• Momentum
• Change in momentum
• Newton’s Second Law of Motion in terms of momentum
• Impulse
• The principle of conservation of linear momentum
• Problem types
• Elastic and inelastic collisions

Summary

Problem types:

1. Two objects collide and continue to move as separate objects after the collision:
   Σp_i = Σp_f
   m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f
2. Two objects collide and unite:
   Σp_i = Σp_f
   m₁v_1i + m₂v_2i = (m₁ + m₂)v_f
3. Two moving objects that are initially joined, then separate:
   Σp_i = Σp_f
   (m₁ + m₂)v1 = m₁v_1f  + m₂v_2f
4. Two stationary objects that are initially joined, separate (e.g. during an explosion):
   Σp_i = Σp_f
   (m₁ + m₂)
   v1 = m₁f_1f + m₂v_2f
   0 = m₁f_1f + m₂v_2f
    m₁f_1f = – m₂v_2f
5. An object falls vertically onto another object that is moving horizontally below it:
   Σpi = Σpf
   m₁v_1i+ m₂v_2i = (m₁ + m₂)vf
   m₁v_1i+ m₂(0) = (m₁ + m₂)vf
   m₁v_1i = (m₁ + m₂)vf

2.1 Momentum

Momentum is a vector quantity with the same direction as the object’s velocity.
You need to remember the differences between speed and velocity.
Speed

• Speed is the distance covered per unit time
• Speed is a scalar quantity (thus has magnitude and no direction).
• Symbol: v

Velocity

• Velocity is the rate at which an object is displaced.
• Velocity is a vector (thus has magnitude and direction).
• Symbol: v with an arrow on it

N.B The formula for momentum is: p = mv
where
p = momentum
m = mass
v = velocity

• mass is measured in kilograms (kg)
• velocity is measured in m⋅s^–1
• the unit of momentum is: kg⋅m⋅s^–1

The momentum of an object is defined as the product of its mass and velocity(p = mv)
a) Momentum is also directly proportional to the mass of the object: 

b) Momentum is directly proportional to the velocity of the object:

Conservation means to keep things (forces) the same (in a closed system).
Linear momentum refers to the momentum of objects in a straight line.
A closed system is a system that does not experience any external forces.

Examples of momentum:
The motion that is a result of objects colliding with each other, an object exploding, or a bullet being fired is described by momentum.
2. Impulse:
Impulse is the product of the net force acting on an object and the time that the force is applied to an object. (Impulse = FΔt). Think of the term “impulsive” or “having an impulse to do something”. This might help you to remember what it means.
3. Newton’s second law of motion in terms of momentum:
The net (resultant) force acting on an object is equal to the object’s rate of change of momentum. In a formula: Fnet = ^Δp/_Δt
4. The law of conservation of linear momentum:
The total linear momentum of an isolated (closed) system remains constant (is conserved).

2.2 Change in Momentum

When an object’s velocity changes in magnitude (size) or direction, its momentum will also change. Since an object's mass remains constant during a collision (assuming it does not break up or approach light speed), it follows that the change in its velocity is what causes a change in its momentum.
We only study objects moving in straight lines, for example, backwards and forwards, left and right or up and down.

Definitions and principles or laws have certain key words that should not be left out.
These are written in bold in each of the above definitions.

Steps to follow when solving problems

1.  Make a sketch (on your rough work page) of the situation.
2.  Always choose and indicate direction and write it down clearly. It is recommended that you choose a positive direction (e.g. to the right is positive).
3.  Write down the information in symbols. Remember to include the correct signs for the directions of the initial and final velocity.
4.  Choose the correct formula from the information sheet.
5.  Substitute the values into the formula.
6.  Solve for the unknown variable.

Worked example 1

1. A car has a momentum of 20 000 kg·m·s^–1. What will the car’s new momentum be if its mass is doubled (by adding more passengers and a greater load) and it travels at the same velocity?
2. What will the velocity be if the momentum is 60 000 kg·m·s^–1 and the mass of the car is 2 000 kg?
3. A truck has a mass of 6 000 kg and travels at 80 km·h^–1. How does the momentum change if the truck is loaded with 1 200 kg and then travels at 60 km·h^–1?

