Educator: Mr NB MAEBELA
Conduct: 0764259778
EMAIL: blessingmaebela@gmail.com
For more information, please contact me using the above details
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1. MOMENTUM
In this topic, we will eventually express Newton’s second law in terms of momentum. In this subtopic
we learn about the momentum vector and that the momentum of an object will change
when a net force acts on an object.
CONCEPT EXPLANATION AND CLARIFICATION
Definition of momentum (p): The product of an object’s mass and its velocity.
The momentum of an object (p) is calculated by using: p = mv, where m is the mass of the
object (measured in kg), and v is the velocity of the object (measured in m.s−1).
Remember that velocity (v ) is a vector quantity. Momentum (p) is also a vector quantity.
The momentum vector should be correctly expressed with a magnitude, a unit and a
direction.
N.B: A vector is a quantity that has both magnitude and direction.
The direction of the momentum vector is the same as the direction of the velocity vector. In
other words, the direction in which an object is moving IS the direction of the momentum
vector.
The unit of momentum is : p =mv kg m s 1 = = kg. m.s1  (stated as “kilograms metres per second”)
N.B: the velocity of the object is directly proportional to the momentum
Change in momentum (Δp)
In everyday situations, almost no object has a constant momentum because its velocity is
usually not constant. Friction slows objects down, objects such as aeroplanes accelerate
down runways and vehicles are involved in collisions.
What happens during a collision?
When a small truck A collides with a large truck B, the small truck exerts a force F on the
large truck. According to Newton’s third law, the large truck will exert a force equal in
magnitude but opposite in direction on the small truck.
The forces that act on each object during a collision only act for a short time interval and
they change the momentum of each object.
Remember: According to Newton’s second law, when a net force acts on an object, it will
accelerate.
When an object accelerates, its velocity will CHANGE. If the velocity of the object is
changing, then its momentum will CHANGE.
The change in momentum (Δp) of an object is calculated by subtracting its initial
momentum pi from its final momentum pf :
Δp = pf  pi
Δp = mvf  mvi
where:
m= is the mass of the object (measured in kg)
vi =is the initial velocity of the object immediately before the collision (measured in m.s−1)
vf =is the final velocity of the object immediately after the collision (measured in m.s−1)
Momentum is a VECTOR quantity. CHANGE in momentum Δp is also a VECTOR
quantity.
2. NEWTON’S SECOND LAW EXPRESSED IN TERMS OF
MOMENTUM.
In reality, the net force acting on an object during a collision is not constant. It is therefore
difficult to measure the average net force. It is easier to measure the velocity of an object
before and after the collision. We, therefore, need to express Newton’s second law in terms of
momentum.
CONCEPT EXPLANATION AND CLARIFICATION
Newton’s second law: When a net force is applied to an object of mass, m, it accelerates in
the direction of the net force. The acceleration (a) is directly proportional to the net force
and inversely proportional to the mass.
The equation for Newton’s second law is: Fnet = ma
Fnet=Δp/Δt
FnetΔt=Δp
N.B: FnetΔt is impulse
NB: WATCH THE BELOW VIDEOS TO ENFORCE YOUR UNDERSTANDING OF THE CONCEPTS DISCUSSED
https://www.youtube.com/watch?v=QAURZq43508
https://www.youtube.com/watch?v=GrjMbTN_DVU&t=2301s
ACTIVITY#1
Grade 12
Due date: 12/09/2021
Instruction:
N.B: Please study the notes and watch videos provided in the introduction section, this will assist you in answering this activity.
If you can't explain it better you didn't understand it better " Albert Einstein"
1. A 10 000 kg train travelling at 10 m.s–1 east collides with a 2 000 kg car travelling at
30 m.s1 in the opposite direction. Calculate:
1.1 The momentum of the train before the collision. (3)
1.2 The momentum of the car before the collision. (3)
The train is brought to rest during the collision and the car bounces backwards with
a speed of 20 m.s–1 after the collision.
1.3 Calculate the change in momentum of the train during the collision. (4)
1.4 Calculate the change in momentum of the car during the collision. (4)
1.5 Draw a labelled momentum vector diagram to illustrate the initial, final and change
in momentum vectors for the car. (3)
2. Car A (mass 600 kg) was travelling at 5 m.s–1 north when it was struck from behind
by car B (mass 800 kg) which was travelling at 12 m.s–1 north. Car A travels forward
(north) at 10 m.s–1 after the collision. Car B continues moving forward (north) at
8,25 m.s–1 after the collision.
2.1 Calculate the momentum of car B before the collision. (3)
2.2 Calculate the change in momentum of car B during the collision. (4)
2.3 Calculate the change in momentum of car A during the collision. (4)
2.4 Use Newton’s laws to explain why the momentum of car B decreases during the
collision. (3)
3. A man of mass 85 kg on roller skates, moving horizontally at a constant speed of
5 m.s–1 in a straight line, sees a child of mass 20 kg standing directly in his path. The
man grabs the child and they both continue moving forward at 2 m.s–1. The collision
between the man and the child lasts for 1,3 s.
3.1 Calculate the average net force acting on the man during the collision. (5)
3.2 What is the magnitude and direction of the average net force acting on the child
during the collision? (2)
4. A man of mass 80 kg wearing a seatbelt, is driving a car at 20 m.s–1 which collides
with the back of a stationary truck causing the car to bounce backwards at
2 m.s–1 after the collision. The collision lasts for 0,2 s. Calculate the average force of
the seatbelt on the man during the collision. (5)
Good luck!!
TOTAL: 43
The below table illustrates how learners will be graded!!
N.B: Please refer to this table when answering the questions
Assessment criteria 
Levels 

