Chapter 4.3 - Law of Conservation of Energy

In the previous section we saw basic details about potential energy. In this section we will see the Law of conservation of Energy.

In our day to day life, we see different sources of energy. Let us see some examples:
Example 1: An ordinary torch cell is a source of energy. In it, energy is stored in the form of chemical energy. But we cannot use this chemical energy directly. In the torch, the chemical energy is first converted into electrical energy. This electrical energy is then converted into light energy.

Example 2: In a motor car, we fill petrol. This petrol contains lot of chemical energy. In the car’s engine, this petrol burns, and causes the engine parts to move. The movement of engine parts is transmitted to the wheels of the car. Thus the car moves forward. So the chemical energy in petrol gets converted into kinetic energy of the car.

Example 3: The sun provides heat energy. This energy causes the water in oceans and lakes to evoporate. The vapours thus formed will merge together to form clouds. When the clouds cool, they come down as rain. This rain is collected in reservoirs. The reservoirs are constructed at higher levels. So the water in them have high potential energy. This water flows down through special pipes with great force and turn the turbines of generators, thus producing electricity. So we find that heat energy of the sun is first converted into potential energy of water in the reservoirs. This potential energy is then converted into electrical energy. We can say that, the heat energy from the sun got finally converted into electrical energy

Example 4: The food that we eat contains a lot of chemical energy. We get energy for our daily activities from this chemical energy. When we climb the stairs of a building and reach an upper floor, our muscles have to do a lot of work. Much energy that we received from the food will be used up when we climb steps. But that energy will not be wasted because, when we reach a higher level, our potential energy increases. We can say that the chemical energy from the food, got finally converted into potential energy.

---

• Thus we see a lot of energy conversions taking place around us. 

• But whenever such conversions take place, a peculiar phenomenon occurs. 

• That is., the total energy remains unchanged. This is called the Law of conservation of energy. 

• Let us see how this law applies to the above examples:

Example 1. In the case of torch cells, the chemical energy got converted into light energy. 
• According to the law, total energy derived from the cell is equal to the total light energy produced.
• But such a perfect conversion is not possible. Because, a portion of the chemical energy is lost as heat energy.
• But if we calculate the sum, we will find that the law is valid. That is.,
• Chemical energy derived from the cell = 
[Light energy] + [Heat energy wasted]

Example 2: Chemical energy derived from the petrol 
= [Kinetic energy of the car] + [Heat energy wasted in the engines] + [Energy lost to overcome friction between tyres and road when the car moves] + [Energy lost to overcome air resistance encountered when the car moves]

Example 3: Heat energy derived from the sun =
[Electrical energy produced in the generators] + [Heat energy wasted in the generators and turbines] + [Energy lost to overcome friction between water and inside surface of pipes] + [Potential energy lost due to leakage in pipes]

Example 4: Chemical energy derived from the food =
[Potential energy of the person] + [Heat energy wasted in the body]

• So we see that the converted energy consists of various components. We will learn about those components in detail, in higher classes. At present, all we need to know is that, when we add those components, we will find that the total energy after conversion remains the same. 
• In other words, the total energy before conversion is equal to the total energy after conversion.

---

The law of conservation of energy states that:
■ Energy can only be converted from one form to another; it can neither be created or destroyed.

---

In this chapter, we learned about kinetic energy and potential energy in some detail. So we will learn the details about the conversion between them. We will learn it with the help of an example.

1. An object A of mass of 25 kg is at rest. It is situated at a height of 6 m above the ground. Let this state of rest be 'Stage 0'. It is shown in fig.4.5 below:

Fig.4.5

• Let the acceleration due to gravity be taken as 10 m s^-2.
• At this stage, it’s potential energy E_p is mgh = 25 × 10 × 6 =  1500 J
• At this stage, it’s kinetic energy E_k = zero. Because it has no velocity. It is at rest.
• So we can say that, at stage 0, total energy, E_p + E_k = 1500 + 0 = 1500 J 

2. It is then allowed to fall freely. Consider the 'stage 1' when it travels 1 m from start . 
• That is., At stage 1, the object is at a height of 5 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 5 = 1250 J
    ♦ So loss of potential energy = E_p at stage 0 - E_p at stage 1 = 1500 - 1250 = 250 J 
• At this stage, we want to know it’s kinetic energy. 
• For that, we want it’s velocity at this stage. We know that, when an object is allowed to fall freely, it accelerates towards the earth, and so it’s velocity increases. Using the third equation of motion, we can calculate the velocity at any stage:
• v² = u² + 2as ⇒ v² = 0 + 2gs ⇒ v² = 0 + 2×10×1 ⇒ v² = 20 [u = 0 because, the object falls from rest]
• So the kinetic energy, E_k = ¹⁄₂ × mv² = ¹⁄₂ × 25 × v² = ¹⁄₂ × 25 × 20 = 250 J
    ♦ So gain in kinetic energy = E_k at stage 1 - E_k at stage 0 = 250 - 0 = 250 J 
• Also we can say that, at stage 1, total energy, E_p + E_k = 1250 + 250 = 1500 J 

3. Consider the stage 2 when it travels 2 m from start . 
• That is., At stage 2, the object is at a height of 4 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 4 = 1000 J
    ♦ So loss of potential energy = E_p at stage 0 - E_p at stage 2 = 1500 - 1000 = 500 J 
• At this stage, we want to know it’s kinetic energy. 
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v² = u² + 2as ⇒ v² = 0 + 2gs ⇒ v² = 0 + 2×10×2 ⇒ v² = 40 [Note that in this problem, we do not need to calculate the square root to find the actual 'v'. Because, we will be using 'v²' in the next step]  
• So the kinetic energy, E_k = ¹⁄₂ × mv² = ¹⁄₂ × 25 × v² = ¹⁄₂ × 25 × 40 = 500 J
    ♦ So gain in kinetic energy = E_k at stage 2 - E_k at stage 0 = 500 - 0 = 500 J 
• Also we can say that, at stage 2, total energy, E_p + E_k = 1000 + 500 = 1500 J 

4. Consider the stage 3 when it travels 3 m from start . 
• That is., at stage 3, the object is at a height of 3 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 3 = 750 J
    ♦ So loss of potential energy = E_p at stage 0 - E_p at stage 3 = 1500 - 750 = 750 J 
• At this stage, we want to know it’s kinetic energy. 
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v² = u² + 2as ⇒ v² = 0 + 2gs ⇒ v² = 0 + 2×10×3 ⇒ v² = 60 
• So the kinetic energy, E_k = ¹⁄₂ × mv² = ¹⁄₂ × 25 × v² = ¹⁄₂ × 25 × 60 = 750 J
    ♦ So gain in kinetic energy = E_k at stage 3 - E_k at stage 0 = 750 - 0 = 750 J 
• Also we can say that, at stage 3, total energy, E_p + E_k = 750 + 750 = 1500 J 

5. Consider the stage 4 when it travels 4 m from start . 
• That is., at stage 4, the object is at a height of 2 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 2 = 500 J
    ♦ So loss of potential energy = E_p at stage 0 - E_p at stage 4 = 1500 - 500 = 1000 J 
• At this stage, we want to know it’s kinetic energy. 
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v² = u² + 2as ⇒ v² = 0 + 2gs ