1. Motion 2 Momentum and Energy
Unit two Physics

2. Momentum Momentum can be defined as ‘mass in motion’.
If an object has a mass and is moving, then it has momentum.
Knowing about Momentum helps us to answer questions like:
How difficult is it to stop a moving object?
How difficult is it to make a stationary object move?
The answer to these questions depends on two things:
The mass of the object
How fast the object is going / How fast you want the object to go

3. Momentum is a vector quantity
The product of these two characteristics is called momentum and can be defined as:
𝑝=𝑚𝑣
𝑚 is the mass of the object measured in 𝑘𝑔
𝑣 is the velocity of the object, measured in 𝑚 𝑠 −1 𝑜𝑟 𝑚/𝑠
𝑝 is the momentum of the object, measured in 𝑘𝑔 𝑚 𝑠 −1 or 𝑘𝑔 𝑚/𝑠
Momentum is a vector quantity

4. Momentum
𝑚 is the mass of the object measured in 𝑘𝑔 𝑣 is the velocity of the object, measured in 𝑚 𝑠 −1 𝑜𝑟 𝑚 𝑠 𝑝 is the momentum of the object, measured in 𝑘𝑔 𝑚 𝑠 −1 or 𝑘𝑔 𝑚/𝑠
𝑝=𝑚𝑣
eg1. Find the momentum of a car of mass 1500 𝑘𝑔, moving north at a velocity of 20 𝑚 𝑠 −1 ?
𝑝=𝑚𝑣
=1500 𝑘𝑔 ×20 𝑚 𝑠 −1
=3000 𝑘𝑔 𝑚 𝑠 −1 North

5. Momentum
𝑚 is the mass of the object measured in 𝑘𝑔 𝑣 is the velocity of the object, measured in 𝑚 𝑠 −1 𝑜𝑟 𝑚 𝑠 𝑝 is the momentum of the object, measured in 𝑘𝑔 𝑚 𝑠 −1 or 𝑘𝑔 𝑚/𝑠
𝑝=𝑚𝑣
eg2. A train with a momentum of 20× 𝑘𝑔 𝑚/𝑠 travels west at a velocity of 120 km/hr.
What is the mass of the train?
𝑝=𝑚𝑣
𝑚= 𝑝 𝑣
= 20× 𝑘𝑔 𝑚 𝑠 −1 (120 ÷ 3.6) 𝑚 𝑠 −1 = 600,000 𝑘𝑔 = 6× 𝑘𝑔 west

6. Momentum & Impulse
Making an object stop or go requires a non-zero net force.
Recall Newtons second law, 𝐹 𝑛𝑒𝑡 =𝑚𝑎
Acceleration is change in velocity over a time interval ∆𝑣 ∆𝑡
Combining these together gives: 𝐹 𝑛𝑒𝑡 =𝑚( ∆𝑣 ∆𝑡 )
Rearranging gives: 𝐹 𝑛𝑒𝑡 ∆𝑡=𝑚 ∆𝑣
This gives the impulse, 𝐹 𝑛𝑒𝑡 ∆𝑡 and can be defined as the product of the net force over the time interval which it acts.

7. Momentum & Impulse Impulse = 𝐹 𝑛𝑒𝑡 ∆𝑡 =𝑚 ∆𝑣 =𝑚(𝑣−𝑢) =𝑚𝑣−𝑚𝑢 = 𝑝 𝑓 − 𝑝 𝑖
𝐹 𝑛𝑒𝑡 = ∆𝑝 ∆𝑡
Where 𝑝 𝑓 =𝐹𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑝 𝑖 =𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
So we can also say that impulse
is the change in momentum, ∆𝑝
And we can say Net Force is the rate of change of momentum given by:
∆𝑝= Change in momentum
∆𝑡= Change in time

