state and apply each of Newton’s laws of motion

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Momentum and Newton’s Laws of Motion

Momentum and Newton’s Laws of Motion

Learning Objective

State and apply each of Newton’s laws of motion to solve quantitative problems involving momentum and impulse.

Newton’s First Law – Law of Inertia

A body remains at rest, or moves with constant velocity, unless acted upon by a net external force.

  • Implication: In the absence of a net force, the momentum \$p = mv\$ of a particle is constant.
  • Application: Predict motion of objects on frictionless surfaces or in deep space.

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

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:

\$\mathbf{F}_{\text{net}} = m\mathbf{a}\$

  • Force is a vector; it changes both magnitude and direction of momentum.
  • Units: \$\text{N} = \text{kg·m·s}^{-2}\$.

Newton’s Third Law – Action and Reaction

For every action force there is an equal and opposite reaction force.

\$\mathbf{F}{AB} = -\mathbf{F}{BA}\$

  • Forces act on different bodies; they do not cancel when analysing the motion of a single object.
  • Essential for understanding internal forces in collisions.

Momentum (\$\mathbf{p}\$)

Linear momentum of a particle of mass \$m\$ moving with velocity \$\mathbf{v}\$ is defined as:

\$\mathbf{p} = m\mathbf{v}\$

Momentum is a conserved vector quantity in isolated systems.

Impulse (\$\mathbf{J}\$)

Impulse is the change in momentum produced by a force acting over a time interval \$\Delta t\$:

\$\mathbf{J} = \int{t1}^{t_2} \mathbf{F}\,dt = \Delta\mathbf{p}\$

For a constant force:

\$\mathbf{J} = \mathbf{F}\,\Delta t\$

Conservation of Momentum

In the absence of external forces, the total momentum of a system remains constant:

\$\sum \mathbf{p}{\text{initial}} = \sum \mathbf{p}{\text{final}}\$

This principle underlies the analysis of collisions and explosions.

Types of Collisions

Collision TypeMomentumKinetic EnergyTypical Example
ElasticConservedConservedBilliard balls
InelasticConservedNot conserved (some lost as heat, deformation)Car crash
Perfectly InelasticConservedMaximum loss (objects stick together)Two carts coupling

Applying Newton’s Laws to Momentum Problems

  1. Identify all external forces acting on the system.
  2. Use Newton’s second law in the form \$\mathbf{F}_{\text{net}} = d\mathbf{p}/dt\$ to relate forces to momentum change.
  3. If the net external force is zero, set the total initial momentum equal to the total final momentum.
  4. For collisions, apply conservation of momentum together with an energy condition (elastic or inelastic) to solve for unknown velocities.
  5. Check units and direction (vector nature) of all quantities.

Worked Example – Elastic Collision in One Dimension

Two spheres, \$m1 = 0.5\ \text{kg}\$ moving at \$u1 = 4\ \text{m·s}^{-1}\$ and \$m_2 = 0.8\ \text{kg}\$ initially at rest, collide elastically. Find their velocities after the collision.

Solution steps:

  1. Write conservation of momentum:

    \$m1u1 + m2u2 = m1v1 + m2v2\$

    Since \$u_2 = 0\$:

    \$0.5(4) = 0.5v1 + 0.8v2\$

  2. Write conservation of kinetic energy (elastic collision):

    \$\frac{1}{2}m1u1^{2} + \frac{1}{2}m2u2^{2} = \frac{1}{2}m1v1^{2} + \frac{1}{2}m2v2^{2}\$

    \$\frac{1}{2}(0.5)(4^{2}) = \frac{1}{2}(0.5)v1^{2} + \frac{1}{2}(0.8)v2^{2}\$

  3. Solve the simultaneous equations to obtain:

    \$v1 = \frac{m1 - m2}{m1 + m2}\,u1 = \frac{0.5 - 0.8}{1.3}\times4 = -0.923\ \text{m·s}^{-1}\$

    \$v2 = \frac{2m1}{m1 + m2}\,u_1 = \frac{2(0.5)}{1.3}\times4 = 3.08\ \text{m·s}^{-1}\$

  4. Interpretation: \$m1\$ rebounds in the opposite direction, \$m2\$ moves forward.

Suggested diagram: A horizontal line showing the two masses before and after collision, with arrows indicating velocities \$u1\$, \$v1\$, and \$v_2\$.

Key Summary Table

LawStatementMathematical FormTypical Application
FirstA body at rest or in uniform motion remains so unless acted upon by a net external force.\$\mathbf{F}_{\text{net}} = 0 \;\Rightarrow\; \mathbf{p} = \text{constant}\$Spacecraft drift, frictionless tracks
SecondNet external force equals the rate of change of momentum.\$\mathbf{F}_{\text{net}} = \dfrac{d\mathbf{p}}{dt}\$Calculating acceleration, impulse problems
ThirdFor every action there is an equal and opposite reaction.\$\mathbf{F}{AB} = -\mathbf{F}{BA}\$Collision analysis, rocket propulsion

Practice Questions

  1. A 2.0 kg cart moving at \$3.0\ \text{m·s}^{-1}\$ collides head‑on with a 3.0 kg cart moving at \$-2.0\ \text{m·s}^{-1}\$. The carts stick together. Find their common speed after the collision.
  2. A constant horizontal force of \$10\ \text{N}\$ acts on a 0.5 kg ball for \$0.2\ \text{s}\$. Determine the change in the ball’s momentum and its final speed if it started from rest.
  3. Two ice skaters, masses \$50\ \text{kg}\$ and \$70\ \text{kg}\$, push off each other on frictionless ice. If the lighter skater moves away at \$2.5\ \text{m·s}^{-1}\$, what is the speed of the heavier skater?

Further Reading

Consult the Cambridge International AS & A Level Physics (9702) syllabus sections on “Momentum and Newton’s laws of motion” for detailed exam specifications and additional worked examples.