state and apply each of Newton’s laws of motion

Momentum and Newton’s Laws of Motion

Learning Objective

State and apply each of Newton’s three laws of motion to solve quantitative problems involving momentum, impulse and collisions (both one‑ and two‑dimensional) as required by Cambridge International AS & A Level Physics (9702) – Section 3.1.

Key Concepts

• Mass (inertia) – a scalar quantity that measures an object’s resistance to a change in its state of motion. The larger the mass, the greater the inertia.
• Weight – the force exerted on a mass by gravity. \[ W = mg \] where g = 9.8 m s⁻² (≈10 m s⁻² on Earth). Weight is a force, measured in newtons (N).
• Linear momentum – a vector quantity defined by \[ \mathbf{p}=m\mathbf{v}\qquad\text{(kg·m·s}^{-1}\text{)}
• Impulse – the change in momentum produced by a force acting over a time interval: \[ \mathbf{J}= \int_{t_1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p} \] For a constant force, \(\mathbf{J}= \mathbf{F}\,\Delta t\).
• Net external force – the vector sum of all forces acting on a system. If \(\mathbf{F}_{\text{net}}=0\) the momentum of the system is constant.

Newton’s First Law – Law of Inertia

Statement: A body at rest or moving with constant velocity will continue in that state unless acted upon by a net external force.

• Mathematical form: \(\mathbf{F}_{\text{net}}=0\;\Rightarrow\;\mathbf{p}= \text{constant}\).
• Implication: In the absence of a net force, both speed and direction of motion are unchanged.
• Typical examples: a spacecraft coasting in deep space; a puck on a frictionless air table.

Newton’s Second Law – Relation Between Force, Mass and Acceleration

Statement: The net external force on a body equals the rate of change of its momentum.

\[ \mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt}= \frac{d}{dt}(m\mathbf{v}) \]

• For constant mass this reduces to the familiar form \[ \boxed{\mathbf{F}_{\text{net}} = m\mathbf{a}} \] where \(\mathbf{a}=d\mathbf{v}/dt\) (m s⁻²). This explicit \(F=ma\) formula is the one required for the syllabus.
• Force is a vector; it can change the magnitude *and* the direction of \(\mathbf{p}\).
• Units: \(1\;\text{N}=1\;\text{kg·m·s}^{-2}\).
• Impulse theorem (integrated form): \(\displaystyle \int\mathbf{F}\,dt = \Delta\mathbf{p}\).

Newton’s Third Law – Action and Reaction

Statement: For every action force there is an equal and opposite reaction force acting on a different body.

\[ \mathbf{F}_{AB} = -\mathbf{F}_{BA} \]

• Because the forces act on different bodies they never cancel when analysing the motion of a single object.
• Essential for analysing internal forces in collisions and for rocket propulsion.

Conservation of Momentum

In an isolated system (no external forces), the total vector momentum remains constant:

\[ \sum\mathbf{p}_{\text{initial}} = \sum\mathbf{p}_{\text{final}} \]

• Applies to linear momentum (the focus of the syllabus). Angular momentum is not required for AS/A Level.
• Forms the basis for analysing collisions and explosions.

Collisions

Collision type	Momentum	Kinetic energy	Typical example
Elastic	Conserved	Conserved	Billiard balls
Inelastic	Conserved	Not conserved (transformed to heat, deformation, sound)	Car crash
Perfectly inelastic	Conserved	Maximum loss (objects stick together)	Two carts coupling

Elastic collisions – useful shortcuts

• Relative speed of approach = relative speed of separation: \[ |u_{1}-u_{2}| = |v_{2}-v_{1}| \] This condition is mathematically equivalent to conservation of kinetic energy and is often quicker to apply.
• The coefficient of restitution \(e\) is not required by the syllabus, but \(e=1\) for a perfectly elastic collision.

Two‑dimensional collisions

Momentum must be conserved in each orthogonal direction independently:

\[ \sum p_{x\text{,i}} = \sum p_{x\text{,f}},\qquad \sum p_{y\text{,i}} = \sum p_{y\text{,f}} \]

Thus the same algebraic procedure used in 1‑D can be applied to the \(x\)‑ and \(y\)-components separately.

Algorithm for Solving Momentum Problems

1. Free‑body diagram: Draw all bodies, indicate velocities and external forces.
2. Identify external forces and decide whether \(\mathbf{F}_{\text{net}}=0\).
   • If \(\mathbf{F}_{\text{net}}=0\) → use conservation of momentum.
   • If \(\mathbf{F}_{\text{net}}eq0\) → apply Newton’s second law in the form \(\mathbf{F}_{\text{net}}=d\mathbf{p}/dt\) or \(\mathbf{J}= \Delta\mathbf{p}\).
3. Choose the collision type (elastic, inelastic, perfectly inelastic) and write the appropriate equations:
   • Momentum equations – one for each dimension.
   • Energy equation for elastic collisions or the relative‑speed condition.
4. Solve the simultaneous equations for the unknown velocities (or other required quantities).
5. Check your answer:
   • Units (N, kg·m·s⁻¹, m s⁻¹, etc.).
   • Vector directions (signs).
   • Physical plausibility (e.g., speeds non‑negative where appropriate).

