2. Momentum

3. Momentum is the product of the mass of an object and it’s velocity.
Formula:
p = mv
mass (kg)
velocity (m⋅s-1)
momentum (kg⋅m⋅s-1)
Unit: kg⋅m⋅s-1

4. Momentum is influenced by: - mass of the - velocity of the object
Momentum is a vector; therefore it has
magnitude and direction
Always choose a DIRECTION as positive
and opposite directions as negatives.
Objects at rest have zero momentum because they have zero velocity.
Objects at rest have zero momentum
because they have zero velocity.

5. 1.1 Calculate the momentum of the cyclist. p = mv = (83)(8)
Example:
A cyclist with a mass of 83 kg rides on his bicycle with a mass of 37 kg at a velocity 8 m∙s-1 in a westerly
direction
Choose direction
1.1 Calculate the momentum of the cyclist.
p = mv
= (83)(8)
= 664 kg∙m∙s-1 west
Choose direction

6. 1.2 Draw and label the cyclist’s velocity and momentum vectors.
v = + 2,5 m∙s-1 west
p = 237,5 kg∙m∙s-1 west
1.3 Calculate the momentum of both the cyclist
and that of his bicycle.
p = mv
= (83+37)(8)
= 960 kg∙m∙s-1 west
1.3 Calculate the momentum of both the cyclist and that of his bicycle.

7. 1.4 If the cyclist’s velocity doubles, with which
factor will his momentum increase. Confirm
your answer with a calculation.
p = mv
= (83)(16)
= 1328 kg∙m∙s-1 west
His momentum will double if his velocity doubles, because pα v
1.4 If the cyclist’s velocity doubles, with which factor will his momentum increase. Confirm your answer with a calculation.

8. Collisions
A collision is an isolated event in which two or more moving bodies exert forces on each other over a relatively short time.
A collision is an isolated event in which two or more moving bodies exert forces on each other over a relatively short time.

9. Change in momentum
If the velocity of an object changes, then the momentum of the object will also change.
If the velocity of an object changes, then the momentum of the object will also change.

10. Change in momentum initial momentum ∆p = pf - pi = mvf - mvi
∆v = vf - vi
∆p = pf - pi
= mvf - mvi
change in momentum
(kg⋅m⋅s-1)
final momentum (kg⋅m⋅s-1)
initial momentum
final velocity (m⋅s-1)
initial velocity
(m⋅s-1)
change in velocity

11. Example:
A cricket ball with a mass of 70 g is travelling horizontally at 35 m⋅s-1 . The ball is struck by Amla and moves horizontally at 50 m⋅s-1 in the opposite direction.
1.1 Calculate the change in momentum.
1.2 Draw a vector diagram to illustrate the relationship
between the initial, final and change in momentum.

12. = 5,95 kg⋅m⋅s-1 away from the bat
1.1 Calculate the change in momentum.
Choose direction: +
m= 0,07 kg (remember to change g to kg)
vi = + 35 m.s-1
vf = - 50 m.s-1
Δp = mvf - mvi
= (0,07)(-50) – (0,07)(+35)
= - 5,95
= 5,95 kg⋅m⋅s-1 away from the bat
m= 0,0057 kg (remember to change g to kg)
Vi = +35 m.s-1
vf = - 50 m.s -1

13. Exercise 1 (p.39) 1.2 Draw a vector diagram to illustrate the
relationship between the initial, final and change
in momentum.
mVi = +2,45 kg⋅m⋅s-1
mVf = - 3,5 kg⋅m⋅s-1
∆p = - 5,95 kg⋅m⋅s-1
Do exercise 1.1 (p.18)
Exercise 1 (p.39)

14. Newton’s second law in terms
of momentum

15. Expressing Newton’s second law in terms of momentum
If a net force (Fnet) acts on an object it will accelerate. This means its velocity will change (increase ,decrease or change direction)
If a net force (Fnet) acts on an object it will accelerate. This means its velocity will change (increase, decrease or change direction) Expressing Newton’s second law in terms of momentum

16. rate of change in the velocity of an object
Acceleration:
rate of change in the velocity of an object
final velocity (m⋅s-1)
initial velocity (m⋅s-1)
time (s)
acceleration (m⋅s-2)
vf – vi
a = Δt

17. Fnet Δt = Δp Fnet = ma Fnet = m Fnet = Δv Δt Δp Δt vf – vi (Δv) a = Δt
Formula change in momentum and NII
Formula change in momentum and NII Δt
Fnet Δt = Δp