8. What is the change in momentum of the netball ball?
eg1. A 200g netball hits a wall horizontally at a speed of 12 m/s and bounces back in the opposite direction at speed of 10 m/s. The ball comes into contact with the wall for 10 ms.
What is the change in momentum of the netball ball?
Assign initial direction on ball as positive
b) What is the impulse on the netball?
Impulse on the tennis ball = change in momentum of the tennis ball =−4.4 𝑘𝑔 𝑚 𝑠 −1
c) What is the magnitude of the force exerted by the wall on the netball?
𝐹 𝑛𝑒𝑡 = ∆𝑝 ∆𝑡 = 4.4 𝑘𝑔 𝑚 𝑠 − = 22 𝑁
∆𝑝=𝑚∆𝑣
= 𝑚 𝑣−𝑢 = × −10−12 = ×−22 =−4.4 𝑘𝑔 𝑚 𝑠 −1
u = 12 m/s
v = 10 m/s

9. then…..Text Book Problems
Now Try
Momentum Worksheet
then…..Text Book Problems
Chapter 6 Questions 46, 47, 48, 49, 50, 51

10. What is the initial momentum of the ball?
Basketball Scenario
What is the initial momentum of the ball?
Assign initial direction on ball as positive
What is the final momentum of the ball?
Assign final direction on ball as negative
What is the Impulse of the ball?
d) What is the magnitude of the Net force exerted by the wall on the ball?
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =𝑚𝑣
= 𝑘𝑔×10 𝑚𝑠 −1
=0.20 𝑘𝑔 𝑚 𝑠 −1
Vinitial = +10 m/s
𝑝 𝑓𝑖𝑛𝑎𝑙 =𝑚𝑣
= 𝑘𝑔×−7 𝑚𝑠 −1
=−0.14 𝑘𝑔 𝑚 𝑠 −1
Vfinal = -7 m/s
𝐼𝑚𝑝𝑢𝑙𝑠𝑒= ∆𝑝= 𝑝 𝑓𝑖𝑛𝑎𝑙 − 𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −0.14 −0.20
=− 0.34 𝑘𝑔 𝑚 𝑠 −1
𝐹 𝑛𝑒𝑡 = ∆𝑝 ∆𝑡 = 𝑘𝑔 𝑚 𝑠 −1 10× 10 −3 𝑠 = 34 𝑁

11. What is the initial momentum of the ball?
eg. A 20g ping pong ball hits a wall horizontally at a speed of 10 m/s, before bouncing off the wall at a speed of 7 m/s. The ball is in contact with the ball for 10 mS
What is the initial momentum of the ball?
Assign initial direction on ball as positive
What is the final momentum of the ball?
Assign final direction on ball as negative
What is the Impulse of the ball?
d) What is the magnitude of the Net force exerted by the wall on the ball?
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =𝑚𝑣
= 𝑘𝑔×10 𝑚𝑠 −1
=0.20 𝑘𝑔 𝑚 𝑠 −1
Vinitial = +10 m/s
𝑝 𝑓𝑖𝑛𝑎𝑙 =𝑚𝑣
= 𝑘𝑔×−7 𝑚𝑠 −1
=−0.14 𝑘𝑔 𝑚 𝑠 −1
Vfinal = -7 m/s
𝐼𝑚𝑝𝑢𝑙𝑠𝑒= ∆𝑝= 𝑝 𝑓𝑖𝑛𝑎𝑙 − 𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −0.14 −0.20
=− 0.34 𝑘𝑔 𝑚 𝑠 −1
𝐹 𝑛𝑒𝑡 = ∆𝑝 ∆𝑡 = 𝑘𝑔 𝑚 𝑠 −1 10× 10 −3 𝑠 = 34 𝑁

12. Impulse using graphs
The force calculated in the previous example is the Average Force on the ball – The force acting on the ball is changing, reaching maximum value when the ball is at its smallest distance from the wall.
𝐼𝑚𝑝𝑢𝑙𝑠𝑒=𝐹 𝑎𝑣 ∆𝑡
We can use the real Force v Time graphs to find the impulse.
This is given by the area under the graph.

