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

Force, work, power and energy

Newton’s first law of motion

Newton’s second law of motion 

Weight of an object

Work done

Kinetic energy

Gravitational Potential energy

Gravitational potential energy to kinetic energy and vice versa

Power

Gravitational potential energy to kinetic energy and vice versa.

 

As we have seen in earlier post an object can have either gravitational potential energy or kinetic energy. However it is always possible for the object to have both kinetic energy and gravitational potential energy. Think of a plane flying at a certain height above the ground.

It is going to have kinetic energy due to its speed and gravitational potential energy due to its height.

Now what happens to a body that is either falling toward the ground or rising to a certain height. 

As you can see in fig 1 the object is rising and as a result its height is also rising, hence its potential energy is also increasing. Its kinetic energy is however decreasing. (see Principle of conservation of energy).

However the object in fig 2 is falling towards the ground. Since its height is decreasing, its gravitational potential energy must also be decreasing and its kinetic energy increasing.

Example 1

In this example a ball is released from rest from a height h.

If the ball is initially at rest then

v = 0  ms^-1 hence E_k = 0 J

However since the object is at a height h the E_p = mgh

When the ball is released it fall under the effect of the force of gravity, as a result it will accelerate downward and as a result the velocity of the object will increase  while the height of the object decreases.

Hence as the velocity decreases the kinetic energy decreases while the gravitational potential energy decreases as the height decreases.

However at all time    E_k +  E_p = Total energy and total energy is constant.

As the object reaches the ground the height becomes zero so does the gravitational potential energy while the kinetic energy reaches its maximum value. At this point all the gravitational potential energy would have been converted to kinetic energy.

When it reaches the ground h = 0 m hence E_p = 0 J

While E_k = 0.5 mv²

Example 1

A man of mass 64 kg jumps from a bridge 25 m high into a river.

(a) Calculate the gravitational potential energy of the man when he is on the bridge.

(b) What is his speed of entry into the water.

Now the man is on the bridge at a height of 25 m. It means that he has gravitational potential energy.

(a)  Gravitational potential energy E_p = mgh

                                                                = 64*9.81*25

                                                                = 15696J

                                                                 = 1.5 *10⁴J                                        

(b) When the person jumps the gravitational potential energy decreases as his height decreases. However as the person fall to the ground his speed increases and as a result the kinetic energy is completely converted to kinetic energy.

Hence what he reaches the river all the gravitational potential energy has been converted to kinetic energy.

Kinetic energy at surface of river =  1.5 *10⁴J

E_k = 0.5 mv²

1.5 *10⁴= 0.5 *64*v²

v = (1.5 *10⁴/0.5/64)^0.5

    = 21.65 m s^-1

=22 m s^-1

It is now time for a question. I will give you an answer to do. I will give the answer when some of you have given the answers.

Good luck.

A girl of mass 50 kg is trying to jump over a bar. She ran at a speed of and leaves the ground and successfully jumped over the bar.

(a) Calculate the kinetic energy that she has when she is running.

(b) Deduce the gravitational potential energy of the girl when she is at her maximum height.

(c) Calculate the height of the bar.

Kinetic energy

Kinetic energy is one if the eight forms of energy. It is simply the energy that a body possesses due to its motion.

The kinetic energy of a body is simply calculated using the equation

Kinetic energy Ek = ½ mv²

Where m is the mass of the mass of the object in kg
v is the velocity of the object in ms^-1

As you can see the kinetic energy of the object depends on both the velocity and the mass of the object. So it is possible for an elephant with a large mass to have the same kinetic energy as a small object moving at a high velocity.

Now let us see an example where the kinetic energy of an object is calculated.

Example 1

An elephant of mass _{3.00 x 10}³ kg is moving at a velocity of 3.00 ms^-1. What is the kinetic energy of the elephant?

So let us recall that the kinetic energy of the elephant is calculated using the equation

Ek = ½ mv²

Hence the kinetic energy Ek = ½ mv²
= ½(_{3.00* 10}³)*(3.00)²
= 13500 J
= _{1.35 *10}⁴ J

Now you can be given the kinetic energy and be asked to calculate the mass or the velocity. Let us look at a second example.

Example 2

An object has a mass of 4.0 kg and a kinetic energy of 16 J. Determine the velocity of the object.

Ek = ½ mv²
16 = ½ (4.0)*v²

Making v the subject of formula

v² = (16 *2)/4.0
v = (16*2/4.0)^½
v = 2.8 ms^-1

Now to test your newly acquired you can do the following questions

1. Calculate the kinetic energy possessed by an aeroplane of mass _{3.4 * 10}⁵ kg flying at 110 ms.
2. If a body has _{3.4 x 10}⁵ J of kinetic energy and mass _{3.2* 10}² kg, what is its velocity?

What is power?

Power is the rate of doing work.

From the definition we can deduce the following equation

Power = Work done /Time taken

The unit of power is the Watt (symbol W) or the Joule/second (J/s)

Example 1

A boy pushes a box and as a result does 120 J of work in 10 s. What is the power developed by the boy?

Power = Work done / time taken

= 120 /10

= 12 W or J/s

There is another definition for power that is often used. It is

Power is the rate of dissipation of energy or the rate of change of energy conversion.

Power = Energy dissipated / Time taken

Example 2

A girl climbs a staircase gaining 500 J of gravitational potential energy in 10 s.

What is the power developed by the girl?

Power = Energy conversion / Time taken

= 500 / 10

= 50 W or J/s

Example 3

During the boiling of some water 4000 J of heat energy is dissipated in the kettle’s heater in a time of 8 s. What is the power of the heater?

Power = Energy dissipated/Time taken

= 4000/8

= 500 W or J/s

It is now time to do some questions. It will give these questions after a few of you have supplied your answers.

1. A trains of mass 50000 kg accelerated form rest and reaches a velocity of 50 ms^-1 in 60 s.

(a) Calculate the kinetic energy gained by the train.

(b) Calculate the power of the train engine.

2. A lamp is rated 80 W. If it is switched on for two hours, how much light energy is dissipated.

Work done

Work done is a very important concept in Physics as it is used in fields, in deriving gravitational potential energy and kinetic energy, etc.

So what is work done?

Let us look at the diagram above. A force F acts on the box at A. During the time the force is acting, the box moves in the direction of the force. When the application of the force stopped the box has moved a distance d.

Using the definition

Work done is the product of the force acting on an object and the distance moved by the object in the direction of the force.

We can deduce that the equation to calculate work done is

Work done = Force * distance moved in the direction of the force

Work done = F *d

Let us have a look at an example.

Example

A man pulls a table by exerting a force of 100 N on it moving it by a distance of 3.9 m.

Calculate the work done by the man?

Work done = Force * distance moved in the direction of the force

Work done = F *d

= 100 * 3.9

= 3900 J

If you have understood the concept do the question below.

Question

A braking force of 2.3*10⁴ N is applied to a car and as a result the car stops in a distance of 23 m. Calculate the work done in stopping the train.

Good luck.