Worked Examples

Example 1 – One‑dimensional elastic collision

Two spheres: \(m_{1}=0.5\;\text{kg}\) moving at \(u_{1}=4.0\;\text{m s}^{-1}\), \(m_{2}=0.8\;\text{kg}\) initially at rest. Find \(v_{1}\) and \(v_{2}\) after the collision.

1. Conservation of momentum: \[ 0.5(4)=0.5v_{1}+0.8v_{2}\qquad(1) \]
2. Relative‑speed condition (elastic): \[ |u_{1}-0| = |v_{2}-v_{1}|\qquad(2) \]
3. Solving (1) and (2) gives \[ v_{1}= \frac{m_{1}-m_{2}}{m_{1}+m_{2}}\,u_{1}= -0.92\;\text{m s}^{-1} \] \[ v_{2}= \frac{2m_{1}}{m_{1}+m_{2}}\,u_{1}= 3.08\;\text{m s}^{-1} \]
4. Interpretation: \(m_{1}\) rebounds opposite to its original direction, \(m_{2}\) moves forward.

Example 2 – Perfectly inelastic collision (1‑D)

A \(2.0\;\text{kg}\) cart moving at \(3.0\;\text{m s}^{-1}\) collides head‑on with a \(3.0\;\text{kg}\) cart moving at \(-2.0\;\text{m s}^{-1}\). The carts stick together. Find the common speed after the collision.

1. Conservation of momentum (no external horizontal force): \[ 2.0(3.0)+3.0(-2.0)= (2.0+3.0)v_{f} \]
2. Calculate: \[ 6.0-6.0 =5v_{f}\;\Rightarrow\;v_{f}=0\;\text{m s}^{-1} \]
3. The combined system comes to rest – the two momenta cancel.

Example 3 – Two‑dimensional elastic collision

Two identical billiard balls (\(m=0.2\;\text{kg}\)) move on a frictionless table. Ball A travels at \(5.0\;\text{m s}^{-1}\) along the \(x\)-axis, ball B is initially at rest. After an elastic collision ball A is deflected \(30^{\circ}\) above the \(x\)-axis. Find the speed and direction of ball B.

1. Because the masses are equal, the speed of each ball after an elastic collision equals the initial speed of the moving ball: \(|v_{A}|=|v_{B}|=5.0\;\text{m s}^{-1}\).
2. Apply momentum conservation in the two components:
   • \(x\)-direction: \(5.0 = v_{A}\cos30^{\circ}+v_{B}\cos\theta\)
   • \(y\)-direction: \(0 = v_{A}\sin30^{\circ}-v_{B}\sin\theta\)
   Using \(v_{A}=5.0\;\text{m s}^{-1}\) gives \(\theta =150^{\circ}\) (i.e. \(30^{\circ}\) below the negative \(x\)-axis) and \(v_{B}=5.0\;\text{m s}^{-1}\).
3. Thus ball B moves away at \(5.0\;\text{m s}^{-1}\) making an angle \(150^{\circ}\) measured from the positive \(x\)-axis.

Key Summary Table

Law	Statement	Mathematical form	Typical application
First	A body remains at rest or in uniform motion unless acted upon by a net external force.	\(\mathbf{F}_{\text{net}}=0\;\Rightarrow\;\mathbf{p}= \text{constant}\)	Spacecraft drift, frictionless tracks
Second	Net external force equals the rate of change of momentum.	\(\mathbf{F}_{\text{net}} = \dfrac{d\mathbf{p}}{dt}=m\mathbf{a}\) (for constant \(m\))	Impulse problems, acceleration calculations, \(F=ma\)
Third	For every action there is an equal and opposite reaction on a different body.	\(\mathbf{F}_{AB} = -\mathbf{F}_{BA}\)	Collision analysis, rocket thrust

Practice Questions

1. Perfectly inelastic collision (1‑D): A \(2.0\;\text{kg}\) cart moving at \(3.0\;\text{m s}^{-1}\) collides head‑on with a \(3.0\;\text{kg}\) cart moving at \(-2.0\;\text{m s}^{-1}\). The carts stick together. Find their common speed after the collision.
2. Impulse: A constant horizontal force of \(10\;\text{N}\) acts on a \(0.5\;\text{kg}\) ball for \(0.20\;\text{s}\). Determine the change in the ball’s momentum and its final speed if it started from rest.
3. Momentum conservation on a frictionless surface: Two ice skaters, masses \(50\;\text{kg}\) and \(70\;\text{kg}\), push off each other. If the lighter skater moves away at \(2.5\;\text{m s}^{-1}\), what is the speed of the heavier skater?
4. Two‑dimensional elastic collision: Two identical pucks (\(m=0.15\;\text{kg}\)) on a smooth air‑table. Puck A moves at \(4.0\;\text{m s}^{-1}\) along the \(x\)-axis; puck B is initially at rest. After an elastic collision puck A is observed to move at \(3.0\;\text{m s}^{-1}\) making an angle \(40^{\circ}\) above the \(x\)-axis. Determine the speed and direction of puck B.

Further Reading

Consult the Cambridge International AS & A Level Physics (9702) syllabus, especially Section 3.1 “Momentum and Newton’s laws of motion”, for detailed exam specifications, additional worked examples, and marking criteria.