13. Impulse - using graphs
We can use the real Force v Time graphs to find the impulse.
This is given by the area under the graph.
eg. For the Force – Time graph given, calculate the Impulse.
Area under graph = 1 2 𝑏ℎ
= 1 2 ×40×100
=2000 𝑁𝑠

14. Impulse - using graphs
eg. The Force – Time graph shows the force on a 50kg skater as she skates from rest.
a) Calculate the Impulse for the skater for the 2-second interval shown on the graph.
Area under graph =A+B+C
= 𝑏ℎ + 𝑙×𝑤 +( 1 2 𝑏ℎ)
= 1 2 ×1.1× ×200 +( 1 2 ×0.9×200) =
=490 𝑁𝑠
b) Estimate her speed after 2 seconds.
∆𝑝=𝑚∆𝑣
∆𝑝=𝑚 𝑣 𝑓 −𝑚 𝑣 𝑖
490= 50× 𝑣 𝑓 − 50×0
𝑣 𝑓 = =9.8 𝑚𝑠 −1
∆𝑡=2.0 𝑠
∆𝑝=490 𝑘𝑔 𝑚𝑠 −1
𝑚=50 𝑘𝑔

15. Impulse - using graphs = 4.4 1.1 =4 𝑚𝑠 −2
eg. The Force – Time graph shows the force on a 50kg skater as she skates from rest.
c) Estimate the speed of the skater after 1.1 seconds
d) What is her acceleration over the first 1.1 seconds?
∆𝑝=𝑚∆𝑣
∆𝑝=𝑚 𝑣 𝑓 −𝑚 𝑣 𝑖
220= 50× 𝑣 𝑓 − 50×0
𝑣 𝑓 = =4.4 𝑚𝑠 −1
∆𝑡=1.1 𝑠
∆𝑝=220 𝑘𝑔 𝑚𝑠 −1
𝑚=50 𝑘𝑔
a= ∆𝑣 ∆𝑡 =
=4 𝑚𝑠 −2

16. Now Try Worksheet - Impulse using graphs
then…..Text Book Problems
Chapter 6 Questions 46, 47, 48, 49, 50, 51, 52

17. Law of Conservation of Momentum
The law of conservation of momentum states that….
The total momentum of a closed system does not change.
This means that if two objects collide, the total momentum of the objects before the collision is the same as the total momentum of the objects after the collision.
Total Momentum means the combined momentum of Object 1 + Object 2.

18. Law of Conservation of Momentum
After a collision, each object may not have the same momentum as it did initially, but the combined momentum of the 2 objects will still be the same as before.
So if one object loses momentum in a collision, the other object gains momentum.
eg. A 2 kg cart is moving east at a constant velocity of 3 𝑚𝑠 −1 , when a 5kg brick is placed on it (from rest) and continues to move east. What is the velocity of the cart and brick system as it moves away (ignoring resistance forces) ?

19. Law of Conservation of Momentum
eg. A 3 kg cart is moving east at a constant velocity of 8 𝑚𝑠 −1 , when a 5kg brick is placed on it (from rest) and continues to move east. What is the velocity of the cart and brick system as it moves away (ignoring resistance forces) ?
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝 𝑓𝑖𝑛𝑎𝑙
𝑚 1 𝑣 𝑚 2 𝑣 2 = 𝑚 𝑓𝑖𝑛𝑎𝑙 𝑣 𝑓𝑖𝑛𝑎𝑙
2 𝑘𝑔×8 𝑚𝑠 −1 =8 𝑘𝑔 × 𝑣 𝑓
16 =8 𝑣 𝑓
𝑣 𝑓 = =2 𝑚𝑠 −1

20. Modelling Collisions
Collisions between two objects that result in the objects moving off together, at a new combined mass and velocity.
Momentum is conserved.
This means that (as shown)……
Initial Momentum of the objects individually = Final Momentum of the objects together
𝑝 𝑂𝐵𝐽𝐸𝐶𝑇 1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑝 𝑂𝐵𝐽𝐸𝐶𝑇 2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝 𝑂𝐵𝐽𝐸𝐶𝑇 1 & 2 𝑓𝑖𝑛𝑎𝑙

