Newton's third law

Question 1

What is Newton’s third law?

Answer 1

for every action there is an equal and opposite reaction (action-reaction forces)

Question 2

What does the law relate to and what can it be applied to?

Answer 2

it relates to bodies in equilibrium and can be applied to collisions when considering conservation of momentum

Question 3

State the properties of action reaction forces

Answer 3

• they act on different bodies
• they are equal
• they are opposite

Question 4

A book is on a table. There is a reaction force of the table pushing up and the force of gravity acting downwards on the book.
Is this an example of an action reaction force?
Explain your answer.

Answer 4

No, because both forces act on the same object, the book.
There is no force being exerted on the table.

However, the gravitational pull of the Earth on the book and the gravitational pull of the book on the Earth would be an example of Newton’s third law as they act on different objects, and are opposite and equal.

Question 5

What is the conservation of momentum?

Answer 5

In a closed system, the total momentum before an collision is the same as after the collision

Question 6

Describe the forces acting on a truck and a car as they collide

Answer 6

• opposite and equal forces are exerted on both
• they are in contact for the same amount of time
• momentum is conserved

Question 7

A model railway wagon (m) of mass 1 kg travelling at 12 m/s runs into a second stationary wagon (M) of mass 3 kg. After the collision the wagons stay linked together and move together.
Calculate the velocity of the wagons after the collision.

Answer 7

momentum = mass × velocity

1 kg × 12 m/s = (1 + 3) kg × v
12 = 4 × v
12 ÷ 4 = v

v = 3 m/s

Question 8

Explain how Newton’s third law applies to collisions.

Answer 8

For two objects involved in a collision, the forces involved are equal in magnitude but opposite in direction, and as the time of impact is the same for both bodies, the momentum must be equal and opposite.

Question 9

A truck of mass 12 kg moving at 8 m/s collides with a truck of mass 8 kg that is stationary. They collide and move off together. Calculate their shared velocity after the collision.

Answer 9

p = mv

before:
p = 12 × 8
p = 96

after:
96 = (12 + 8) × v
96 = 20 × v
v = 96 ÷ 20

v = 4.8 m/s