9478 / 2027
Motion and Forces overview

Lesson 5 of 6

Apply all three Newton laws

  • I can state and apply all three Newton laws: persistence of velocity, momentum-change rate and simultaneous interaction pairs.Syllabus

    Syllabus 9478, 3(h). State and apply all three Newton laws: persistence of velocity, momentum-change rate and simultaneous interaction pairs.

Make a guess. You are not marked.

A small car runs head-on into a loaded lorry. During the collision, how does the force of the lorry on the car compare with the force of the car on the lorry?

  1. The car pushes harder on the lorry, because the car was moving faster.
  2. They are equal in size and opposite in direction.
  3. The lorry pushes harder on the car, because the lorry has more mass.
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They are equal in size and opposite in direction.

The two forces are a Newton's third law pair: two bodies pushing on each other. The car is damaged more and its velocity changes more because the same size of force acts on a smaller mass, so its deceleration is larger.

Identify the body before applying a law. The resultant force describes the combined effect of forces on that body; an interaction pair acts on two different bodies.

Use an approximately inertial laboratory frame for these models. A free-body diagram helps keep the recipient of every force explicit.

First law: zero resultant preserves velocity

A body at rest stays at rest, and a moving body continues at constant velocity, unless a resultant external force acts on it.

Zero resultant does not mean that no forces act. For example, the weight and support on a gliding object may balance vertically while horizontal resistance is negligible. It can continue moving horizontally without a forward resultant force.

When a vehicle brakes, a passenger tends to continue with the previous velocity. A restraint or another contact supplies the force needed to change the passenger's motion. The tendency to continue is inertia, not evidence of an extra forward force.

Second law: resultant force changes momentum

The rate of change of momentum is proportional to the resultant force and is in its direction. With consistent SI units, the instantaneous relationship is:

Fresultant = dp/dt

Average resultant force = Δp/Δt

Here dp/dt means the local rate of change of momentum. The interval expression gives the average force over the stated interval; it does not by itself establish the force at every instant.

Worked average force

Use the full momentum change

A 1.20 kg body changes velocity from +3.0 to -2.0 m/s in 0.50 s. Its momentum change is 1.20(-2.0 - 3.0) = -6.0 kg m/s.

Average resultant force = -6.0/0.50 = -12 N.

The negative sign gives the force direction. The body initially moved positively, but its momentum changed negatively. The result does not show that the force was -12 N throughout the interval.

Third law: interaction partners are simultaneous

If body A exerts a force on body B, body B simultaneously exerts a force of equal magnitude and opposite direction on body A. The pair describes one interaction and acts on different bodies.

The partner is not a later response. It can exist whether or not either body is moving. When drawing the forces on B, include A's force on B; B's force on A belongs on A's diagram.

The contact pair acts on two different skaters

These are separate horizontal interaction views. Vertical weight and support forces are omitted. Both contact forces exist at the same instant.

Equal contact forces give different accelerationsThe upper view selects the forty-kilogram skater: the sixty-kilogram skater exerts a 120-newton leftward force on this body. The lower view selects the sixty-kilogram skater: the forty-kilogram skater exerts a 120-newton rightward force on this different body. The two purple force arrows have equal lengths. With other horizontal forces neglected, the forty-kilogram body accelerates left at three metres per second squared and the sixty-kilogram body right at two. The green acceleration arrows use a separate common scale, with lengths in the ratio three to two; they are not extra forces.Force from the 60 kg skater120 N40 kg bodya = 3.0 m/s2 leftForce from the 40 kg skater120 N60 kg bodya = 2.0 m/s2 right

Purple arrows compare forces on one scale; green arrows compare accelerations on another. The pair's forces are equal. The accelerations differ because the masses differ.

The two 120 N contact forces act simultaneously on different skaters. This pair-only view omits their vertical forces. Equal contact forces need not produce equal accelerations when the masses differ.

Equal forces, different accelerations

Two skaters push each other

A 40 kg skater is pushed left and a 60 kg skater right by a pair of 120 N contact forces. Other horizontal forces are negligible.

Their accelerations are 120/40 = 3.0 m/s2 left and 120/60 = 2.0 m/s2 right. The different responses follow from their different masses, not from unequal partner forces.

Try it step by step

Work it out in steps: Mass arriving each second in kg/s, then Force = (mass per second) x (change in velocity) in N.

  1. Problem 1. A hose delivers 0.50 kg of water every second at 12 m/s onto a wall, where the water stops. Find the average force on the wall.
  2. Problem 2. A jet delivers 1.2 kg of water every second at 15 m/s onto a plate, where it stops. Find the average force on the plate.
  3. Problem 3. Sand falls from rest onto a conveyor belt at 3.0 kg every 2.0 s, and the belt carries it away at 0.80 m/s. Find the extra force the belt motor must provide.
Show the answer

Problem 1.

Mass arriving each second in kg/s: Mass per second = 0.50 kg/s

Force = (mass per second) x (change in velocity) in N: F = 0.50 x (12 - 0) = 6.0 N

Problem 2.

Mass arriving each second in kg/s: Mass per second = 1.2 kg/s

Force = (mass per second) x (change in velocity) in N: F = 1.2 x 15 = 18 N

Problem 3.

Mass arriving each second in kg/s: Mass per second = 3.0 / 2.0 = 1.5 kg/s

Force = (mass per second) x (change in velocity) in N: F = 1.5 x (0.80 - 0) = 1.2 N

A stationary book's weight and table support both act on the book. They can balance, but they are not a third-law pair. Earth's pull on the book is paired with the book's pull on Earth. The table's force on the book is paired with the book's force on the table.

Equal and opposite is not enough to identify a third-law pair. Name both source and recipient and check that the forces are the two sides of the same interaction.

Check your understanding

A 40 kg skater pushes a 60 kg skater rightward with 120 N. Other horizontal forces are negligible. Which statement describes the simultaneous interaction?

  1. The 60 kg skater pushes the 40 kg skater left with 120 N.
  2. The 60 kg skater pushes back with 180 N because that skater has greater mass.
  3. The 60 kg skater pushes back with 120 N once it has started moving, not before.
Show the answer

The 60 kg skater pushes the 40 kg skater left with 120 N.

The forces act on different skaters with equal magnitude and opposite direction. Their accelerations are 120/40 = 3.0 m/s2 left and 120/60 = 2.0 m/s2 right.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

State Newton's second law of motion. [2]

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