21. Modelling Collisions Momentum is conserved. This means that……
Collisions between two objects that result in the objects bouncing off each other.
Momentum is conserved.
This means that……
Initial total Momentum of the objects = Final total Momentum of the objects 𝑝 𝑂𝐵𝐽𝐸𝐶𝑇 1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑝 𝑂𝐵𝐽𝐸𝐶𝑇 2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝 𝑂𝐵𝐽𝐸𝐶𝑇 1 𝑓𝑖𝑛𝑎𝑙 + 𝑝 𝑂𝐵𝐽𝐸𝐶𝑇 2 𝑓𝑖𝑛𝑎𝑙

22. eg. A 1000kg car travelling at 20 𝑚𝑠 −1 hits a 3000kg stationary truck.
The truck moves forward at a 10 𝑚𝑠 −1 velocity and the car rebounds off the truck, in the direction opposite to what it was initially travelling in.
What is the velocity of the car after the collision?
𝑝 𝑐𝑎𝑟 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + 𝑝 𝑡𝑟𝑢𝑐𝑘 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 𝑝 𝑐𝑎𝑟 (𝑓𝑖𝑛𝑎𝑙) + 𝑝 𝑡𝑟𝑢𝑐𝑘 (𝑓𝑖𝑛𝑎𝑙)
20 𝑚𝑠 −1 ×1000 𝑘𝑔 =1000 𝑣 𝑐𝑎𝑟 (𝑓𝑖𝑛𝑎𝑙) 𝑚𝑠 −1 ×3000 𝑘𝑔
20000=1000𝑣+30000
−10000=1000𝑣
𝑣 𝑐𝑎𝑟 (𝑓𝑖𝑛𝑎𝑙) = − =−10 𝑚𝑠 −1

23. Elastic Collisions vs Inelastic Collisions
Elastic Collisions: Kinetic Energy is conserved
Kinetic Energy before collision = Kinetic Energy after collision
Inelastic Collisions: Kinetic Energy is not conserved – the energy is LOST
Kinetic Energy before collision > Kinetic Energy after collision

24. For more examples click the link:

25. Cart Simulation Activity
Now Try
Cart Simulation Activity
Click the link to access the simulator:
Have a play with the simulator then use the worksheet to investigate and answer the problems

26. Now Try
Chapter 7 Questions 2a, 3, 4, 5

27. Conservation of Momentum
Eg1. A 10kg cart is travelling at 5m/s before a 7kg mass (initially at rest) is carefully placed on top of it. Ignoring friction:
What is the initial momentum (combined) before the collision?
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 10×5 +0=50 𝑘𝑔 𝑚/𝑠
What is the combined momentum after the collision?
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝 𝑓𝑖𝑛𝑎𝑙 =50 𝑘𝑔 𝑚/𝑠
c) What is the velocity of the cart after the collision?
𝑝=𝑚𝑣→ 50=17𝑣 → 𝑣= =2.94 𝑚/𝑠

28. 𝑝 𝑓𝑖𝑛𝑎𝑙 = 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 ∆𝑝=𝑚∆𝑣=3× 5−4 =3×1=3 𝑘𝑔 𝑚/𝑠
Eg. A ball of mass 3kg moving at 5 m/s, collides elastically with a ball of mass 1kg, moving at 3m/s in the same direction. The speed of the 3kg ball after the collision is 4m/s.
What is the total combined momentum of the two balls before the collision.
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 3×5 + 1×3 =15+3 =18 𝑘𝑔 𝑚/𝑠
What is the combined momentum after the collision?
What is the velocity of the 1kg ball after the collision?
d) What is the change in momentum of the 3 kg ball?
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝 𝑓𝑖𝑛𝑎𝑙 =18 𝑘𝑔 𝑚/𝑠
𝑝 𝑓𝑖𝑛𝑎𝑙 = 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2
18= 3×4 + 1𝑣
18=12+𝑣
𝑣=18−12 =6 𝑚/𝑠
∆𝑝=𝑚∆𝑣=3× 5−4
=3×1=3 𝑘𝑔 𝑚/𝑠

