1. Force
A force is something which causes an object to accelerate.

2. Force
A force may cause an object to slow down, speed up, stop or change direction.
a force always acts in a certain direction
ex. if you push something, the force is in the direction of the push
Sometimes is seems there is no force because they cancel each other out as in a tug of war and no team is moving.
(show lever forces each side no movement)

3. Measuring Force
Force is measured using a spring balance (Newton-meter)
Force is measured in Newtons (N)
1N is weight of 100g.
Definition of the newton
1N is the force that gives a mass of 1 kg an acceleration of 1 m/s2 .

4. T ypes of forces
Pushing force
Pulling force
Force of tension

5. Types of Forces
Weight
Friction
Magnetic
Electric

6. Types of Forces Stretching force Twisting force Compression force
Reaction force

7. Force Force = Mass x acceleration or F = m a This  a  F a  1/m
How might we prove a  F using a linear track?
What quantity must we keep constant in such an experiment?

8. Experiment: Acceleration  Force
Lights gates
Connect to
A timer
Mass must be kept constant. Why?

9. Weight and Mass
Weight is a force due the pull of gravity. As it is a force it is measured in Newtons (N)
Weight is a force. Mass is the amount of matter in a substance measured in Kg
Weight depends on where you are but your mass always stays the same

10. Weight and mass difference
Your weight on Earth may for example be 600 N about 60 kg of mass
On the moon your mass is still 60 kg but your weight is only 100 N
Out in the vacuum of space your weight would be 0 N and your mass still 60 kg.

11. Calculating Weight
On Earth objects fall with increasing speed in fact 9.8 m/s every second.
This is the acceleration due to gravity which is 9.8 m/s2
This is different depending on gravity, on the moon it is only 1.6 m/s2 and in space is 0 m/s2.
On Earth the weight of an object:
weight = mass (kg) x 9.8

12. Calculating Weight (Hons)
Find the weight of a block of mass 7.5 kg.
ANSWER
wt.= mass x 9.8
= 7.5 x 9.8
73.5 N
What is the weight of the same block on the moon.
wt. = mass x 1.6
=7.5 x1.6
= 12 N

13. Weight examples (hons)
What is the weight of the same block in space?
ANSWER
wt. = mass x 0
= 7.5 x 0
0 N
What is mass of a block that weighs 20 N ?
wt. = mass x 9.8
20 = mass x 9.8
Mass = 2.04 kg

14. Weight examples What is the weight of a 500g bag of sugar? ANSWER
wt. = mass (kg) x 9.8
=0.5 kg x 9.8
= 4.9 N
Note the grams must be converted to kg first.

15. Problems using F = m a Example 1
A body of mass 105 kg has a velocity of 10 m/s. What force is required to stop it in 0.5 s?
Answer
u = 10, v = 0, t = 0.5 F = ?
1st we must find a: v = u + a t
0 = 10 + a 0.5
a = - 20 m/s2
Now f = m a
f = 105 x 20
f = 2100 N

16. Problems using F = m a Example 2
Find the acceleration due to a force of 112 N to the left and a force 96 N to the right on a mass of 8 kg.
Answer
F = 112 – 96 = 16 to the left. (rem if the forces are in opposite directions you must give one a negative value.)
Now f = m a
16 = 8 x a
a = 2 m/s2 to the left

17. What is friction? Friction
force that opposes the motion between two objects in contact.
the force of friction works in the opposite direction of the force of motion

18. What causes Friction?
the source of friction is the contact between two surfaces, which are not perfectly smooth. The tiny humps and hollows catch off each other opposing the motion.

19. Advantages of Friction
Used to stop or slow things down, e.g brakes on a car or a bicycle.
Without friction we could not walk or kiss, car tyres would have no grip on the road.
Without friction you could not pick things up with your fingers.

20. Disadvantages of Friction
Friction slows things down, ships travel much slower than planes because of the friction due to the water.
Energy is often wasted e.g in car engine energy is wasted as heat and sound due to friction
Friction wears away surfaces, brake pads wear, tyres lose their thread.

