1. Momentum: Definition 2
Newton used the concept of momentum to explain the results of collisions
Momentum = mass x velocity
p = m v
Units :
p (kg m/s) = m (kg) x (m/s)
Note since velocity is a vector quantity, (both magnitude and direction) then momentum is also a vector quantity

2. Conservation of Momentum 1
When objects collide, assuming that there are no external forces, then momentum is always conserved.... Definition :
When two or more objects interact, the total momentum remains constant provided that there is no external resultant force
Mass 75 kg
Velocity 4m/s
Mass 125 kg
Velocity ??? m/s
Mass 50 kg
Velocity 0m/s

3. Types of Collision 1
When objects hit each other the resulting collision can be considered to be either elastic or inelastic. Momentum and total energy are always conserved in both cases.
Elastic :
momentum conserved, kinetic energy conserved, total energy conserved
Inelastic :
momentum conserved, kinetic energy NOT conserved, total energy conserved
In an Inelastic collision some of the kinetic energy is converted to other forms of energy (often heat & Sound)

4. Types of Collision 2 Is the following collision elastic or inelastic?
A toy lorry of mass 2kg is travelling at 3m/s and crashes into a smaller a toy car with a mass of 800g which is travelling at 2 m/s in the same direction as the lorry. The velocity of the lorry after the collision is 2.6 m/s. What is the velocity of the car after the collision and is the crash elastic or inelastic?

5. Types of Collision 3 Mass 800g Velocity ? m/s Mass 800 g
Mass 2 kg
Velocity 2.6 m/s
Mass 2 kg
Velocity 3 m/s
Total momentum after= 7.6 kgm/s
Lorry Momentum = 2 x 2.6
Lorry Momentum = 5.2 kgm/s
Car Momentum = Total - Lorry
Car Momentum = 7.6 – 5.2
Car Momentum = 2.4 kgm/s
Car Momentum = 0.8v
 v = 2.4 / 0.8 = 3 m/s
Lorry Momentum = 2 x 3
Lorry Momentum = 6 kgm/s
Car Momentum = 0.8 x 2
Car Momentum = 1.6 kgm/s
Total momentum before = 7.6 kgm/s

6. Types of Collision 3 Mass 800g Velocity 3 m/s Mass 800 g
Mass 2 kg
Velocity 2.6 m/s
Mass 2 kg
Velocity 3 m/s
Lorry Ek = ½ x 2 x (3)2
Lorry Ek = 9J
Car Ek = ½ x 0.8 x (2)2
Car Ek = 1.6J
Total Ek before = 10.6J
Lorry Ek = ½ x 2 x (2.6)2
Lorry Ek = 6.76J
Car Ek = ½ x 0.8 x (3)2
Car Ek = 3.6J
Total Ek after= 10.36J
Total Ek before  Total Ek after
Inelastic Collision
Where is the energy likely to have gone?

7. Problems 1
Two trolleys on an air track are fitted with repelling magnets. The masses are 0.1kg and 0.15kg respectively. When they are released the lighter trolley moves to the left at 0.24m/s. What is the velocity of the heavier trolley
A ball of 0.6kg moving at 5m/s collides with a larger stationary ball of mass 2kg. The smaller ball rebounds in the opposite direction at 2.4m/s
Calculate the velocity of the larger ball
Is the Collision elastic or inelastic. Explain your answer

8. Momentum Learning Objectives
Book Reference : Pages 4-17
To revisit Newton’s 2nd law in terms of momentum.
To define the impulse of a force and connect it to the change in momentum
To understand the significance of the area under a force v time graph
To be able to complete “impulse of a force” calculations

9. Newton’s 2nd Law Newton’s 2nd law :
The rate of change of momentum of an object is proportional to the resultant force on it. The resultant force is proportional to the change in momentum per second
At AS we simply considered this to be F=ma
We will now revisit this is terms of momentum

10. Newton’s 2nd Law
Consider an object of constant mass m acted on by a constant force F.
The force causes an acceleration from the initial speed of u to the final speed v.
Therefore initial momentum is mu and the final momentum is mv. So the change in momentum is mv –mu
F  change in momentum mv –mu
time taken t

11. Newton’s 2nd Law F  mv –mu can be rewritten as t
F  m(v – u) From SUVAT a = (v-u)/t
F  ma after defining a suitable constant of proportionaility
F = kma
We can make k=1 by defining the unit of force.....