29. Conservation of Momentum
A ball of mass 3kg moving at 5 m/s, collides elastically with a ball of mass 1kg, moving at 3 m/s in the same direction. The speed of the 3kg ball after the collision is 4 m/s.
What is the total combined momentum of the two balls before the collision.
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 3×5 + 1×3 =15+3 =18 𝑘𝑔 𝑚/𝑠
What is the combined momentum after the collision?
𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝 𝑓𝑖𝑛𝑎𝑙 =18 𝑘𝑔 𝑚/𝑠
c) What is the velocity of the 1kg ball after the collision?
𝑝 𝑓𝑖𝑛𝑎𝑙 = 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 → 18= 3×4 + 1𝑣 → 18=12+𝑣
𝑣=18−12 =6 𝑚/𝑠

30. Prac Activity Brick and Cart collision
Now Do
Prac Activity Brick and Cart collision

31. Work
When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object.
There are three key ingredients to work - force, displacement, and cause.
In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement.
Examples where work is done…
- A father pushing a grocery cart down the aisle of a grocery store
- A weightlifter lifting a barbell above his head
- A horse pulling a plough through the field
In each case described here there is a force exerted upon an object to cause that object to be displaced.

32. Work No YES YES - A rocket accelerates through space
Work requires a force to be exerted upon an object to cause that object to be displaced in the direction of the force.
Is work done in the following scenarios?
- A teacher applies a force to a wall and becomes exhausted
The wall is not displaced – A force must cause
displacement for work to be done
- A falls off a table and free falls to the ground
Gravity (force) acts on the book causing it to be displaced downward
- A rocket accelerates through space
The force (expelled gases) push the rocket, causing displacement through space No
YES
YES

33. Work 𝑊=𝐹𝑥 Calculating Work: 𝑊−𝑊𝑜𝑟𝑘 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐽𝑜𝑢𝑙𝑒𝑠 𝐽
𝑊−𝑊𝑜𝑟𝑘 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐽𝑜𝑢𝑙𝑒𝑠 𝐽
𝐹−𝐹𝑜𝑟𝑐𝑒, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑁𝑒𝑤𝑡𝑜𝑛𝑠 𝑁
𝑥−𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑚𝑒𝑡𝑟𝑒𝑠 (𝑚)

34. Work
𝑊=𝐹𝑥
Eg1. A brick is lifted from the ground with a force of 100N, to a platform 2 metres above. What work was done on the brick?
Eg2. A wheelbarrow is pushed with a force of 400N, to a point 10 metres away.
What work was done on the wheelbarrow?
𝑊=𝐹𝑥
=100×2
=200 𝐽
𝑊=𝐹𝑥
=400×10
=4000 𝐽

35. Work
𝑊=𝐹𝑥
eg3. A mum pushes a pram horizontally with a force of 200 N to move the pram a distance of 8 metres. If the friction force opposing the motion is 120 N,
How much work is done on the pram by:
The force applied by the mum to push the pram?
The net force?
The shopper to oppose the friction force?
𝑊=𝐹𝑥
=200×8
=1600 𝐽
𝐹 𝑛𝑒𝑡 =200𝑁−120𝑁=80 𝑁
𝑊=𝐹𝑥
=80×8
=640 𝐽
𝑊=𝐹𝑥
=120×8
=960 𝐽

36. Work
Sometimes when a force acts, the displacement does not have the same direction as the force. In these problems, we use:
𝑊=𝐹𝑥 cos 𝜃
eg. If the cart pictured was pulled with a 80 N force, and the rope was inclined at a 40° angle, how much work was done to pull the cart 10 metres?
𝑊=𝐹𝑥 cos 𝜃
=80×10× cos 40
=800 cos 40
=612.8 𝐽

37. Energy All matter possesses energy Energy comes in various forms…..
Nuclear Energy
Chemical Potential Energy
Kinetic Energy
Etc…..
Gravitational Potential Energy
Light Energy
Electrical Energy

38. Energy The law of conservation of Energy
Energy cannot be created or destroyed
Energy can be stored, transferred to other matter, or transformed from one form to another.
Some energy transfers and transformations can be seen, felt, heard, smelt, however many are not observable and cannot be measured.

