CSEC Chemistry
Teacher: Ms. Kristi Mohammed
Miss Shazara Kristi Mohammed
Newton’s Laws of Motion
SECTION A - DYNAMICS
Newton’s Laws of Motion
Students should be able to:
1) State Newton’s Three Laws of Motion.
2) Use Newton’s Laws to explain dynamic systems.
3) Define the term Linear Momentum.
4) Apply the Law of Conservation of Momentum to Elastic & Inelastic
Collisions.
Newton’s Laws of Motion
Sir Isaac Newton published three laws in the 17th century:
 Newton’s First Law of Motion is an introduction to motion of the object and the force acting on it.
 Newton’s Second Law of Motion pertains to the behavior of objects for which all existing forces are unbalanced.
 Newton’s Third Law of Motion signifies a particular symmetry in nature: forces always occur in pairs, and one body
cannot exert a force on another without experiencing a force itself.
Newton’s Laws of Motion
Newton’s First Laws of Motion
Newton’s First Law of Motion states that a body remains in the state of rest or uniform motion in a straight line unless
and until an external force acts on it.
Putting Newton’s 1st law of motion in simple words, a body will not start moving until and unless an external force acts
on it. Once it is set in motion, it will not stop or change its velocity until and unless some force acts upon it once more.
The first law of motion is sometimes also known as the law of inertia.
Newton’s First Laws of Motion
There are two conditions on which the 1st law of motion is dependent:
Objects at rest: When an object is at rest velocity (v= 0) and acceleration (a = 0) are zero. Therefore, the object continues
to be at rest.
Objects in motion: When an object is in motion, velocity is not equal to zero (v ≠ 0) while acceleration (a = 0) is equal to
zero. Therefore, the object will continue to be in motion with constant velocity and in the same direction.
Newton’s Third Laws of Motion
Newton’s Third Law of Motion states that “When one body exerts a force on the other body, the first body experiences a
force which is equal in magnitude in the opposite direction of the force which is exerted”.
The above statement means that in every interaction, there is a pair of forces acting on the interacting objects. The
magnitude of the forces are equal and the direction of the force on the first object is opposite to the direction of the force
on the second object.
Newton’s Second Laws of Motion
Newton’s Second Law of Motion states that the acceleration of an object depends upon two variables – the net force
acting on the object and the mass of the object. The force applied to a body is directly proportional to its mass and its
resultant acceleration.
𝐹𝑜𝑟𝑐𝑒 = 𝑀𝑎𝑠𝑠 × 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝐹 = 𝑚𝑎
Where: F – Force (N)
m – Mass (kg)
a – Acceleration due to gravity (ms-2)
Momentum
Newton’s Second Law of Motion also states that the rate of change of momentum of a body is proportional to the applied
force and takes place in the direction of the force.
A sustained force, however, results in change of momentum.
Momentum can be defined as the product of the mass of an object and its velocity. Momentum is a vector quantity.
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑀𝑎𝑠𝑠 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑝 = 𝑚𝑣
Where: p – Momentum (kgms-1)
m – Mass (kg)
v – Velocity (ms-1)
Impulse
The Impulse an object experiences is equal to the change of Momentum. Impulse can be defined as the product of the
force exerted by an object and the time taken to exert this force.
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚
1
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑉
(2)
𝐼𝑚𝑝𝑙𝑢𝑠𝑒 = 𝐹𝑡
(3)
Substituting (1) and (3) into (2), we obtain:
𝐹𝑡 = 𝑚𝑣
Making Force the subject of the formula, we get the following equation:
𝐹=
𝑚𝑣
𝑡
OBJECTIVE 1.1
Particulate Theory of Matter
Students should be able to:
1) Define the term ‘Matter’.
2) State the four (4) main ideas behind the Particulate Theory of Matter.
3) Show evidence to support the Particulate Theory of Matter.
4) Describe and explain experiments showing Diffusion and Osmosis.
5) Use of salt or Questions
sugar to control garden pests and as a preservative.
Practice
Example 1: Momentum & Force
A free runner with a mass of 50 kg, jumped off a building and landed on the ground with a speed of 15 ms -1 in a time of 0.1 s.
(a) Calculate the momentum at the end of his fall.
(b) Calculate the force which he impacts the ground with.
Hint: (i) State your data. Check to make sure no conversions are necessary.
(ii) State and equation for Momentum and Force based on the data provided in the question.
(iii) Substitute the data into the formula. Ensure to state the magnitude & unit.
Data: m = 50 kg
v = 15 ms-1
Equation for Momentum:
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑀𝑎𝑠𝑠 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
t = 0.1 s
p=?
kgms-1
Equation for Impulse:
𝑝 = 𝑚𝑣
𝑚𝑣
𝐹=
𝑡
Example 1: Momentum & Force
Solution: (a) Calculate the momentum at the end of his fall.