18. Fnet Δt = Δp Newton’s second law (in terms of momentum)
The net force acting on an object is equal to the rate of change of momentum
Fnet Δt = Δp
Fnet is the net force acting on the object (N)
∆p is the change in momentum (kg⋅m⋅s-1 )
∆t is the time interval over which the momentum of an object is changed in seconds (s)
The net force acting on an object is equal to the rate of change of momentum
Fnet is the net force acting on the object (N)
∆p is the change in momentum (kg⋅m⋅s-1 )
∆t is the time interval over which the momentum of an object is changed in seconds (s)

19. Newton’s second law Fnet = ma only applies to situations in which the mass of an object is constant.
If we use the concept of momentum it is possible to use the law to situations where both the mass and velocity of an object are changing.
Newton’s second law Fnet = ma only applies to situations in which the mass of an object is constant.
If we use the concept of momentum it is possible to use the law to situations where both the mass and velocity of an object are changing
While the rocket engines are fired, the rocket fuel is being burned – mass of rocket is decreasing and there is a change in velocity.

20. Applying Newton’s second law in terms of momentum
Example 1
In a test crash, a car with a mass of 1500kg crashes into
a wall with velocity of 15 m⋅s-1 and bounces back at
2,6 m⋅s-1 . In the course of the crash the wall exerts a
force of N on the car.
Example 1

21. 1.1 Draw a vector diagram to indicate how the
change in momentum can be illustrated.
initial momentum (pi)
final momentum (pf)
change momentum (Δp)
1.2 Calculate the change in momentum of the car +
Δp = mvf - mvi
= (1500)(-2,6) – (1500)(+15) =
= kg⋅m⋅s-1 right/away from wall
Draw a vector diagram to indicate how the change in momentum can be illustrated.
Initial momentum

22. 1.3 Calculate the time that the car was in contact with the wall.
Fnet ∆t = ∆p
∆t =
= 0,15 s
Why would car manufactures, design a car that
would come to rest during an accident rather
than to bounce back.
The net force undergone by a car is α to the car’s
change in momentum. Cars which bounce back
undergo a greater change in momentum and
therefore experience a greater force – more
damage and serious injuries.
Why would car manufactures, design a car that would come to rest during an accident rather than to bounce back.

23. A 65 kg Olympic springboard diver jumps into the air
Example 2
A 65 kg Olympic springboard diver jumps into the air
at the end of a diving board. The diver lands on the
springboard, travelling downward at 8 m⋅s-1 and
leaves the springboard moving upward at 12m⋅s-1
2.1 Draw a free-body diagram of the forces acting on
the diver while in contact with the springboard.
Force of
springboard F
Weight Fg
A 65 kg Olympic springboard diver jumps into the air at the end of a diving board. The diver lands on the springboard, travelling downward at 8 m⋅s-1 and leaves the springboard moving upward at 12m⋅s-1
2.1 Draw a free-body diagram of the forces acting on the diver while in contact with the springboard.
2.2 Calculate the driver’s change in momentum while she is in contact with the springboard?
2.3 The driver’s feet are in contact with the springboard for 0,8s. Calculate the force that the springboard exerts on her.
Weight Fg

24. while she is in contact with the springboard?
2.2 Calculate the diver’s change in momentum
while she is in contact with the springboard?
Choose direction Δp = mvf - mvi
= (65)(+12) – (65)(-8)
= kg⋅m⋅s-1 upwards +
NB there are

25. 2.3 The diver’s feet are in contact with the
springboard for 0,8s. Calculate the force
that the springboard exerts on her.
Calculate divers weight
Fg = mg
= (65)(-9,8)
= = 637 N down
Apply Newton’s second law (NB remember there
are 2 forces)
Fnet ∆t = ∆p
(F + Fg) ∆t = ∆p
F + (-637) 0,8 = +1300
= N upwards

26. Impulse in terms
of momentum

27. Impulse A force must cause an object to move in order
for the object to have momentum.
The force is exerted on the object for a certain
amount of time.
A force must cause an object to move in order for the object to have momentum.
The force is exerted on the object for a certain amount of time.