39. Kinetic Energy
Is the energy associated with the movement of an object, given by the equation:
𝐸 𝐾 = 1 2 𝑚 𝑣 2
𝐸 𝐾 −𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐽𝑜𝑢𝑙𝑒𝑠 𝐽
𝑚−𝑚𝑎𝑠𝑠 𝑘𝑔
𝑣−𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ( 𝑚𝑠 −1 )

40. Kinetic Energy 𝐸 𝐾 = 1 2 𝑚 𝑣 2 = 1 2 ×70×36 =1260 𝐽 = 1 2 ×1000×225
A 70kg runner moves with a velocity of 6 m/s. What is his kinetic energy?
𝐸 𝐾 = 1 2 ×70× 6 2
= 1 2 ×70×36
=1260 𝐽
A 1000 kg car moves with a velocity of 15 m/s. What is the kinetic energy of the car?
𝐸 𝐾 = 1 2 ×1000× 15 2
= 1 2 ×1000×225
= 𝐽

41. Kinetic Energy
The Kinetic Energy of an object can only change as a result of a non-zero net force acting on it in the direction of the motion.
The change in Kinetic Energy of an object is equal to the work done on it by the net force acting on it.
So we Kinetic Energy can be expressed as:
Change in Kinetic Energy = Work Done by the force acting on the object
∆𝐸 𝐾 = 𝐹 𝑁𝑒𝑡 𝑥

42. Now Do
Chapter 7
Questions 7, 8, 11, 12, 13

43. Potential Energy
Energy that is stored is called Potential Energy. These include: Chemical Potential Energy, Gravitational Potential Energy, etc…
Our course covers Gravitational Potential Energy.
Gravitational potential energy is the energy stored in an object as a result of its position and its attraction to the earth by the force of gravity.
So if something can fall down from somewhere, then it has Gravitational Potential energy.

44. Gravitational Potential Energy
Is the energy associated with its position, given by the equation:
𝐸 𝑔𝑝 =𝑚𝑔ℎ
𝐸 𝑔𝑝 −𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐽𝑜𝑢𝑙𝑒𝑠 𝐽
𝑚−𝑚𝑎𝑠𝑠 𝑘𝑔
ℎ−ℎ𝑒𝑖𝑔ℎ𝑡 𝑓𝑟𝑜𝑚 𝑒𝑎𝑟𝑡ℎ 𝑜𝑟 𝑎 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(𝑚)

45. Gravitational Potential Energy
𝐸 𝑔𝑝 =𝑚𝑔ℎ
Gravitational Potential Energy
A 40kg air conditioner is mounted to a wall 3 metres above the ground. What gravitational potential energy does the air conditioner have?
𝐸 𝑔𝑝 =𝑚𝑔ℎ
=40×10×3
=1200 𝐽
A skater with mass 60kg stands at the top of a 4 metre skate ramp. What gravitational potential energy does the skater have?
𝐸 𝑔𝑝 =𝑚𝑔ℎ
=60×10×4
=2400 𝐽

46. Conservation of Mechanical Energy
Kinetic (KE) and Gravitational Potential Energy (GPE) are referred to as Mechanical Energy.
One type of energy can transform from Potential to Kinetic, or vice versa, if the object with the energy, interacts with a force.
Eg. A tennis ball sits at a height of 2 metres.
What type of mechanical energy does it possess?
If its dropped to the ground, during this motion, its GPE is transformed throughout the journey, from GPE to KE.
At the bottom of the 2 metres, just before it hits the ground, all GPE is transformed into KE.