21. Friction friction makes motion possible
friction also makes it hard to move objects
reducing friction makes it easier to move objects

22. How can friction be reduced?
by polishing and making surfaces smoother
by using lubricants
Lubricants – substances that reduce friction

23. Momentum Momentum = Mass x Velocity
Which has more momentum: a 80 kg football player moving at 5m/s or a 60 kg player moving at 10m/s?

24. Answer: Momentum of the 80 kg player is 80 kg x 5m/s = 400 kgm/s
Momentum of the quarterback is
60 kg x 10m/s = 600 kgm/s
The quarterback has a greater momentum!
10/3/2017

25. elastic collisions

26. Elastic collision

27. Momentum problems
2 cars are heading east, car A is traveling 30m/s, car B is traveling 60 m/s. Each car weighs 2000 kg.
What is the momentum of car A?
What is the momentum of car B?
If car B crashes into car A, what is the total momentum?

28. Answers: P=mv Car A’s momentum = 30m/s x 2000 kg
PA = 60,000 kgm/s east
Car B’s momentum = 60 m/s x 2000 kg
PB = 120,000 kgm/s east
Total momentum = PA + PB
= 60, ,000
= 180,000 kgm/s east
10/3/2017

29. Another momentum problem!
Car X is traveling 30m/s east, car Y is traveling 60 m/s west. Each car weighs kg.
What is the momentum of car X?
What is the momentum of car Y?
If car X crashes into car Y, what is the total momentum?

30. Answers: P=mv Car X’s momentum = 30mi/hr x 2000lbs
PA = 60,000 m/s east
Car Y’s momentum = 60m/s x 2000 kg
PY = 120,000 m/s west
Total momentum = PY - PX
= 120, ,000
= 60,000 m/s west
10/3/2017

31. Newton's Laws of Motion
NEWTON'S 1ST. LAW OF MOTION – states a body will remain in a state of rest or travelling with a constant velocity unless an external force acts ón it.
NEWTON'S 2ND. LAW OF MOTION – states the rate of change of momentum is directly proportional to the force and takes place in the direction of the force.
NEWTON'S 3RD. LAW OF MOTION – states if body A exerts a force on body B, then body B exerts an equal but opposite force on body A. i.e. To every action there is an equal and opposite reaction.

32. F = m a F = m a is a special case of Newton's 2nd. Law.
F  rate of change of Momentum
F  change in momentum
Time taken
F  final momuntum – initial momentum
F  mv – mu  F = km (v-u) but a = v – u
t t t
F = k m a since 1 newton = 1 kg at 1 m/s2 k= 1
Therefore f = m a which means a  F which we can prove by experiment.

33. Practical effects of Newton's 3rd. Law
Newton's 3rd. Law tells forces always occur in pairs.
e.g. car crashes into a wall, the wall exerts a force on the car and car exerts an equal force on the wall in the opposite direction.
For a rocket to accelerate forward is expels a mass of gas backwards. This gas exerts and equal and opposite force on the rocket.
The recoil of a rifle when fired.
Friction always opposes the foward
motion.

34. Practical effects of Newton's 3rd. Law
Air resistance is an opposing force to objects falling downwards. When the force due to the air resistance = the force due to the weight of the object then the object has a constant velocity called terminal velocity. Under Newton's 1st. Law the ojbect continues downward with a constant velocity. (no resultant force = no acceleration)

35. Example Problems
A crane lifts a block of mass 20 kg from the ground to the top of a building by means of a cable.
(i) What force does the cable exert on the block if the block is accelerating upwards a 1.5 m/2 ?
(ii) What force does the cable exert on the block if it is moving upwards at a constant speed?
Answer
There are two forces on the block the tension T acting
Upwards and the weight W acting downwards. T
20 kg W

36. Example Problems
(i) if the block accelerates upwards then T must be greater than the W. The net upward force is T – W
Applying F = m a gives
T – W = m a  T – (20)(9.8) = (20)(1.5)
 T = 226 N
(ii) Constant velocity  acceleration = 0 m/s2. Applying F = m a
T – W = m a  T – (20)(9.8) = 0
T = W = (20)(9.8) = 196 N

37. Example Problems
A man of 80 kg stands on a weighing scales in a lift. If the scales reads newtons find the reading on it when the lift:
(i) is at rest
(ii) moves upwards with a steady speed of 3 m/s.
(iii) moves downwards with a steady speed of 4 m/s.
(iv) accelerates upward at 2 m/s2.
(v) accelerates downwards at 2 m/s2.
(vi) accelerates downwards at 9.8 m/s2.
Answer
There are two forces on the man the normal reaction N acting
Upwards due to the floor of the lift and the weight W acting downwards. We apply F = m a in the direction of the acceleration.