Solutions

1. The formula for momentum is p = mv , so the momentum will double and will be equal to 40 000 kg·m·s^–1 in the same direction as before.
2. p = mv
   60 000 = 2 000 × v
   ∴ v = 30 m·s^–1 in the same direction as the momentum.
3. Convert both velocities to m·s^–1:
   vi = 22,22 m·s^–1 & vf = 16,67 m·s^–1
   pi = m_iv_i = (6000 × 22,22) = 133 320 kg·m·s^–1
   pf = m_f v_f = (7200 × 16,67) = 120 024 kg·m·s^–1
   Δ p = p_f – p_I = 133 320 – 120 024 = 13296 kg·m·s^–1

Momentum is a vector quantity, so you must give the direction in all your answers.

2.3 Newton’s Second Law of Motion in Terms of Momentum

Newton’s Second Law of Motion can be used to find the object’s acceleration due to the net force, and the object’s change in momentum due to the net force.
We know that the object’s change in momentum is always:

• directly proportional to the net force acting on the object Δp ∝ Fnet
• directly proportional to the time that the net force acts on the object Δp ∝ Δt in the direction of the net force acting on the object.

Worked example 2
Why is it less painful for a high jumper to land on foam-rubber carpet than on the ground?

Solution
F_net = Δp
           Δt
The F_net needed to bring the jumper to rest (vf = 0 m·s⁻¹) depends on Δp and Δt. When he lands on the foam-rubber, he comes to rest over a longer period of time (Δt) than if he lands on the ground.

• So time taken Δt to change his momentum increases
• F_net decreases (F_net ∝  t )
                                       Δ1
• The magnitude of F_net determines the amount of pain experienced, so it is less painful to land on foam-rubber.

Worked example 3

1. A spaceship has a mass of 1 000 kg. The rocket engines discharge for 5 s and increase the rocket’s velocity from 25 to 30 m·s^–1. Calculate the force exerted by the engines to cause this change in momentum.
2. Assume the direction of the initial velocity is positive and the answer you obtain in the above problem is negative, what would be the direction of the exerted force?

Solutions

1. Let the direction of the initial velocity be positive.
   F_net = Δp = mv_f – mv_i = (1000)(30) – (1000)(25)
             Δt           Δt                           5
   = 1 000 N in the initial direction of motion.
2. The same i.e. in the initial direction of motion.

NOTE: The same reasoning explains why a cricket player would draw his hands back to catch a fast ball and why modern motor vehicles are designed with air bags and crumple zones.

NOTE:

• As the question is asking for a vector quantity (force) the answer must have both magnitude and direction.
• Since the answer is positive and it was decided the direction of the initial velocity is positive the direction of the force exerted is the same as that of the initial velocity.

Newton’s Second Law of Motion states that: The resultant/net force acting on object is equal to the rate of change of momentum and the change is in the direction of the resultant/net force.

Activity 1
Study the diagrams below showing the movement of a 150 g baseball thrown at a wall at right angles.

Diagram A:
m = 150 ÷ 1000 = 0,15 kg
vi = + 18 m·s⁻¹
vf = −12 m·s⁻¹
Diagram B:
m = 0,15 kg
vi = + 18 m·s⁻¹
vf = 0 m·s⁻¹

Let the direction towards the wall be positive.

1. Calculate the momentum of the baseball as it strikes the wall in Diagram A. (3)
2. Calculate the momentum of the baseball as it leaves the wall in Diagram A. (3)
3. Calculate the change in momentum during the collision in Diagram A. (3)
4. Calculate the force exerted by the wall on the baseball in Diagram A and in Diagram B, if each collision lasts 0,1 s. (6)
5. Draw a vector diagram to illustrate the relationship between the initial momentum (pi), the final momentum (pf) and the change in momentum (Δp = pf – pi) for the baseball in Diagram A. (6)
   [21]
   Convert mass to kg!

2.4 Impulse

Impulse is another way to define momentum. Impulse is a measure of the amount of force applied to an object, for a certain period of time. Think of it as a measure of the shock experienced by an object when another object collides with it.
The formula for Impulse is: Impulse = FΔt where F is force in newtons N Δt is change in time in seconds Impulse and momentum are in fact the same thing. We can show this by dimensional analysis, that is, by working out what the units of impulse are, and comparing the units to the units of momentum.
F = ma
∴ a = F 
         m
a = Δs  
       t²
∴ F = Δs  × m
          t²
Impulse = Δt × m × Δt
                 t²
Impulse = Δs × m
                 t
p = mv = vm
p = Δs × m
       t
∴ p = Impulse

