
Inertia: The tendency of an object to resist a change in its state of motion. Mass is a measure of inertia — the greater the mass, the greater the inertia.
Momentum: A measure of the motion of an object, calculated as mass × velocity. Momentum is a vector quantity (p = mv).
📖 Newton's First Law
If there is no resultant force on an object:
- If it is stationary, it stays stationary
- If it is moving, it continues moving at the same speed in the same direction (constant velocity)
Newton's 1st Law explains why objects keep moving: In everyday life, things slow down because of friction. But in space (with no friction), an object would keep moving forever at constant velocity with no force needed.
📐 Newton's Second Law — F = ma
F = ma
F = resultant force (N), m = mass (kg), a = acceleration (m/s²)
The greater the resultant force, the greater the acceleration. The greater the mass, the smaller the acceleration for the same force.
Inertia and Newton's 2nd Law: A larger mass has greater inertia — it needs a larger force to produce the same acceleration. This is why a car accelerates faster than a lorry with the same engine force.
Worked Example 1: Calculating force
A car of mass 1200 kg accelerates at 3 m/s². Calculate the resultant force.
F = ma = 1200 × 3 = 3600 N
Worked Example 2: Calculating acceleration
A resultant force of 500 N acts on a mass of 100 kg. Calculate the acceleration.
a = F / m = 500 / 100 = 5 m/s²
Worked Example 3: Calculating mass
A resultant force of 2400 N produces an acceleration of 4 m/s². Calculate the mass.
m = F / a = 2400 / 4 = 600 kg
🔄 Newton's Third Law
When object A exerts a force on object B, object B exerts an equal and opposite force on object A. The forces are always equal in magnitude and opposite in direction.
Newton's 3rd Law force pairs: The two forces must act on different objects. They must be the same type of force (e.g. both gravitational, both contact). They do NOT cancel each other because they act on different objects.
Examples:
- You push on the ground → the ground pushes back on you (normal contact force)
- Rocket pushes gas downwards → gas pushes rocket upwards
- Earth pulls you down (gravity) → you pull Earth up (gravity on Earth)
⚖️ Weight
W = mg
W = weight (N), m = mass (kg), g = gravitational field strength (N/kg)
On Earth, g ≈ 9.8 N/kg
Weight vs mass: Mass is the amount of matter in an object (scalar, measured in kg — does not change with location). Weight is the gravitational force on the object (vector, measured in N — changes with gravitational field strength).
🎯 Momentum
p = mv
p = momentum (kg m/s), m = mass (kg), v = velocity (m/s)
Momentum is a vector — it has direction. An object at rest has zero momentum (v = 0).
Worked Example 4: Calculating momentum
A car of mass 1000 kg travels at 20 m/s. Calculate its momentum.
p = mv = 1000 × 20 = 20 000 kg m/s
🔄 Conservation of Momentum
Conservation of momentum: In a closed system, the total momentum before an event equals the total momentum after the event. Momentum is always conserved.
Worked Example 5: Conservation of momentum
A trolley of mass 2 kg moving at 5 m/s collides with a stationary trolley of mass 3 kg. They join together. Calculate the velocity after the collision.
Momentum before = momentum after
(2 × 5) + (3 × 0) = (2 + 3) × v
10 + 0 = 5v
v = 10 / 5 = 2 m/s
⚡ Force and Momentum
F = Δp / Δt = (mv − mu) / t
F = force (N), Δp = change in momentum (kg m/s), Δt = time (s)
This explains safety features: Crumple zones, airbags and seatbelts all increase the time over which the momentum changes. Since F = Δp/Δt, a longer time means a smaller force for the same change in momentum, reducing injuries.
❓ Practice Questions
Q1: Foundation State Newton's three laws of motion.
Q2: Foundation A resultant force of 1500 N acts on a car of mass 750 kg. Calculate the acceleration.
Q3: Higher A trolley of mass 4 kg moving at 6 m/s collides with a stationary trolley of mass 2 kg. They join together. Calculate the velocity after the collision.
Q4: Higher Explain, using F = Δp/Δt, how a seatbelt reduces the force on a passenger during a crash.
Q5: Foundation Calculate the weight of a person with mass 70 kg on Earth (g = 9.8 N/kg).
✅ Answers
- 1st: If the resultant force is zero, an object stays stationary or moves at constant velocity. 2nd: F = ma, acceleration is proportional to force and inversely proportional to mass. 3rd: When two objects interact, they exert equal and opposite forces on each other.
- a = F / m = 1500 / 750 = 2 m/s²
- Momentum before = (4 × 6) + (2 × 0) = 24 kg m/s. After: (4 + 2) × v = 6v. So 6v = 24, v = 4 m/s
- During a crash, the passenger's momentum changes from a high value to zero. The seatbelt stretches, increasing the time over which this momentum change happens. Since F = Δp/Δt, a longer time means a smaller force acts on the passenger, reducing injuries.