28. Impulse
Is the product of the net force on an object and the time for which it acts
FnetΔt
The art of catching a cricket ball painlessly.
When a fielder catches a cricket ball he pulls his hands back.
Reason: The force (Fnett) necessary to stop the ball is dependant ∆p and ∆t.
The ball approaches the fielder with a specific momentum. In order to stop the ball, the fielder must change the momentum of the ball to zero. By pulling back his hands the fielder increases the time (∆t) in which the ball’s momentum changes. According to the equation F(nett) = ∆p/∆t the force exerted on the fielder’s hand is reduced (gets smaller)

29. Impulse
Fnet∆t = m∆v

30. A golf ball with a mass of 45 g rests on a tee. Mr Uys
Example 1
A golf ball with a mass of 45 g rests on a tee. Mr Uys
hits the ball and it leaves the head of the club at
41 m⋅s-1.
Determine:
1.1 the change in the momentum of the ball
Use the direction of movement of the club to the right and +
Δp = mvf - mvi
= (0,045)(41) – (0,045)(0)
= 1,85 kg⋅m⋅s-1 Right

31. 1.2 the net force on the ball if the ball and head of
the club are in contact for 0,01 seconds
Fnet ∆t = ∆p
Fnet (0.01) =1,85
= 185 N Right
1.3 the net force on the head of the golf club.
From Newton’s third law: 185 N left

32. Example 2
Simon drops a rubber ball, mass 1kg, vertically downwards. It hits the ground and bounces back up.
vi vf
The graph shows the average net force exerted on the ball by the ground as a function of time.
140
Fnet (N)
0, t(s)

33. What physical quantity is represented by the area under the graph?
If the area under the graph is calulated, the
units will be seonds x newton. This is the
unit for impulse.
Calculate the change in the momentum of the
ball.
∆p= impulse= area under the graph=(1/2bh)
= ½(0,1)(140)
= 7 kg⋅m⋅s-1 upwards

34. change in momentum = impulse Fnet∆t = m∆v mvf - mvi = Fnet∆t
If the ball hits the ground at 4 m⋅s-1 , calculate the speed at which it leaves the ground.
change in momentum = impulse
Fnet∆t = m∆v
mvf - mvi = Fnet∆t
(1)(v) – (1)(-4) = 7
vf = 3 m⋅s-1 upwards +

35. APPLIACATION OF IMPULSE IN EVERY DAY LIFE
A very important application of impulse is the improvement of safety and lessening of injuries.
If the time during which the momentum changes can be increased, the exerted force will be less and cause less damage.
Air bags
Safety belts
Crumple zones in vehicles
Bending your knees when you jump from a height.
A very important application of impulse is the improvement of safety and lessening of injuries.

36. Sometimes we must use impulse principles to exert a greater force.
Tennis, golf, cricket players use a follow through.
Sometimes we must use impulse principles to exert a greater force.

37. The art of catching a cricket ball painlessly.
When a fielder catches a cricket ball he pulls his hands back.
Reason: The ball approaches the fielder with a specific
momentum. In order to stop the ball,
the fielder must change the momentum of the ball to
zero. By pulling back his hands the fielder increases
the time (∆t) in which the ball’s momentum changes.
According to the equation F(nett) = ∆p/∆t the force
exerted on the fielder’s hand is reduced (gets smaller)

38. When the player does not pull his hands back, the following happens:
Fnet Δt = Δp
Remains the same: ball stops.
Player does not pull back hand. Therefore time of contact is very short.
Thus: Δp stays same
Δt is small to change velocity
Therefore the force which the ball exerts
on the player’s hand, increases.
He will feel the pain.
Tennis players use a follow through

39. Tennis players use a follow through with their arms
when hitting the ball.
The tennis player use the follow through with the arm to increase the contact time between the ball and racquet. If the time increase the force will be greater. This causes the change in momentum of the ball to increase and the ball undergoes a greater change in velocity if the contact time is shorter. The ball leaves the racquet with a greater velocity.
The tennis player use the follow through with the arm to increase the contact time between the ball and racquet. If the time increase the force will be greater. This causes the change in momentum of the ball to increase and the ball undergoes a greater change in velocity if the contact time is shorter. The ball leaves the racquet with a greater velocity.