Activity 2
A cricket ball of mass 175 g is thrown horizontally towards a player at 12 m·s⁻¹. It is hit back in the opposite direction with a velocity of 30 m·s⁻¹.
The ball is in contact with the bat for a period of 0,05 s. Calculate:

1. The impulse of the ball. (4)
2. The force exerted on the ball by the bat. (3)
3. The force exerted on the bat by the ball. Motivate your answer by referring to a Law of Motion. (5)
   [12]

2.5 The Principle of Conservation of Linear Momentum

Steps for solving problems on conservation of linear momentum
Step 1. Choose a direction as positive.
Step 2. Sketch the situation – draw a block to represent each object.
Step 3. Write down the equation for the Conservation of Momentum: Ʃp i = Ʃpf
Step 4. Expand this equation according to the type of collision.
Step 5. Substitute the known values into the equation. Remember to check the direction of the objects’ velocities and to use the correct signs for the directions.
Step 6. Calculate the answer.
Step 7. Write the answer, include units and indicate the direction.

Very Important

• Always remember to include units in your answer
• Remember that the +/- signs represent direction

Remember:
1N·s = 1 kg·m·s^–1 1 newton second
= 1 kilogram metre per second
The principle of conservation of linear momentum states that: The total linear momentum in a closed system remains constant (is conserved)

We can solve problems about the conservation of linear momentum according to the nature of the collision or separation (explosion) of the objects involved. We usually solve problems in which two objects are involved.

2.6 Problem Types

1. Two objects collide and continue to move as separate objects after the collision:
   Σ_pi = Σp_f
   m₁v_1i + m₂ v_2i = m₁v_1f + m₂v _2f
2. Two objects collide and unite:
   Σ_pi = Σp_f
   m₁v_1i + m₂
   v2i = (m₁ + m₂)vf
3. Two moving objects that are initially joined, then separate:
   Σ_pi = Σp_f
   (m₁ + m₂)v1 = m₁v_1f + m₂v_2f
4. Two stationary objects that are initially joined, separate (e.g. during an explosion):
   Σ_pi = Σp_f
   (m₁ + m₂)v1 = m₁v_1f + m₂v_2f
   0 = m₁v_1f+ m₂v_2f
   m₁v_1f = –m₂v_2f
5. An object falls vertically onto another object that is moving horizontally below it:
   Σ_pi = Σp_f
   m₁v_1i+ m₂v_2i = (m₁ + m₂)vf
   m₁v_1i + m2(0) = (m₁ + m₂)vf
   m₁v_1i = (m₁ + m₂)vf

This looks scary, but it’s not really! It’s saying that the sum of the momenta remains the same, that is, before collision and after collision, the total momentum before and after is the same. So, Σpi is the sum of all the initial momenta.
Σpf is the sum of all final momenta. To calculate the initial momentum sum, just add up the momenta of all the objects.

Problem Type 1: Two objects collide and continue to move as separate objects after the collision
Activity 3

In a railway shunting yard, a locomotive (train engine) of mass 4 000 kg, travels due east at a velocity of 1,5 m·s⁻¹. The train driver tries to link it to a stationary wagon of mass 3 000 kg by letting them collide. Instead, the wagon moves due east with a velocity of 2,8 m·s⁻¹.
Calculate the magnitude and direction of the velocity of the locomotive immediately after the collision. (11)

Remember: If an object is stationary, its velocity is 0 m·s^-1; "stationary" means "not moving".

Problem Type 2: Two objects collide and unite
Activity 4
A boy of mass 40 kg runs at 5 m·s–1 east and jumps onto a skateboard of mass 2 kg moving at 3 m·s–1 east.
Calculate the speed at which the boy and skateboard move together. (7)

Problem Type 3: Two moving objects that are initially joined, then separate
Activity 5
Hendrik is an amateur rocket builder. He launches a two-stage rocket as shown in the diagram. Section A (stage 1) contains the rocket engine and fuel. Section B (stage 2) has a mass of 2 kg.

1. Hendrik says that Newton’s Third Law of Motion is used to explain why the rocket moves upwards during flight. Identify one actionreaction pair of forces involved with the rocket’s motion. (1)
   At a certain height, when the rocket has a velocity of 5 m·s^–1 upwards, the last fuel is used up, and section A has a mass of 3 kg.
   To get section B even higher, a small explosion separates section B from section A at this point and increases the upwards velocity of section B to 8 m·s^–1.
2. State the Law of Conservation of Linear Momentum in words. (4)
3. Calculate the velocity of section A after the explosion. (11)
   
   [16]

Remember: The question asks for the speed of the skateboard. Speed is a scalar, so you don’t need the direction of the motion in the answer.

