Conservation of kinetic energy in collisions

As we have seen in this post during a collision between two object the velocity of the two colliding objects changes. Hence you would agree that he kinetic energy of the objects would change.

We have also seen in this post that in all collisions the sum of linear momentum is a constant. This is the principle of conservation of linear momentum.  However as we are going to see in some collisions, elastic collisions, the sum of kinetic energy is also constant. That is the sum of the kinetic energies of the objects before the collision is the same as the sum of the kinetic energies after the collision.

We are now going to see an example of how to use this “principle” which i am going to call the “principle of conservation of kinetic energy”.

Fig 1 below shows two objects travelling towards each other and fig 2 shows the two objects separating after the two objects have separated.

So if the principle of conservation of kinetic energy applies then it means that Sum of kinetic energy before collision = sum of kinetic energy after collision

Fig 1 Before collision

Before the collision

Sum of kinetic energy before collision = kinetic energy of object 1 + kinetic energy of object 2

= ¹/₂m₁u₁² + ¹/₂m₂v₁²

Fig 2 After collision

 

After the collision

Sum of kinetic energy after collision = kinetic energy of object 1 + kinetic energy of object 2

= ¹/₂m₁u₂² + ¹/₂m₂v₂²

If you refer back to the principle then

Sum of kinetic energy before collision = sum of kinetic energy after collision

 ¹/₂m₁u₁² + ¹/₂m₂v₁²   =   ¹/₂m₁u₂² + ¹/₂m₂v₂²

Example

Two balls of mass 1.0 kg are travelling towards each other at a speed of 5.0 m/s towards the right (ball 1)  and 5.0 m/s towards the left (ball 2) respectively. If after separation one of them moves moves off a speed of 5 m/s towards the right (ball 2 ) calculate the speed of the other ball.

Sum of kinetic energy before collision = sum of kinetic energy after collision

 ¹/₂m₁u₁² + ¹/₂m₂v₁²   =   ¹/₂m₁u₂² + ¹/₂m₂v₂²

 ¹/₂*1.0*5.0² + ¹/₂*1.0*(-5.0)²   =   ¹/₂*1.0*u₂² + ¹/₂*1.0*5.0²

Solving for u₂ the velocity of object 1 after the collision

12.5+ 12.5 = 0.5*u₂²+12.5

12.5= 0.5*u₂²

12.5 /0.5 = 25 =u₂²

u₂² = 25

There can be two solution to u₂

First u₂=-5 m/s and secondly u₂= 5 m/s

Since the second ball is already travelling towards the right and according to the principle of conservation of linear momentum then the only solution is

u₂=-5 m/s

Conservation of linear momentum in collisions

As we have seen in this previous post, in a close system the sum of linear momentum is a constant. Hence in any collision if the sum of linear momentum of the objects before the collision would be equal to the sum of linear momentum after the collision.

Fig 1 below shows two spherical objects travelling toward each other. The mass of the two objects are 1.0 kg each. Object 1 is travelling at a velocity of 5.0 m/s  and object 2 is travelling at a velocity of –5.0 m/s.

Fig 1

If  p_1b is the linear momentum of object 1 before the collision and p_2b

is the linear momentum of  object 2 before the collision then

the sum of linear momentum = p_1b +p_2b

= m₁u₁ + m₂v₁

=1.0*5.0 + 1.0*(-5.0)

= 5.0 - 5.0

=0 Kg m/s

As you can see before the collision the sum of the linear momentum is 0 kg m/s. According to the principle of conservation of linear momentum the sum of linear momentum at all time will thus be 0 kg m/s before and after the collision.

Fig 2 below shows the spherical objects after the collision while they are separating. According to the principle of conservation of linear momentum after the collision the sum of linear will thus be 0 kg m/s.

Question

If after the collision the two objects separate such that object 2 moves with a velocity of  5.0 m/s, calculate the velocity of object 1.

Fig 2

If  p_1a is the linear momentum of object 1 after the collision and p_2a

is the linear momentum of  object 2 after the collision then

the sum of linear momentum = p_1a +p_2a

= m₁u₂ + m₂v₂

=1.0*u₂  + 1.0 *5.0

= u₂  +5.0

Now if we use the principle of  of conservation of linear momentum

sum of linear momentum before the collision = sum of linear momentum before the collision

0 =   u₂  +5.0

u₂  = - 5.0 m/s

Newton’s second law of motion

We have seen in a previous post the Newton’s first law of motion.

Today we are going to see the second law of motion.

From our discussion in the first law of motion and in the post on acceleration when a force acts on an object either velocity increases due to a change in speed or the velocity changes due to a change in direction. Remember velocity is a vector quantity hence it will change if either the direction or the magnitude changes.

Hence we can deduce that if a force acts on an object then the velocity of the object changes.The more force is applied the greater the change in velocity. This must mean that if the mass remain constant then the momentum of the object changes. And hence the more force is applied the more the linear momentum will change.

Let us look at the situation below:

A force F acts on an object of mass m for a time t such that before the application of the force the object has a velocity v₁ hence a linear momentum of p₁ and after the force is applied the velocity of the object is v₂ hence the linear momentum is p₂.

Before the force is applied velocity = v₁ and linear momentum = p₁

After the force is applied velocity =va₂  and linear momentum is p₂.

Now the Newton’s second law of motion states that the force applied is directly proportional to the rate of change of linear momentum of the object.