1(Adequate) 
2(moderate) 
3(substantial) 
4(Meritorious) 
5 (Outstanding) 

Strategic Approach (S) 
Little or no understanding of how to approach the problem. 
An invalid approach that demonstrates little understanding of the problem. 
Valid approach with multiple errors that impede understanding. 
Valid approach with minor errors that don’t disrupt understanding. 
The approach chosen is clearly shown, clearly written & all elements are valid. 
Physics Concepts (P) 
Little or no understanding of physics concepts. 
At least one concept was identified but unable to demonstrate understanding. 
Appropriate concepts identified, but not employed or understood. 
Appropriate concepts that are mostly understood but employed with errors.

Appropriate concepts that are fully understood (symmetries, conserved quantities, etc.), clearly stated & employed correctly 
Mathematical Concepts (M) 
Incorrect equations; demonstrates little or no understanding of mathematical concepts involved. 
Can identify at least one equation, but unable to apply them 
Correct starting equations. The mathematical steps are hard to follow and errors begin to impede application. 
Correct starting equations. All mathematical steps are clearly shown but minor errors yield the wrong answers. OR Correct starting equations with correct final result but the mathematical steps are hard to follow. 
Correct starting equations; All mathematical steps are clearly shown and they flow easily toward the correct answer 
Answer (A) 
No answer. 
Unable to reach a correct answer on this path. 
Incorrect answer, but on the right path. 
Correct answer analytically (IA), but not numerically (IA). 
100% correct answer – analytically (IA) numerically (IA) & conceptually (IA). 
Mark: /43 
REMARKS:
In this lesson, we covered:
For more understanding, you can read the following:
1. PHYSICAL SCIENCES GRADE 12 PHYSICS TERM II By’ (2020), pp. 0–74. 2. Objectives, L. et al. (no date) ‘Grade 12 Physical Sciences Lesson Plans Physical Sciences Frames of reference Lesson SUMMARY FOR : DATE STARTED : DATE COMPLETED : Term 1 Page 1 © Gauteng Department of Education (ver. 1 ) Grade 12 Physical Sciences Lesson Plans Term 1 Page 2’, pp. 1–141. 3. Support, T. (no date) ‘Interesting Science fact # 9 When Helium is cooled to almost absolute zero ( 460 ° F or 273 ° C ), the lowest temperature possible, it becomes a liquid with surprising properties : it flows against gravity and will start running up PHYSICAL Content Booklet : Targeted Support’. 4. The, B. (2016) Study Guide 12. More videos to watch:
Good luck!! If I have seen further than others, it is by standing upon the shoulders of giants "Isaac Newton" 
It gives me great pleasure to express my sincere appreciation to the following people:
1. PHYSICAL SCIENCES GRADE 12 PHYSICS TERM II By’ (2020), pp. 0–74.
2. Objectives, L. et al. (no date) ‘Grade 12 Physical Sciences Lesson Plans Physical Sciences Frames of reference Lesson SUMMARY FOR : DATE STARTED : DATE COMPLETED : Term 1 Page 1 © Gauteng Department of Education (ver. 1 ) Grade 12 Physical Sciences Lesson Plans Term 1 Page 2’, pp. 1–141.
3. Support, T. (no date) ‘Interesting Science fact # 9 When Helium is cooled to almost absolute zero ( 460 ° F or 273 ° C ), the lowest temperature possible, it becomes a liquid with surprising properties : it flows against gravity and will start running up PHYSICAL Content Booklet : Targeted Support’.
4. The, B. (2016) Study Guide 12.
https://www.youtube.com/watch?v=N5s6rbHdQVw
https://www.youtube.com/watch?v=Foi4zeWJTcs
https://www.youtube.com/watch?v=QAURZq43508&t=867s
Educator: Mr NB MAEBELA
Conduct: 0764259778
EMAIL: blessingmaebela@gmail.com
For more information, please contact me using the above details
Regards