Equation for Momentum:
𝑝 = 𝑚𝑣
Step 1: Substitute the data into the equation for Momentum.
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑀𝑎𝑠𝑠 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
Data: m = 50 kg
v = 15 ms-1
t = 0.1 s
𝑝 = 𝑚𝑣
𝑝 = (50)(15)
𝑝 = 750 𝑘𝑔𝑚𝑠 −1
p = ? kgms-1
(b) Calculate the force which he impacts the ground with.
Equation for Impulse:
𝑚𝑣
𝐹=
𝑡
Example 1: Momentum & Force
Solution: (b) Calculate the force which he impacts the ground with.
Equation for Momentum:
𝑝 = 𝑚𝑣
Data: m = 50 kg
Step 1: Substitute the data into the equation below to determine the Force, F.
𝐹=
𝑚𝑣
𝑡
v = 15 ms-1
t = 0.1 s
50 15
𝐹=
0.1
p = 750 kgms-1
F=?N
𝐹 = 7500 𝑁
Example 2: Momentum & Force
A boy catches a ball of mass 0.14 kg moving with a velocity of 20 ms -1. Calculate the following:
(a) The momentum of the ball.
(b) The force used to stop the ball in: I. A time of 0.5 s,
II. A time of 0.01 s.
Hint: (i) State your data. Check to make sure no conversions are necessary.
(ii) State and equation for Momentum and Force based on the data provided in the question.
(iii) Substitute the data into the formula. Ensure to state the magnitude & unit.
Data: m = 0.14 kg
v = 20
ms-1
Equation for Momentum:
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑀𝑎𝑠𝑠 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑝 = 𝑚𝑣
p = ? kg ms-1
Equation for Impulse:
𝑚𝑣
𝐹=
𝑡
Example 2: Momentum & Force
Solution: (a) Calculate the momentum of the ball.
Equation for Momentum:
𝑝 = 𝑚𝑣
Step 1: Substitute the data into the equation for Momentum.
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑀𝑎𝑠𝑠 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
Data: m = 0.14 kg
𝑝 = 𝑚𝑣
v = 20 ms-1
𝑝 = (0.14)(20)
p = ? kg ms-1
𝑝 = 2.8 𝑘𝑔𝑚𝑠 −1
Example 2: Momentum & Force
Solution: (b) The force used to stop the ball in: I. A time of 0.5 s,
Equation for Momentum:
𝑝 = 𝑚𝑣
𝐹=
𝐹=
Data: m = 0.14 kg
v = 20
ms-1
𝑚𝑣
𝑡
(0.14) 20
0.5
𝐹 = 5.6 𝑁
𝑡1 = 0.5 s
𝑡2 = 0.01 s
p = 2.8 kgms-1
F=?N
Solution: (b) The force used to stop the ball in: I. A time of 0.01 s
𝑚𝑣
𝐹=
𝑡
(0.14) 20
𝐹=
0.01
𝐹 = 2800 𝑁
Law of Conservation of Momentum
The law of conservation of linear momentum states that, in the absence of external forces, the total momentum of a system
of bodies is constant.
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚3 𝑣3
Example 3: Velocity after the collision, 𝑣3
A toy van with mass 500 g, moving with a velocity of 0.25 ms-1 collides with a stationary toy truck with a mass of 700 g.
Determine the velocity after collision assuming the wreck stays together as one.
Hint: (i) State your data. Check to make sure no conversions are necessary.
(ii) State the law of Conservation of Momentum.
(iii) Substitute the data into the formula, solving for the unknown. Ensure to state the magnitude & unit.
Data: m1 = 500 g
v1 = 0.25
ms-1
m2 = 700 g
v2 = 0 ms-1 [stationary]
m3 = m1 + m2
The Law of Conservation of Momentum states that:
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚3 𝑣3
Example 3: Velocity after the collision, 𝑣3
Solution: Calculate Velocity after the collision, 𝑣3 .
The Law of Conservation of
Step 1: Substitute the data into the equation and solve for 𝑣3 .
Momentum:
The Law of Conservation of Momentum states that:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚3 𝑣3
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
Data: m1 = 500 g
v1 = 0.25 ms-1
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚3 𝑣3
500 0.25 + 700 0 = (500 + 700)𝑣3
m2 = 700 g
125
𝑚𝑠 −1
1200
v2 = 0 ms-1 [stationary]
𝑣3 =
m3 = m1 + m2
𝑣3 = 0.104 𝑚𝑠 −1
Elastic & Inelastic Collisions
There are two types of collisions:
 Elastic Collisions.
 Inelastic Collisions.
Example 4: Elastic Collision
Two balls, moving with velocities, 0.7 ms-1 and 0.01 ms-1 respectively, collide with one another. After the collision the first ball
remains stationary, whereas the second ball continues moving with an unknown velocity. Determine the velocity, 𝑣3 , that the
second ball moves with after the collision. Assume that both balls have equal mass of 160 g.