If = a then

If k = 1

Then this will reduce the equation to

F= ma

This where the famous equation F = ma come from. This why it is also considered as the Newton’s second law of motion

Newton’s first law of motion

In order for an object to accelerate it needs to be acted upon by a force. As we have seen in an earlier post on acceleration.

From the post on acceleration we can deduce that there are four changes that can take place if a force acts on an objects:

1. The speed of the object will increase from zero if it was at rest and the object will move in the direction in which the force is acting.
2. The speed of the object will increase if the force is acting in the same direction as the the direction of motion of the object until the force no longer acts on the object.
3. The speed of the object will decrease if the force is acting in a direction opposite to the direction of motion of the object. The speed of the object might decrease to zero if the force acts for long enough.
4. If the object is moving and the force is acting in a direction perpendicular to the direction of motion of the object then the object will move in a circular path. When the application of the force stops the object will leave the circular path and would move in a straight line in a direction tangential to the circular path.

These four changes can be combined together to produce the Newton’s first law of motion as shown below:

Unless a force is applied an object will continue in its state of rest or uniform motion in a straight line.

Another way to write this law:

If a force is applied (i) an objects that was at rest will no longer be at rest or (ii)if moving with a uniform  will no longer have a uniform or constant speed or lastly (iii) if moving in a straight line will no longer move in  the same direction

This law is thus explaining that if a force is acting on an object then the object will experience an acceleration. However we are officially learn this only when we see the Newton’s second law of motion.

What is a collisions?

A collision is a phenomenon that can be described in three distinct steps.

1. Firstly the two objects that would collide must approach each other. The two objects may be travelling towards each other as shown below in fig 1. They may also be moving in the same direction as in Fig 2 but if the first one is slower than the second one they would meet at a certain time.

Fig 1 Two objects moving towards each other directly

 

Fig 2 Two objects moving in the same direction but about moving closer to each other.

2. The two objects that are travelling towards each other must meet. That is they make contact. During this contact they interact by exerting forces on each other. The forces that they exerts on each other must be equal to each other according to Newton’s third law of motion.

Fig 3 The two objects make contact and exert forces on each other

3. The two objects then separate and move away from each other in opposite as shown in fig 4 or moves in the same direction but  with increasing distance of separation as shown in fig 5.

Fig 4 The two objects move away from each other

Fig 5 Two objects move in the same direction but the distance between them increases with time

 

Homogeneity of equations

As we have seen in the chapter on derived units some quantities have units that is composed of two or more base units.

The physics students will meet with different equations. Now some of these equations are correct while some of them are not.

The test to check whether an equation is correct is through the homogeneity test as shown in fig 1 below. 

Fig 1

So the question is how to check whether an equation is homogenous?

Example 1

Check whether the following equation is homogenous?

P =hdg

Where p is pressure. h is height, d is density and g is the acceleration due to gravity.

As you can see the equation has two sides. The left-hand side and the right hand side.

Now you  have to check that both side of the equation has the same unit.

Pressure = Force / Surface area

Unit of pressure = Unit of force/ unit of area

                                    = kg m s^-2 / m²

                                    = kg m^-1 s^-2

unit of hdg = unit of h* unit of d* unit of g

                       = m * kg m^-3 * m s^-2

                        = kg m^-1 s^-2

Since both side of the equation has the same unit then the equation is homogenous.

Please note that if their is a dimensionless constant in the equation like k, 1/2 or something like that then the dimensionless constant, by definition, has no unit and as a result will not be involved when the test for homogeneity is performed.

Example 2

Check whether the following equation is homogenous.

v² = u² + 2as

In this case the equation is composed of three groups of quantities: v² , u² and 2as. You should check that the three groups of quantities have the same unit.

If the three group have the same unit the equation is homogenous otherwise it is not homogenous.

Note: Please see chan’s comment below for a more detailed  answer to example 2

What are atoms?

Atoms are the basic constituent of matter. Hence matter is made of atoms arranged in a particular way . Of course there are different types of atoms and the type of atoms and the type of arrangement will determine the type of matter.

We are now going to see a simple model of the atom and what are the different constituents of the atom.

Constituents

The atom is composed of three basic constituents.

1. The proton
2. The neutron
3. The electron

All the different types of atoms will have these three constituent particles. However the number of each will differ from one type of atom to another. We can thus deduce that it is the combination of these three constituent particles that determine the nature of the atom.

However in neutral atoms the number of proton is the same as the number of electron. If the two differs it means that the atom is charged. If there are more protons that electrons then the atom is positively charged and if the number of electron is greater than the number of proton then it means that the atom is negatively charged.

 

The table below lists the properties of the different constituent particles of the atom.

 

Property	Proton	Neutron	Electron
Mass	1.67 x 10–27 kg	1.67 x 10–27 kg	9.10 x 10-31 kg
	The proton and the neutron is on average 10000 times more massive that the electron
charge	+1.60 x1019 C	No charge This is because as you will see later on in radioactivity it is composed of one proton and an electron. The two opposite charges thus cancel each other	-1.60 x10-19 C
Location	In the nucleus	In the nucleus	In orbits around the nucleus

As we have seen in the table above the neutron and the proton are present in the nucleus while the electron is present in orbits around the nucleus (They are in reality different  but we assume that they are planet-like in this model).

A model of the atom is shown below in fig 1

Fig 1

This model is a mere representation of the atom. You will learn later on how the electrons would be arranged if there are more than four. The size of the nucleus is also quite small. So small that its diameter is 10000 times smaller than the radius of the atom. Thus to represent it truly on fig 1 a tiny dot would have been more appropriate.