47. Conservation of Mechanical Energy
A ball has a mass of 500g and is dropped from a height of 2 metres.
What total mechanical energy does the ball have at the top of the path, before being dropped?
𝐸 𝑔𝑝 =𝑚𝑔ℎ
=0.5 𝑘𝑔×10 𝑚/ 𝑠 2 ×2
=10 𝐽
What type of energy the ball possess when it’s 1 metre from the ground?
GPE and KE.
What total energy does the ball have when it’s 1 metre from the ground?
Total Energy is still 10 𝐽 – But now this energy consists of BOTH KE and GPE.
𝐸 𝑔𝑝 =𝑚𝑔ℎ
=0.5×10×1
=5 𝐽
The remaining 5 J is Kinetic Energy

48. Conservation of Mechanical Energy
How fast is the ball travelling at the 1 metre point?
𝐸 𝐾 =5 𝐽
𝐸 𝐾 = 1 2 𝑚 𝑣 2
5= 1 2 ×0.5× 𝑣 2
𝑣 2 = =20
𝑣= 20 =4.47 𝑚/𝑠

49. Energy Skate Park

50. Now Do
Chapter 7
Questions 14, 16, 19, 20, 23

51. Strain Potential Energy - Hooke’s Law
The potential energy stored in a spring is known as Strain Potential Energy.
The force needed to extend or compress a spring by some distance is proportional to the distance in which it moves.
𝐹∝∆𝑥
Each spring has an individual stiffness associated with it, which is a measure of how strong the spring is. This is called the spring constant represented by ‘ k ’
So, the Force required to extend or compress a spring can be found by:
𝐹=𝑘∆𝑥
F (Force applied) measured in N (Newtons)
k (Spring Constant) measured in N/m (Newtons per metre)
x (displacement of the spring) measured in m (metres)

52. Strain Potential Energy - Hooke’s Law
The Strain Potential Energy stored in a spring, whether compressed or extended, can be found by calculating the area under a Force versus displacement graph for the spring.
The spring constant k , can be found by calculating the gradient of the line.

53. Strain Potential Energy - Hooke’s Law
Eg. The graph below models a spring being stretched.
a) Find the Potential Energy gained by stretching the spring 25 cm
b) Calculate the spring constant.
a) 𝐸 𝑃𝑂𝑇 =𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑎 𝐹−𝑥 𝑔𝑟𝑎𝑝ℎ
= 1 2 𝑏ℎ
= 1 2 ×0.25𝑚×20𝑁
=2.5 𝐽
b) 𝑘=𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
= 20−0 0.25−0 =80 𝑁/𝑚

54. Strain Potential Energy - Hooke’s Law
We can also calculate the Strain Potential Energy stored in a spring using the following equation:
𝑆𝑡𝑟𝑎𝑖𝑛 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 1 2 ×𝑘× (∆𝑥) 2
eg. A spring a stretched 15 cm. If the spring constant for the spring is 40 N/m, find:
The force applied to the spring
𝐹=𝑘∆𝑥
=40×0.15
=6 𝑁
The strain potential energy stored in the extended spring
𝑆𝑃𝐸= 1 2 𝑘( ∆𝑥) 2
= 1 2 ×40×(0.15 ) 2
=0.45 J

55. Strain Potential Energy - Hooke’s Law
eg. A ball with mass 400g, on a pinball machine is released by compressing & releasing a spring. If the spring with a spring constant of 200 N/m is pulled back 15 cm, find:
The strain potential energy of the spring just before the ball is released.
𝑆𝑡𝑟𝑎𝑖𝑛 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 1 2 ×𝑘× (∆𝑥) 2 = 1 2 ×200× =2.25 𝐽
The kinetic energy of the ball just after it has been released.
All SPE converted into KE, so 𝐸 𝐾 =𝑆𝑃𝐸=2.25 𝐽
The speed of the pinball just after it has been released.
𝐸 𝐾 = 1 2 𝑚 𝑣 → = 1 2 ×0.4× 𝑣 2
𝑣 2 = → 𝑣= 50 =3.35 𝑚/𝑠

56. Now Do
Chapter 7
Questions 18, 26, 27, 28