38. Example Problems For (i), (ii) and (iii) the acceleration = 0.
 the weight = the normal reaction i.e. W = N
 N = 80 g = 80 x 9.8 = 784 N
(iv) N has to be greater than W so N – W = m a
N – 80 x 9.8 = 80 x 2  N = = 944 N
(V) W has to greater that N so W – N = m a
80 x 9.8 – N = 80 x 2  N = 784 – 160 = 624 N
(vi) going downwards so W – N = m a
80 x 9.8 – N = 80 x 9.8  N = 0
This means the man is weightless. For training a plane flies to a certain height and then is allowed to freefall. As the plane falls the occupents experience weightlessness.

39. Principal of Conservation of Momentum
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
Momentum before = momentum after the collision
m1u1 + m2u2 = m1v1 + m2v2

41. Experiment to Prove the Principal of Conservation of Momentum
What must we measure to prove the principal of conservation of momentum?
What apparatus could we use to measure the momentum.
What is the momentum of the trolley 2 when it is at rest?
What do you expect to happen when the two trollies collide?
Will the velocity of trolley 1 before the collision be greater than after the collision and why?
What measurements are made in this experiment?

42. Experiment to Prove the Principal of Conservation of Momentum
How are the initial and final velocity calculated?
What would be the equation be for the momentum before and momentum after the collision?

43. Principal of Conservation of Momentum
Example 1
A body of mass 20 kg moving at a speed of 4 m/s2collides with body of mass 14 kg which is at rest. If the two bodies stick together on collision find the combined velocity after the collision?
Answer
Momentum before = momentum after the collision
m1u1 + m2u2 = (m1 + m2)v
20 x x 0 = ( ) v
34v = 80
v = 2.35 m/s

44. Principal of Conservation of Momentum
Example 2
A body of mass 20 kg moving at a speed of 3 m/s collides with body of mass 15 kg moving at 6 m/s in the opposite direction. If the two bodies stick together on collision find the combined velocity after the collision?
Answer
Momentum before = momentum after the collision
m1u1 + m2u2 = (m1 + m2)v
20 x x -6 = ( ) v
35v =
35v = -30
V = m/s
v is in the direction of 15 kg mass before the collision
Why – 6 m/s

45. Principal of Conservation of Momentum
Example 3
A gun of mass 3 kg fires a bullet of mass 10 g with a speed of 500 m/s collides. Calculate the recoil velocity of the gun. Remember the gun will move in the opposite direction of the bullet.
Answer
Momentum before = momentum after the collision
m1u1 + m2u2 = m1v1 + m2v2
0 = 3v x 500
3v = - 5
V = m/s
The gun recoils backwards at a speed of 1.67 m/s

46. Principal of Conservation of Momentum
A spacecraft of mass 400 kg moving at 1000 m/s ejects an object of mass 20 kg at 2000 m/s at right angles to the direction in which the craft is moving.
Calculate the resultant velocity of the craft in magnitude and direction.
Answer
Since no forces act in the original direction of motion the velocity of the craft remains the same 1000 m/s. Let x = the recoil velocity acquired by the craft which is at right angles to the 1000 m/s.
2000 m/s
Recoil
velocity
Resultant
velocity

47. Principal of Conservation of Momentum
Answer (cont.)
Applying conservation of momentum perpendicular to the original direction of motion
m1u1 + m2u2 = m1v1 + m2v2
0 = 380x +20 x
380x = 4000
X = m/s.
Why – 2000 m/s
1000 m/s 
v2 = (1000)2 + (105.26)2
v = m/s
Tan  =   = 60
1000
v = m/s at 60 to the original direction of motion.
Recoil
velocity
Resultant
Velocity v
m/s.