recall F = ma and solve problems using it, understanding that acceleration and resultant force are always in the same direction

Momentum and Newton’s Laws of Motion

Learning Objective

Recall the relationship \$F = ma\$ and apply it to solve quantitative problems. Understand that the direction of the resultant force is always the same as the direction of the acceleration.

Newton’s First Law – Inertia

An object remains at rest or moves with constant velocity unless acted upon by a non‑zero resultant force.

Newton’s Second Law – \$F = ma\$

The resultant (net) force \$\mathbf{F}_{\text{net}}\$ on a particle is proportional to its mass \$m\$ and to the acceleration \$\mathbf{a}\$ it acquires:

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

Key points:

• Both \$\mathbf{F}_{\text{net}}\$ and \$\mathbf{a}\$ are vectors; they have the same direction.
• Mass \$m\$ is a scalar (always positive).
• If \$\mathbf{F}_{\text{net}} = 0\$, then \$\mathbf{a}=0\$ and the motion is uniform (including rest).

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}\$

These forces act on different bodies, so they do not cancel in the \$F=ma\$ equation for a single object.

Momentum

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

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

Momentum is a vector quantity and has the same direction as the velocity.

Impulse

The impulse \$\mathbf{J}\$ delivered to an object is the integral of the net force over the time interval \$\Delta t\$:

\$\mathbf{J} = \int{t1}^{t2}\mathbf{F}{\text{net}}\,dt \approx \mathbf{F}_{\text{avg}}\Delta t\$

Impulse changes the momentum:

\$\mathbf{J} = \Delta\mathbf{p} = \mathbf{p}f - \mathbf{p}i\$

Conservation of Momentum

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

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

This principle is frequently used to analyse collisions and explosions.

Key Equations Summary

Problem‑Solving Strategy Using \$F = ma\$

1. Read the question carefully; identify what is known and what is required.
2. Draw a clear diagram, indicating forces, directions, and any given distances or times.
3. Choose a convenient coordinate system (positive direction).
4. Write down the relevant equations:

   • Newton’s second law \$F = ma\$ for each object.
   • Kinematic relations if needed (e.g., \$v^2 = u^2 + 2as\$).
   • Momentum or impulse equations for collisions.

5. Substitute known values, solve algebraically for the unknowns.
6. Check that the direction of the resultant force matches the direction of the calculated acceleration.
7. State the final answer with correct units and appropriate significant figures.

Common Misconceptions

• Force and acceleration are not always “upward” or “downward”. Their direction is determined by the vector sum of all forces acting on the object.
• Action–reaction forces cancel. They act on different bodies, so they do not affect the \$F = ma\$ calculation for a single object.
• Mass changes during motion. In classical mechanics the mass of a particle is constant; only velocity changes.

Worked Example

Question: A 2.0 kg cart is pulled horizontally by a constant force of 10 N. The cart starts from rest. Calculate the speed of the cart after it has moved 5.0 m.

Solution:

1. Identify known quantities:

   • Mass \$m = 2.0\ \text{kg}\$
   • Force \$F = 10\ \text{N}\$ (horizontal, so direction is positive \$x\$)
   • Initial speed \$u = 0\ \text{m·s}^{-1}\$
   • Displacement \$s = 5.0\ \text{m}\$

2. Find the acceleration using \$F = ma\$:

   \$a = \frac{F}{m} = \frac{10\ \text{N}}{2.0\ \text{kg}} = 5.0\ \text{m·s}^{-2}\$

3. Use the kinematic equation \$v^{2}=u^{2}+2as\$ to obtain the final speed \$v\$:

   \$v^{2}=0^{2}+2(5.0\ \text{m·s}^{-2})(5.0\ \text{m}) = 50\ \text{m}^{2}\!\!/\!\text{s}^{2}\$

   \$v = \sqrt{50}\ \text{m·s}^{-1} \approx 7.1\ \text{m·s}^{-1}\$

4. Result: The cart’s speed after 5.0 m is \$7.1\ \text{m·s}^{-1}\$ (to three significant figures).

Practice Questions

1. A 0.5 kg ball is thrown vertically upward with an initial speed of 12 m·s⁻¹. Ignoring air resistance, calculate the maximum height reached. (Use \$g = 9.8\ \text{m·s}^{-2}\$.)
2. A 3.0 kg sled is pulled across a frictionless ice surface by a horizontal force of 15 N. Determine the time required for the sled to travel 10 m starting from rest.
3. Two ice skaters, A (mass 60 kg) and B (mass 80 kg), push off each other. If skater A moves away with a speed of 2.5 m·s⁻¹, what is the speed of skater B? (Assume no external horizontal forces.)
4. A 0.2 kg bullet traveling at 400 m·s⁻¹ embeds itself in a 2.0 kg block initially at rest on a frictionless surface. Find the common speed immediately after impact.
5. A 10 N constant force acts on a 4 kg crate for 3 s. Determine the change in momentum of the crate and its final speed if it started from rest.

Summary

Newton’s second law, \$F = ma\$, links the vector quantities of force and acceleration. Because they share the same direction, any problem that provides a net force can be solved directly for acceleration, and vice‑versa. Momentum (\$p = mv\$) and impulse (\$J = \Delta p\$) are powerful tools for analysing situations where forces act over short time intervals, such as collisions. Mastery of these concepts enables systematic problem solving in A‑Level physics.