40. Seat belts and airbags Crumple zones
Vehicle safety devices are designed
to increase collision times and so
reduce the net force acting on motorist during
collision.
Crumple zones
Modern cars are design with crumple zones. It ensured that the car will not bounce back.
If it bounce back the ∆p will be very large and the force exerted will also be very large.
Seat belts and airbags

41. Arrestor beds It is a sand or gravel pathway.
An arrestor bed decrease the truck’s momentum to zero over a long period of time and therefore decrease the force.
Arrestor beds

42. In all of the above situations the following will happen:
Fnet∆t = ∆p
∆p = stays constant
∆t = contact time increase
Fnet = force will be smaller F
Fnet inversely proportional to time.
In all of the above situations the following will happen:

43. Graphic representation of Impulse
The area below the force-time graph gives us the impulse on the object as well as the change in the object’s momentum.
The area below the force-time graph gives us the impulse on the object as well as the change in the object’s momentum
Exercise 2 (p49)

44. EXAMPLE
The following graph shows the force exerted on a hockey ball over time. The hockey ball is initially stationary and has a mass of 150g. Calculate the magnitude of the impulse (change in momentum) of the hockey ball.
REMEMBER:
If graph given, show reference to graph in calculations: AREA or GRADIENT

45. Conservation of momentum, elastic and inelastic collisions.

46. System Two moving objects each have their own momentum.
This physical interaction/collision of object is called the “system”.
Examples of system:
Colliding billiard balls on a pool table
Two cars involved in collisions
Hunter firing a bullet from his rifle
System

47. A closed system is a system on which no
external forces, from outside the system,
are exerted on the system.
Example of external forces:
air resistance
friction between wheels and road
brakes
drive of engine
When two objects collide, they exert forces on each other called internal forces.
A closed system is a system on which no external forces, from outside the system, are exerted on the system.

48. Definition system
A system is a small part of the complete scenario that is observed when a specific problem is solved. Everything outside the system is called the environment.
A system is a small part of the complete scenario that is observed when a specific problem is solved. Everything outside the system is called the environment.

49. Ʃpbefore collision= Ʃp after collision Ʃpi = Ʃpf
Law of conservation of linear momentum:
The total linear momentum in a closed
system remains constant
Symbol format:
Ʃpbefore collision= Ʃp after collision
Ʃpi = Ʃpf
Both in magnitude and direction.

50. The principle of the conservation of momentum is especially used in collision-type of problems:
Collisions
Explosions
Gunshots
Jumping on
The principle of the conservation of momentum is especially used in collision-type of problems:

51. Collisions Move off together Collide and deflect
Objects can collide and move off separately
When objects collide and move off together, their masses can be added as one object

52. Collisions
Object dropped vertically
Linear momentum= momentum along one axis. A dropped object has a horizontal velocity of zero, ∴viB= 0m∙s－1

53. Explosions
Explosions
Springs
Objects that experience the same explosion will experience the same force.
The spring will exert the same force on both objects (Newton’s Third Law).

54. Explosions
Firearms
The gun and bullet will experience the same force. The acceleration of the weapon is significantly less than the bullet due to mass difference Recoil can be reduced by increasing the mass of the weapon.

55. Examples:
A toy truck with mass 1,6kg approaches another toy truck B at 10 m⋅s-1 eastwards. Truck B has a mass of 2kg and is stationary. After the collision, truck A has a velocity of 4 m⋅s-1 eastwards. Calculate the velocity of truck B after the collision.

56. Ʃpbefore collision= Ʃp after collision
(1,6)(10) + (2)(0) = (1,6)(4) + (2)vfB
VfB= 4,8 m⋅s-1 eastwards

57. A boy with a mass of 60kg stands on a moving skateboard with mass of 20kg that moves at a velocity of 5 m⋅s-1 eastwards on a straight, frictionless track. Calculate the velocity of the skateboard after the boy jumped off backwards in such a way that his horizontal velocity is 3 m⋅s-1 westwards when he lands.
A boy with a mass of 60kg stands on a moving skateboard with mass of 20kg that moves at a velocity of 5 m⋅s-1 eastwards on a straight, frictionless track. Calculate the velocity of the skateboard after the boy jumped off backwards in such a way that his horizontal velocity is 3 m⋅s-1 westwards when he lands.

58. Ʃpbefore collision= Ʃp after collision
(60+20)(5) = (60)(-3) + (20)vf
VfB= 29 m⋅s-1 eastwards

59. A trolley with mass 4kg moves at 3 m⋅s-1 to the right, when a brick with mass of 1 kg falls vertically on top of it. Calculate the trolley’s velocity after the collision.
A trolley with mass 4kg moves at 3 m⋅s-1 to the right, when a brick with mass of 1 kg falls vertically on top of it. Calculate the trolley’s velocity after the collision.

