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Physics

Motion and Dynamics

PDF
Matthew Williams
|May 20, 2026|6 min read
CSEC PhysicsMomentumMotionNewton's LawsPaper 01Paper 02Section AVelocity

Distance, displacement, speed, velocity, acceleration, motion graphs, Newton's three laws, linear momentum, and conservation of momentum in collisions.

Distance, Displacement, Speed, Velocity

These four quantities form two scalar-vector pairs. Distinguishing them precisely is important in physics.

ScalarVectorDefinition
DistanceDisplacementDisplacement is the straight-line separation from start to finish, with direction. Distance is the total path length.
SpeedVelocityVelocity is the rate of change of displacement. Speed is the rate of change of distance.

velocity=displacementtimev=st\text{velocity} = \frac{\text{displacement}}{\text{time}} \qquad v = \frac{s}{t}velocity=timedisplacement​v=ts​

speed=distancetime\text{speed} = \frac{\text{distance}}{\text{time}}speed=timedistance​

Both speed and velocity use units of m s⁻¹.

Acceleration is the rate of change of velocity:

a=ΔvΔt=v−uta = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}a=ΔtΔv​=tv−u​

where uuu is initial velocity and vvv is final velocity. Units: m s⁻².

Motion Graphs

Displacement-Time Graphs

The gradient of a displacement-time graph equals velocity.

  • Horizontal line: stationary (zero velocity)
  • Straight line with positive gradient: constant positive velocity
  • Curve with increasing gradient: acceleration

Velocity-Time Graphs

The gradient of a velocity-time graph equals acceleration. The area under a velocity-time graph equals displacement.

  • Horizontal line: constant velocity (zero acceleration)
  • Straight line with positive gradient: constant acceleration
  • Negative gradient: deceleration
Velocity-time graph: shaded area = total displacement
Exam Tip

To find displacement from a v-t graph, calculate the area of each region geometrically (rectangle for constant velocity, triangle for acceleration/deceleration). Areas below the time axis represent motion in the opposite direction and count as negative displacement.

Newton's Three Laws of Motion

First Law

An object remains at rest or continues moving at constant velocity unless acted on by a resultant external force. This is the law of inertia, objects resist changes to their state of motion.

The practical consequence: if something is accelerating, there must be a non-zero resultant force. If it moves at constant velocity, the forces are balanced.

Second Law

The resultant force on an object equals its mass multiplied by its acceleration:

F=maF = maF=ma

where FFF is in newtons, mmm in kilograms, and aaa in m s⁻². The acceleration is in the same direction as the resultant force.

Third Law

Every force has an equal and opposite reaction force, acting on a different object. The two forces in an action-reaction pair are equal in magnitude, opposite in direction, and act on different bodies, they cannot cancel each other.

Object AObject BForce A on BForce B on AEqual in magnitude, opposite in direction, acting on different objects
Newton's 3rd Law: every action has an equal and opposite reaction
Example/Newton's laws applied, crash test (2015 Paper 02, Q4)

A 70 kg test dummy is in a car travelling at 26 m s⁻¹. The car crashes and stops in 0.1 s.

Part (i), Initial momentum:

p=mv=70×26=1 820N sp = mv = 70 \times 26 = 1\,820 \text{N s}p=mv=70×26=1820N s

Part (ii), Force from seatbelt:

The dummy decelerates from 26 m s⁻¹ to 0 in 0.1 s.

a=ΔvΔt=0−260.1=−260m s−2a = \frac{\Delta v}{\Delta t} = \frac{0 - 26}{0.1} = -260 \text{m s}^{-2}a=ΔtΔv​=0.10−26​=−260m s−2

F=ma=70×260=18 200NF = ma = 70 \times 260 = 18\,200 \text{N}F=ma=70×260=18200N

The seatbelt exerts a force of 18,200 N on the dummy.

Linear Momentum

Momentum is the product of mass and velocity:

p=mvp = mvp=mv

Units: kg m s⁻¹ (equivalent to N s).

Momentum is a vector, it has the same direction as the velocity.

Newton's Second Law in Terms of Momentum

Force equals the rate of change of momentum:

F=ΔpΔt=mv−mutF = \frac{\Delta p}{\Delta t} = \frac{mv - mu}{t}F=ΔtΔp​=tmv−mu​

This form is more general than F=maF = maF=ma because it works even when mass changes.

Conservation of Linear Momentum

In a closed system (no external forces), the total momentum before a collision equals the total momentum after:

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2m1​u1​+m2​u2​=m1​v1​+m2​v2​

This holds for all collisions, elastic and inelastic, as long as no external forces act.

Example/Conservation of momentum, collision (2016 Paper 02, Q2)

An 8 kg ball moving east at 10 m s⁻¹ collides with a 2 kg ball moving west at 5 m s⁻¹. After the collision they stick together. Find their common velocity.

Take east as positive.

Total momentum before:

pbefore=(8×10)+(2×−5)=80−10=70kg m s−1p_{\text{before}} = (8 \times 10) + (2 \times -5) = 80 - 10 = 70 \text{kg m s}^{-1}pbefore​=(8×10)+(2×−5)=80−10=70kg m s−1

After collision (combined mass = 10 kg):

70=10v70 = 10v70=10v

v=7m s−1eastv = 7 \text{m s}^{-1} \text{east}v=7m s−1east

Exam Tip

Assign a positive direction at the start of every momentum problem and stick to it. A velocity in the opposite direction gets a negative sign. If your final velocity comes out negative, the object is moving in the direction you called negative.

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Statics, Forces, and Moments
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Energy, Work, and Power