Hint: (i) State your data. Check to make sure no conversions are necessary.
(ii) State the law of Conservation of Momentum.
(iii) Substitute the data into the formula, solving for the unknown. Ensure to state the magnitude & unit.
Data: m1 = 160 g
v1 = 0.7 ms-1
m2 = 160 g
v2 = 0.01 ms-1
v4 = 0 ms-1 [The first ball becomes stationary after collision]
v3 = ? ms-1
Example 4: Elastic Collision
Solution: Calculate Velocity after the collision, 𝑣3 .
Data: m1 = 160 g
v1 = 0.7 ms-1
m2 = 160 g
v2 = 0.01 ms-1
Step 1: Determine an appropriate equation to use for this situation.
The Law of Conservation of Momentum states that:
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑚1 𝑣1 + 𝑚2 𝑣2 (1)
v4 = 0 ms-1
v3 = ?
ms-1
This is an elastic collision, and it was stated that the first ball becomes stationary
after the collision whereas the second ball continues moving.
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 𝑚2 𝑣3 + 𝑚1 𝑣4
(2)
Substituting equations (1) and (2) into (3):
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣4 + 𝑚2 𝑣3
Example 4: Elastic Collision
Solution: Calculate Velocity after the collision, 𝑣3 .
Data: m1 = 160 g
v1 = 0.7 ms-1
m2 = 160 g
v2 = 0.01 ms-1
Step 2: Substitute the data into the equation to solve for the unknown.
The Law of Conservation of Momentum states that:
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣4 + 𝑚2 𝑣3
v4 = 0 ms-1
v3 = ? ms-1
160 0.70 + 160 0.01 = 160 0 + (160)𝑣3
112 + 1.6 = 160 𝑣3
𝑣3 =
113.6
𝑚𝑠 −1
160
𝑣3 = 0.71 𝑚𝑠 −1
Example 5: Inelastic Collision
Two toy cars, moves towards one another with a velocity of 3.1 ms-1 and 1.1 ms-1 respectively. The first car has a mass of 650 g
and the second car has a mass of 1000 g respectively. When both cars collide, they stick together. Determine the velocity after
collision and the direction in which the cars would move.
Data: m1 = 650 g
v1 = 3.1 ms-1 [The positive sign indicates that it to the right]
m2 = 1000 g
v2 = - 1.1 ms-1 [The negative sign indicates that it moves in the opposite direction, left]
Example 5: Inelastic Collision
Solution: Calculate Velocity after the collision, 𝑣3 .
Data: m1 = 650 g
v1 = 3.1 ms-1
m2 = 1000 g
Step 1: Determine an appropriate equation to use for this situation.
The Law of Conservation of Momentum states that:
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
v2 = - 1.1 ms-1
m3 = m1 + m2
v3 = ? ms-1
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑚1 𝑣1 + 𝑚2 𝑣2 (1)
This is an inelastic collision, both objects stick together.
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 𝑚3 𝑣3
Substituting equations (1) and (2) into (3):
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚3 𝑣3
(2)
Example 5: Inelastic Collision
Solution: Calculate Velocity after the collision, 𝑣3 .
Data: m1 = 650 g
v1 = 3.1 ms-1
Step 2: Substitute the data into the equation to solve for the unknown.
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
m2 = 1000 g
v2 = - 1.1
ms-1
m3 = m1 + m2
v3 = ? ms-1
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚3 𝑣3
650 3.1 + 1000 −1.1 = (650 + 1000)𝑣3
2015 − 1100 = (1650)𝑣3
915
𝑣3 =
𝑚𝑠 −1
1650
𝑣3 = 0.55 𝑚𝑠 −1
The toy cars move with a velocity of 0.55 ms-1 to the right.
OBJECTIVE 1.1
Particulate Theory of Matter
Students should be able to:
1) Define the term ‘Matter’.
2) State the four (4) main ideas behind the Particulate Theory of Matter.
3) Show evidence to support the Particulate Theory of Matter.
4) Describe and explain experiments showing Diffusion and Osmosis.
5) Use of salt or sugar
to control garden pests and as a preservative.
Exam-Style
Question
Question 1
Question 1
Question 2
Question 3
OBJECTIVE 1.1
Particulate Theory of Matter
Students should be able to:
1) Define the term ‘Matter’.
2) State the four (4) main ideas behind the Particulate Theory of Matter.
3) Show evidence to support the Particulate Theory of Matter.
4) Describe and explain experiments showing Diffusion and Osmosis.
Answers
5) Use of salt or sugar to control garden pests and as a preservative.
Question 1 Solutions
Question 2 Solutions
Question 3 Solutions
CSEC Physics
End of Lecture