Problem Type 4: Two stationary objects that are initially joined, and then separated (e.g. during an explosion)
When two objects are forced apart by an explosion or as a result of a compressed spiral spring released between them, they move in opposite directions after the explosion e.g. when a gun fires a bullet, the bullet moves forwards and the gun moves backwards.

Activity 6
A gun of mass 1 kg is attached to a trolley of mass 4 kg and is loaded with a bullet of mass 2 g. The system is at rest on a frictionless horizontal surface. The gun is fired by remote control and the bullet has a muzzle velocity of 350 m·s⁻¹.
Calculate the velocity of the trolley and gun after the bullet has been fired. (8)

Problem Type 5: An object falls vertically onto another object that is moving horizontally below it

Activity 7
A trolley of mass 3 kg moves at 4 m·s⁻¹, west along a frictionless horizontal path. A brick of mass 1 kg drops vertically onto it. The brick lands on the trolley at a vertical velocity of 0,5 m·s⁻¹. Calculate the velocity of the brick and trolley system after the collision. [13]

2.7 Elastic and Inelastic Collisions

2.7.1 Revision
Linear momentum is always conserved in a closed system. Kinetic energy is, however, always conserved, and is often transformed into other forms of energy, like heat and sound, or potential energy.
2.7.2 Differentiating between elastic and inelastic collisions
Elastic Collisions:

• linear momentum is conserved
• colliding objects remain separate and are not changed in any way
• total kinetic energy is conserved: ΣE_ki = ΣE_kf
• E_k before collision = E_k after collision
• the initial kinetic energy is not transformed into any other forms of energy.

Inelastic Collisions:

• linear momentum is conserved
• colliding objects are joined or change their shapes
• total kinetic energy is not conserved:
  ΣE_ki > ΣE_kf
• Ek before collision > E_k after collision
• some of the initial kinetic energy is transformed into other forms of energy e.g. heat, light, sound.

Remember that for objects moving much below the speed of light (e.g. bullets, trains, people, bricks),
E = ½ mv² thus, if p = mv, then E =½ pv

Steps for solving problems on elastic and inelastic collisions
Step 1. Calculate the sum of the kinetic energies of all the objects before the collision
ƩE_ki =½ m₁v_1i 2 + ½ m₂v_2i
Step 2. Calculate the sum of the kinetic energies of all the objects after the collision
Ʃ Ekf = ½ m₁v_1f 2 + ½ m₂v_2f
Step 3. Compare the total kinetic energy of the system before the collision to the total kinetic energy of the system after the collision.
Step 4. If ƩE_ki= E_kf ∴ ƩE_k before the colliison = ƩE_k after the collision therefore the collision was elastic
If E_ki≠ ƩE_kf ∴ ƩE_k before the collision ≠ ƩE_k after the collision therefore the collision was inelastic

Collisions are classified as either elastic collisions or as inelastic collisions.

Activity 8
Collisions between vehicles take place on the roads in our country daily. In one of these collisions, a car of mass 1 600 kg, travelling at a speed of 30 m·s^–1 to the left, collides head-on with a minibus of mass 3 000 kg, travelling at 20 m·s^–1 to the right. The two vehicles move together as a unit in a straight line after the collision.
1. Calculate the velocity of the two vehicles after the collision. (8)
2. Do the necessary calculations to show that the collision was inelastic. (13)
3. New cars have a crumple zone to help minimise injuries during accidents. Air bags and padded interiors can also help to reduce the chance of death or serious injury. Use principles in Physics to explain how crumple zones and air bags can reduce the chance of death or injury. (9)
[30]

Activity 9
A bullet of mass 10 g, moving at a velocity of 300 m·s^–1, strikes a wooden block of mass 1,99 kg resting on a flat horizontal surface as shown in the diagram below. The bullet becomes embedded in the block. Ignore the effects of air friction.

1. Write down in words the principle of conservation of linear momentum. (2)
2. Calculate the speed of the block-bullet system immediately after the collision. (4)
3. Is this collision elastic or inelastic? Give a reason for the answer. (2)
   The floor exerts a constant frictional force of 8 N on the block-bullet system as it comes to rest.
4. Calculate the distance that the block-bullet system moves after the collision. (5)
   [13]