- W = mg = 70 × 9.8 = 686 N
🎯 Exam Tips
- Newton's 3rd Law forces act on different objects — they don't cancel out
- For F = ma, make sure you use the resultant force (not just one force)
- In conservation of momentum questions, remember that momentum is a vector — objects moving in opposite directions have opposite signs
- When asked about safety features, always link back to F = Δp/Δt — increase time to decrease force
- Weight changes with location (different g), mass does not — this is commonly tested
- Inertia = resistance to change in motion. A larger mass = greater inertia = harder to accelerate
🔢 Maths Skills
Mathematical Skills
Use F = ma to find force, mass or acceleration. Calculate momentum using p = mv. Apply conservation of momentum: total pbefore = total pafter. Use F = Δp/Δt to find force from a change in momentum. Remember momentum is a vector — use positive and negative signs for opposite directions.
Maths Example
A 1500 kg car travelling at 20 m/s crashes into a wall and stops in 0.3 s. Calculate the force: Δp = mv − mu = 1500 × 0 − 1500 × 20 = −30 000 kg m/s. F = Δp/Δt = 30 000 / 0.3 = 100 000 N. If the crumple zone increases the time to 0.8 s: F = 30 000 / 0.8 = 37 500 N — a much smaller force.
⚠️ Common Misconceptions
Watch Out!
1. Wrong: If a car moves at constant speed, there must be a resultant force driving it forward. Correct: At constant speed, the resultant force is ZERO (Newton's 1st Law). The driving force equals the resistive forces (friction + drag).
2. Wrong: In Newton's 3rd Law, the forces cancel out so nothing happens. Correct: The forces act on DIFFERENT objects, so they don't cancel. The Earth accelerates towards you just as you accelerate towards it, but the Earth's mass is so large its acceleration is negligible.
3. Wrong: Momentum is always conserved only in collisions. Correct: Momentum is conserved in ANY closed system interaction — collisions, explosions, and recoil situations. In an explosion, total momentum before = 0, so the fragments have equal and opposite momenta after.
✍️ 6-Mark Question
Extended Answer
6 marks: A car of mass 1200 kg travelling at 25 m/s crashes into a barrier. Explain, using the equation F = Δp/Δt, how crumple zones and seatbelts reduce the force on the driver. Compare this to a rigid car with no seatbelt.
The driver has a momentum of p = mv = 1200 × 25 = 30 000 kg m/s before the crash. During the crash, this momentum must be reduced to zero. The force on the driver depends on the time over which the momentum changes: F = Δp/Δt. In a rigid car with no seatbelt, the driver would hit the dashboard or windscreen and stop in a very short time (perhaps 0.01 s), giving a very large force: F = 30 000 / 0.01 = 3 000 000 N. Crumple zones at the front of the car deform plastically, increasing the time over which the car decelerates. Seatbelts stretch slightly, increasing the time over which the driver decelerates relative to the car. Airbags also spread the force over a larger area and increase the stopping time. With these safety features, the stopping time might be 0.5 s, giving F = 30 000 / 0.5 = 60 000 N — a much smaller force that is less likely to cause serious injury.
Mark scheme: 1 mark for calculating initial momentum; 1 mark for stating F = Δp/Δt; 1 mark for rigid car gives very short stopping time → very large force; 1 mark for crumple zones increase stopping time; 1 mark for seatbelts increase stopping time (and spread force); 1 mark for quantitative comparison showing reduced force. (6 marks total)
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
Two trolleys collide on a track. Trolley A (mass 0.8 kg) moves at 3 m/s to the right. Trolley B (mass 1.2 kg) is stationary. After the collision, Trolley A moves at 0.6 m/s to the left.
(a) Calculate the velocity of Trolley B after the collision.
(b) Verify that momentum is conserved.
(c) Calculate the total kinetic energy before and after the collision. Explain what this tells you about the type of collision.
Answers: (a) Momentum before = (0.8 × 3) + (1.2 × 0) = 2.4 kg m/s. After = (0.8 × −0.6) + (1.2 × v) = −0.48 + 1.2v. So 1.2v = 2.4 + 0.48 = 2.88, v = 2.4 m/s to the right. (b) Before: 2.4 kg m/s. After: (0.8 × −0.6) + (1.2 × 2.4) = −0.48 + 2.88 = 2.4 kg m/s. Momentum is conserved ✓. (c) KE before = ½ × 0.8 × 3² = 3.6 J. KE after = ½ × 0.8 × 0.6² + ½ × 1.2 × 2.4² = 0.144 + 3.456 = 3.6 J. Total KE is the same before and after, so this is an elastic collision (no energy is lost to heat/sound/deformation).
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