Dynamics
Mass and inertia, Newton's three laws, force as rate of change of momentum, weight, drag and terminal velocity, and conservation of momentum in elastic and inelastic collisions in one and two dimensions.
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In 30 seconds
- Mass is the property of an object that resists changes in its motion. Weight is the effect of a gravitational field on a mass: W = mg.
- Force is the rate of change of momentum, F = Δp/Δt. For constant mass this is F = ma, and the acceleration is in the direction of the resultant force.
- Momentum is always conserved in an interaction when no resultant external force acts. Kinetic energy is conserved only in an elastic collision.
- In an elastic collision, the relative speed of approach equals the relative speed of separation.
- Objects falling through air speed up until drag equals weight: then the resultant force is zero and they move at terminal velocity.
By the end you can
- State and apply Newton's three laws, with correct third-law pairs.
- Define linear momentum and force as rate of change of momentum.
- Describe qualitatively how drag affects falling objects and explain terminal velocity.
- Apply conservation of momentum to elastic and inelastic collisions in one and two dimensions.
1Newton's laws
The laws
Newton's first law — an object stays at rest or moves with constant velocity unless a resultant force acts on it.
Newton's second law — the rate of change of momentum of an object is proportional to the resultant force acting on it, and takes place in the direction of the force (F = Δp/Δt; F = ma for constant mass).
Newton's third law — if body A exerts a force on body B, then B exerts a force on A that is equal in magnitude, opposite in direction, and of the same type.
Linear momentum — the product of mass and velocity, p = mv (vector, unit kg m s⁻¹ or N s).
Using F = ma
- Draw a free-body diagram of one object, with a ruler, every force labelled ("weight", not "gravity"). Then resultant force = mass × acceleration.
A lift of mass 800 kg accelerates upwards at 1.5 m s⁻². Find the tension in the cable.
- resultant force = T − mg = ma C1
- T = 800 × (9.81 + 1.5) = 9048 N = 9.0 kN A1
A tennis ball of mass 0.15 kg hits a wall at 20 m s⁻¹ and rebounds at 15 m s⁻¹ along the same line. Contact lasts 0.050 s. Find the average force on the ball.
- taking the rebound direction as positive: Δp = 0.15 × 15 − 0.15 × (−20) = 5.25 N s C1
- F = Δp/Δt = 5.25/0.050 = 105 N A1 (in the rebound direction)
2Drag and terminal velocity
Non-uniform motion
- Friction opposes relative motion between surfaces. Drag (air resistance) opposes motion through a fluid and increases with speed.
- Falling from rest: at first, drag is zero, so a = g. As speed rises, drag rises, the resultant force falls, and so the acceleration falls. When drag = weight, a = 0: terminal velocity.
- A ball thrown upwards with air resistance: drag and weight both act downwards on the way up, so the deceleration is greater than g; on the way down they act in opposite directions, so the acceleration is less than g. Don't reuse the falling-from-rest graph for upward motion, as some 2024 candidates did.
3Conservation of momentum
The principle
Principle of conservation of momentum — for a system of interacting objects, the total momentum remains constant, provided no resultant external force acts on the system.
- Why it follows from Newton's laws: during a collision A pushes B and B pushes A with equal and opposite forces for the same time. So the momentum gained by one equals the momentum lost by the other.
- Write "total" or "sum of" in the statement: examiners in 2024 penalised "momentum is conserved" with no reference to the total.
A 2.0 kg trolley moving at 3.0 m s⁻¹ hits a 1.0 kg trolley at rest and they stick together. Find their common velocity and the kinetic energy lost.
- 2.0 × 3.0 + 0 = 3.0v ⇒ v = 2.0 m s⁻¹ C1 A1
- KE before = ½ × 2.0 × 3.0² = 9.0 J; after = ½ × 3.0 × 2.0² = 6.0 J C1
- 3.0 J lost (inelastic) A1
Elastic and inelastic collisions
| Momentum | Kinetic energy | Relative speed | |
|---|---|---|---|
| Elastic | conserved | conserved | approach = separation |
| Inelastic | conserved | not conserved (some becomes heat, sound, deformation) | separation < approach |
| Explosion (from rest) | total stays zero | increases (from chemical or stored energy) | — |
A 1.0 kg ball moving at 4.0 m s⁻¹ collides elastically head-on with a 3.0 kg ball at rest. The 1.0 kg ball rebounds at 2.0 m s⁻¹. Find the velocity of the 3.0 kg ball and show the collision is elastic.
- momentum: 1.0 × 4.0 = 1.0 × (−2.0) + 3.0v ⇒ v = 2.0 m s⁻¹ C1 A1
- relative speed of approach = 4.0 m s⁻¹; of separation = 2.0 − (−2.0) = 4.0 m s⁻¹, equal ⇒ elastic B1
- (check: KE before 8.0 J, after 2.0 + 6.0 = 8.0 J)
Collisions in two dimensions
- Momentum is conserved in each direction separately. Resolve every momentum into two perpendicular components, write one equation for each direction, then combine with Pythagoras and tan.
A 2.0 kg puck moving at 5.0 m s⁻¹ hits a stationary 3.0 kg puck. Afterwards the 2.0 kg puck moves at 3.0 m s⁻¹ at 60° to its original direction. Find the velocity of the 3.0 kg puck.
- along the original line: 2.0 × 5.0 = 2.0 × 3.0 cos 60° + 3.0v_x ⇒ 3.0v_x = 7.0 C1
- perpendicular: 0 = 2.0 × 3.0 sin 60° − 3.0v_y ⇒ 3.0v_y = 5.20 C1
- 3.0v = √(7.0² + 5.20²) = 8.72 ⇒ v = 2.9 m s⁻¹ A1
- direction: tan⁻¹(5.20/7.0) = 37° to the original line, on the other side A1
Test yourself — cover the answers first
Quick check
- Define linear momentum.
- State Newton's second law in terms of momentum.
- Is kinetic energy conserved in an inelastic collision?
- A gun of mass 4.0 kg fires a 20 g bullet at 400 m s⁻¹. Find the recoil speed.
- What is the resultant force on an object at terminal velocity?
Exam-style
- Water leaves a hose at 15 m s⁻¹ with mass flow rate 2.0 kg s⁻¹ and hits a wall, stopping. Find the force on the wall.
- Sketch and explain the velocity–time graph of a skydiver from leaving the plane until after the parachute opens.
- A 0.50 kg ball at 6.0 m s⁻¹ collides with a 1.5 kg ball moving at 2.0 m s⁻¹ in the opposite direction. They stick together. Find their velocity.
Show answers
- the product of mass and velocity
- the resultant force is proportional to (equal to) the rate of change of momentum, in the same direction
- no (momentum is)
- 0.020 × 400/4.0 = 2.0 m s⁻¹
- zero
- F = Δp/Δt = 2.0 × 15 = 30 N
- speed rises with decreasing gradient to terminal velocity; parachute opens: drag > weight, so decelerates (gradient negative, decreasing in size) to a new, lower terminal velocity
- 0.50 × 6.0 − 1.5 × 2.0 = 2.0v ⇒ v = 0
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