12. Newton’s 2nd Law
The Newton is the amount of force that gives an object of mass 1kg and acceleration of 1 m/s2
We can write the 2nd law as:
F = (mv) t
This can be used in two scenarios:
Firstly if the mass is constant then (mv)/t becomes m v/t
Change in velocity i.e. acceleration

13. Newton’s 2nd Law
Secondly if the mass is changes at a constant rate then (mv)/t becomes v m/t
Where m/t is the change in mass per second
This could be applied to a rocket which is losing mass each second in the form of hot exhaust gas

14. Impulse of a Force Definition :
The impulse of a force is defined as the product of the force and the time which the force acts for
The impulse = Ft = mv
The impulse of the force acting upon an object is equal to the change of momentum for the force

15. Force v Time Graphs
An object of constant mass m is acted upon by a constant force F which results in a change of velocity from u to v
From the 2nd law F = (mv – mu )/t
Rearranging : Ft = mv – mu
Graphically..... F
Area under graph “Ft” = change of momentum
force
time t

16. Force v Time Graphs Units of momentum revisited :
From the area under the graph F x t we naturally arrive at units of “Ns” for change of momentum and hence momentum itself. Ns is simply an alternative form of kgm/s

17. Problems 1...
A train of mass 24,000kg moving at a velocity of 15m/s is stopped by a braking force of 6000N. Calculate :
The initial momentum of the train
The time taken for the train to stop
An aircraft with total mass 45,000kg accelerates on the runway from rest to 120m/s at which point it takes off. The engines provide a constant driving force of 120kN. Calculate the gain in momentum and the take of time

18. Problems 2...
The velocity of a car of mass 600kg was reduced from a speed of 15m/s by a constant force of 400N which acted for 20s and then by a constant force of 20N for a further 20s.
Sketch a force v time graph
Calculate the initial momentum of the car
Use your Force v time graph to establish the change in momentum
Show the final velocity of the car is 1m/s

19. Momentum : Real World Examples
Learning Objectives
Book Reference : Pages 4-17
To Summarise what has been learnt about momentum by looking at specific real world examples.
To revisit car safety
Rebound impacts including oblique impacts
Dropped balls
Explosions & Guns

20. Car Safety The impulse = Ft = mv
During the Y11 course of study, it was discussed how many car safety features such as seatbelts, crumple zones and air bags increase safety by making the crash “last longer”
During our Y12 presentations, change in momentum was connected to car safety. Now taking it further and considering the impulse of a force :
The impulse = Ft = mv
For a given crash the mass & velocity of the vehicle are defined. By increasing t we decrease the force acting on the occupants

21. Rebound Impacts 1
We have seen that momentum is a vector quantity since it’s related to velocity which is a vector quantity.  direction is important and therefore we need a “sign” convention to take this into account.
If we consider a ball with mass m hitting a wall and rebounding normally, (i.e. at 90°):
Towards the wall we take as positive
Away from the wall we take as negative
Initial velocity = +u
Initial momentum = +mu

22. Rebound Impacts 2
Final velocity = -u
Final momentum = -mu
If we assume there is no loss of speed after the impact then considering the change in momentum...
Ft = final momentum – initial momentum
Ft = -mu – (+mu)
F = -2mu /t

23. Rebound Impacts 3
When the impact is oblique, (i.e. At an angle, not normally at 90°):  
Initial velocity = +u
Initial momentum = +mu
In this case we use the normal components of the velocity. Initially, this is +(u cos ). Similarly this will give an overall change in momentum of :
Ft = -2mu cos 

24. Dropped Bouncing balls 1
If we drop a perfectly bouncy ball onto a hard surface it should rebound to almost the original height it was dropped from.
Kinetic energy just prior to impact will equal kinetic energy just after impact if it is a perfect elastic collision
In this way we can make a connection between “elastic” and a perfectly bouncy ball

25. Dropped Bouncing balls 2
When approaching “dropped ball” calculations one applies our SUVAT equations in a manner similar to the way we approach projectiles
v = u + at (1)
s = (u + v)t (2) 2
s = ut + ½at2 (3)
v2 = u2 + 2as (4)

26. Explosions 1
Explosion problems are categorised by all the components initially being at rest. Then after some event, two or more objects move apart.
Since initially all objects are at rest, the total initial momentum is zero
We use the “signed” nature of direction to again equate the total final momentum to zero
Common examples include, trolleys or air track vehicles pushed part by springs or by repelling magnets
The recoil in gun barrels are also a good example

27. Explosions 2
Explosion problems can be tested with either sprung trolleys or air track vehicles:
Spring loaded bolt
Block
Block B A
Trolley A
Trolley B
When the sprung bolt is released the two trolleys move apart in opposite directions. The blocks A and B are positioned such that the trolleys strike them at the same moment. From s=d/t, since the time is identical, the ratio of the distances to the blocks is the same as the ratio of the speeds of the trolleys which is turn is the inverse of the mass ratios.