60. Ʃpbefore collision= Ʃp after collision
(4)(3) + (1)(0) = (4+1)vf
Vf= 2,4 m⋅ s-1 to the right

61. A cannon, mass 500kg, shoots a cannon ball, mass 10 kg, horizontally
A cannon, mass 500kg, shoots a cannon ball, mass 10 kg, horizontally. If the velocity of the cannon ball, after it has been shot, is 100 m⋅s-1 to the right. Calculate the recoil velocity of the cannon.
A cannon, mass 500kg, shoots a cannon ball, mass 10 kg, horizontally. If the velocity of the cannon ball, after it has been shot, is 100 m⋅s-1 to the right, calculate the recoil velocity of the cannon.

62. Ʃpbefore collision= Ʃp after collision
Choose a direction for the motion +
100 m⋅s-1
0 m⋅s-1
10 kg
BEFORE
AFTER
10 kg v
500 kg
500 kg
Remember that there is no motion before the shot. Let v be the velocity of the cannon after the shot.
Ʃpbefore collision= Ʃp after collision
Remember that there is no motion before the shot. Let v be the velocity of the cannon after the shot.
(500)(0) + (10)(0) = (10)(100)+(500)vf
Vf = - 2 m⋅s-1
Vf = 2 m⋅s-1 to the left

63. Elastic and Inelastic collisions
If the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision.
During a collision, kinetic energy is converted into other forms of energy such as sound and heat energy.
Elastic collision
If the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision
During a collision, kinetic energy is converted into other forms of energy such as sound and heat energy.

64. EXAMPLE
The velocity of a moving trolley of mass 1 kg is 3 m∙s－1. A block of mass 0,5 kg is dropped vertically on to the trolley. Immediately after the collision the speed of the trolley and block is 2 m∙s－1 in the original direction. Is the collision elastic or inelastic? Prove your answer with a suitable calculation.
Exercise 3 (p67)

66. DEFINITIONS – MOMENTUM
Momentum - Momentum is the product of the mass and velocity of an object.
Newton’s 2nd Law & momentum - Momentum is the product of the mass and velocity of an object.
Impulse - Is the product of the net force on an object and the time for which it acts.
System - A system is a small part of a complete scenario that is observed when a specific problem is solved.
DEFINITIONS - MOMENTUM

67. Closed/Isolated system - a system on which the
resultant / external forces is zero.
Law of conservation of momentum - total linear momentum in a closed system remains constant both in magnitude and direction.
Elastic collision - if the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. Energy is conserved.

68. Exam questions
QUESTION 4 (Start on a new page.) Nov 2015
A bullet of mass 20 g is fired from a stationary rifle of mass 3 kg. Assume that the bullet moves horizontally. Immediately after firing, the rifle recoils (moves back) with a velocity of 1,4 m∙s-1.
4.1 Calculate the speed at which the bullet leaves the rifle. (4)

69. 4.2 Calculate the magnitude of the average force
exerted by the block on the bullet (5)
4.3 How does the magnitude of the force calculated in
QUESTION 4.2 compare to the magnitude of the force
exerted by the bullet on the block? Write down only
LARGER THAN, SMALLER THAN or THE SAME. (1)
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70. QUESTION 3 – Northern Cape Trail exam 2015
A soccer ball of mass 430 g is moving at 20 m·s-1 horizontally towards the head of a waiting soccer player. The ball is “headed” back, in the opposite direction, along the same straight line, at 25 m·s-1. Ignore the effects of air resistance
3.1 Define impulse of a force in words. (2)
3.2 Calculate the impulse exerted on the ball while the head is
in contact with the ball. (3)
3.3 Using the answer in QUETION 3.2, calculate the time for
which the ball must be in contact with the head of the
player in order to experience a force of magnitude 300 N.
3.4 Is the collision of the soccer ball with the head of the player
elastic or inelastic? Give a reason for the answer

71. QUESTION 4 (Start on a new page.) March 2016
The diagram below shows two trolleys, P and Q, held together by means of a compressed spring on a flat, frictionless horizontal track. The masses of P and Q are 400 g and 600 g respectively
When the trolleys are released, it takes 0,3 s for the spring to unwind to its natural length. Trolley Q then moves to the right at 4 m∙s-1.

72. 4.1 State the principle of conservation of linear
momentum in words (2)
4.2 Calculate the:
Velocity of trolley P after the trolleys are released (4)
Magnitude of the average force exerted by the
spring on trolley Q (4)
4.3 Is this an elastic collision? Only answer YES or
NO (1)
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