28. Problems 1...
A squash ball is released from rest above a flat surface. Describe how the energy changes is i) it rebounds to the same height, ii) It rebounds to a lesser height
If the ball is released from a height of 1.20m and rebounds to a height of 0.9m show that 25% of the kinetic energy is lost upon impact

29. Problems 2...
A shell of mass 2kg is fired at a speed of 140m/s from a gun with mass 800kg. Calculate the recoil velocity of the gun
A molecule of mass 5.0 x kg moving at a speed of 420m/s hits a surface at right angles and rebounds at the opposite direction at the same speed. The impact lasted 0.22ns. Calculate:
The change in momentum
The force on the molecule

30. Problems 3...
Repeat the last molecule question. This time the molecule strikes the surface at 60° to the normal and rebounds at 60° to the normal.

31. Uniform Circular Motion

32. Uniform Circular Motion
Learning Objectives
Book Reference : Pages 22-23
To summarise the relationship between degrees and radians
To understand the term angular displacement
To define angular velocity
To connect angular velocity to the period and frequency of rotation
To connect angular velocity to linear speed

33. Radians & Degrees Angles can be measured in both degrees & radians :
Arc
length r
The angle  in radians is defined as the arc length / the radius
For a whole circle, (360°) the arc length is the circumference, (2r)
 360° is 2 radians
Common values :
45° = /4 radians
90° = /2 radians
180° =  radians
Note. In S.I. Units we use “rad”
How many degrees is 1 radian?

34. Angular Displacement
Angular velocity, for circular motion, has counterparts which can be compared with linear speed s=d/t.
Time (t) remains unchanged, but linear distance (d) is replaced with angular displacement  measured in radians.
Angular displacement  r
Angular displacement is the number of radians moved  r

35. Angular Displacement Problems
For a watch calculate the angular displacement in radians of the tip of the minute hand in
One second
One minute
One hour
Each full rotation of the London eye makes takes 30 minutes. What is the angular displacement per second?

36. Angular Velocity : Definition
Consider an object moving along the arc of a circle from A to P at a constant speed for time t: 
Arc length r P A
Definition : The rate of change of angular displacement with time
“The angle, (in radians) an object rotates through per second”
 =  / t
Where  is the angle turned through in radians, (rad), yields units for  of rad/s
This is all very comparable with normal linear speed, (or velocity) where we talk about distance/time

37. Angular Velocity : Period & Frequency
The period T of the rotational motion is the time taken for one complete revolution (2 radians).
Substituting into :  = /t
 = 2 / T
T = 2 / 
From our earlier work on waves we know that the period (T) & frequency (f) are related T = 1/f
f =  / 2

38. Angular Velocity : linear speed
Considering the diagram below, we can see that the linear distance travelled is the arc length 
Arc length r P A
Linear speed (v) = arc length (AP) / t
v = r /t
Substituting... ( =  / t)
v = r

39. Angular Velocity : Worked example
A cyclist travels at a speed of 12m/s on a bike with wheels which have a radius of 40cm. Calculate:
The frequency of rotation for the wheels
The angular velocity for the wheels
The angle the wheel turns through in 0.1s in
i radians ii degrees

40. Angular Velocity : Worked example
The frequency of rotation for the wheels
Circumference of the wheel is 2r
= 2 x 0.4m = 2.5m
Time for one rotation, (the period) is found using
s =d/t rearranged for t
t = d/s = T = circumference / linear speed
T = 2.5/12 = 0.21s
f = 1/T = 1/0.21 = 4.8Hz

41. Angular Velocity : Worked example
The angular velocity for the wheels
Using T = 2 / , rearranged for 
 = 2 /T
 = 2 /0.21
 = 30 rad/s

42. Angular Velocity : Worked example
The angle the wheel turns through in 0.1s in
i radians ii degrees
Using  =  / t re-arranged for 
 = t
 = 30 x 0.1
 = 3 rad
= 3 x (360°/2) = 172°

43. Centripetal acceleration & Force

44. Centripetal acceleration & Force
Learning Objectives
Book Reference : Pages 24-25
To consider speed & velocity around a circle
To consider acceleration as a change in velocity
To define an equation for centripetal acceleration
To define an equation for centripetal force

45. Centripetal Acceleration : Introduction
If an object is moving in a circle with a constant speed, it’s velocity is constantly changing....
Because the direction is constantly changing....
If the velocity is constantly changing then by definition the object is accelerating
If the object is accelerating, then an unbalanced force must exist
Velocity v
acceleration

46. Centripetal Acceleration : Proof 1
Consider an object moving in circular motion with a speed v which moves from point A to point B in t seconds
Velocity vB B
Velocity vA  C A
(From speed=distance/time), the distance moved along the arc AB, s is vt v
The vector diagram shows the change in velocity v :
(vB – vA)
Velocity vA 
Velocity vB

47. Centripetal Acceleration : Proof 2
The triangles ABC & the vector diagram are similar
Velocity vB B
Velocity vA  C A
If  is small, then v/v = s/r
Substituting for s = vt
v/v = vt /r
(a = change in velocity / time)
a = v/ t = v2/r v
Velocity vA 
Velocity vB

48. Centripetal Acceleration : angular
We can substitute for angular velocity....
a = v2/r
From the last lesson we saw that:
v = r (substituting for v into above)
a = (r)2/r
a = r2

49. Centripetal Force
In exactly the same way as we can connect force f and acceleration a using Newton’s 2nd law of motion, we can arrive at the centripetal force which is keeping the object moving in a circle
f = mv2/r or
f = mr2
Any object moving in a circle is acted upon by a single resultant force towards the centre of the circle. We call this the centripetal force

50. Centripetal Force : Gravity & Orbits
Gravity which keeps satellites in orbit around Earth and the Earth in orbit around the sun is a classic example of a centripetal force.
satellite
Gravity
Planet

51. Worked example 1
The wheel of the London Eye has a diameter of 130m and takes 30mins for 1 revolution. Calculate:
The speed of the capsule
The centripetal acceleration
The centripetal force on a person with a mass of 65kg

52. Worked example 1 The speed of the capsule : Using v = r
we know that we do a full revolution (2 rad) in 30mins (1800s)
v = (130/2) x (2 / 1800)
v = 0.23 m/s

53. Worked example 1 The centripetal acceleration: Using a = v2/r
a = x 10-4 m/s2
The centripetal force:
Using f = ma
F = 65 x x 10-4
F = N

54. Problem 1
An object of mass 0.15kg moves around a circular path which has a radius of 0.42m once every 5s at a steady rate. Calculate:
The speed and acceleration of the object
The centripetal force on the object

55. Centripetal Force & the Road

56. Centripetal Force & The Road
Learning Objectives
Book Reference : Pages 26-27
To show that the centripetal force is provided by real world forces such as tension, gravity & friction
To consider three particular cases of motion:
Over the top of a hill or humped back bridge
Around flat curves (roundabouts)
Around banked curves

57. Centripetal Acceleration : Recap
During the last lesson we saw that an object moving in a circle has a constantly changing velocity, it is therefore experiencing acceleration and hence a force towards the centre of rotation.
We called this the centripetal force: The force required to keep the object moving in a circle. In reality this force is provided by another force, e.g. The tension in a string, friction or the force of gravity.

58. Over the top 1
Consider a car with mass m and speed v moving over the top of a hill... S mg r

59. Over the top 2
At the top of the hill, the support force S, is in the opposite direction to the weight (mg). It is the resultant between these two forces which keep the car moving in a circle
mg – S = mv2/r
If the speed of the car increases, there will eventually be a speed v0 where the car will leave the ground (the support force S is 0)
mg = mv02/r v0 = (gr)½
Any faster and the car will leave the ground

60. Around a Roundabout 1
On a level road, when a car travels around a roundabout the centripetal force required to keep the car moving in a circle is provided by the friction between the road surface and tyres
velocity
friction
Force of Friction F
F = mv2/r

61. Around a Roundabout 2
To avoid skidding or slipping, the force of friction F0 must be less than the point where friction is overcome which occurs at speed v0
Friction is proportional to weight and can be given by the coefficient of friction ():
F  mg F = mg
At the point of slipping:
F0 = mv02/r  mg = mv02/r
 v0 = (gr)½

62. Banked Tracks 1
For high speed travel, race tracks etc have banked corners. In this way a component of the car’s weight is helping friction keep the car moving in a circle N1 N2  
Towards centre of rotation mg 

63. Around a Roundabout 2
Without any banking the centripetal force is provided by friction alone. Banked corners allows greater speeds before friction is overcome
The centripetal force is provided by the horizontal components of the support forces
(N1 + N2) sin  = mv2/r
and the vertical components balance the weight
(N1 + N2) cos  = mg

64. Around a Roundabout 3 Rearranging sin  = mv2/ (N1 + N2) r
cos  = mg / (N1 + N2)
and since tan  = sin  / cos 
tan  = mv2/ (N1 + N2) r x (N1 + N2) / mg
tan  = mv2 / mgr v2 = gr tan 
Thus there is no sideways frictional force if the speed v is such that v2 = gr tan 

65. Problems 1
A car with mass 1200kg passes over a bridge with a radius of curvature of 15m at a speed of 10 m/s. Calculate:
The centripetal acceleration of the car on the bridge
The support force on the car when it is at the top
The maximum speed without skidding for a car with mass 750kg on a roundabout of radius 20m is 9m/s. Calculate:
The centripetal acceleration of the car on the roundabout
The centripetal force at this speed

66. Problems 2
A car is racing on a track banked at 25° to the horizontal on a bend with radius of curvature of 350m
Show that the maximum speed at which the car can take the bend without sideways friction is 40m/s
Explain what will happen if the car takes the bend at ever increasing speeds

67. Centripetal Force & the Fairground

68. Centripetal Force & the Fairground
Learning Objectives
Book Reference : Pages 28-29
To apply what we have learnt about circular motion to rides at the fairground
To consider three particular cases of motion:
The big dipper
The long swing
The “centrifuge” wall of death

69. Centripetal Force: Big Dippers 1
At the bottom of a big dipper you are pushed into your seat and feel “heavier”...
Centre of curvature S
velocity mg

70. Centripetal Force: Big Dippers 2
At the bottom of the dip at speed v with radius r, Resolving the vertical forces:
S – mg = mv2/r
S = mg + mv2/r
The “extra” weight you experience when feeling “heavy” is given by the centripetal force

71. Centripetal Force: The Long Swing 1
Consider a person of mass m on a very long swing of length r
To winch r
Fixed point
Initial position S
Velocity mg

72. Centripetal Force: The Long Swing 2
As the swing is released we can consider the conservation of energy, loss in potential energy is the gain in the kinetic energy
mgh = ½mv2  v2 = 2gh
The passenger is on a circular path with radius r. At the bottom of the swing the support force S acts against the person’s weight mg. This provides the centripetal force:
S – mg = mv2/r

73. Centripetal Force: The Long Swing 3
Substituting for v2:
S – mg = mv2/r
S – mg = 2mgh/r
S = mg + 2mgh/r
The person “feels heavier” by 2mgh/r, if the swing drops through 90°, then the extra support force is twice the persons weight (mg)

74. Centripetal Force: The wall of death 1
Consider a fairground ride which spins fast enough to keep you in place even when upside down at the top of the ride....
Velocity
Reaction R & weight mg
Rotation

75. Centripetal Force: The wall of death 2
The wheel turns fast enough to keep the passenger in position as they pass over the top.
At the top, the reaction R acts downwards and together with the weight provides the centripetal force:
mg + R = mv2/r
R = mv2/r – mg
At a certain speed v02 such that v02 = gr, then the reaction from the wall will be zero

76. Problems 1
A car on a big dipper starts from rest and descends though 45m into a dip which has a radius of curvature of 78m. Assuming that air resistance & friction are negligible. Calculate:
The speed of the car at the bottom of the dip
The centripetal acceleration at the bottom of the dip
The extra force on a person with a weight of 600N in the train

77. Problems 2
A swing at a fair has a length of 32m. A passenger of mass 69kg falls from the position where the swing is horizontal. Calculate:
The speed of the person at the lowest point
The centripetal acceleration at the lowest point
The support force on the person at the lowest point

78. Problems 3
A wall of death ride at the fairground has a radius of 12m and rotates once every 6s. Calculate:
The speed of rotation at the perimeter of the wheel
The centripetal acceleration of a person on the perimeter
The support force on a person of mass 72kg at the highest point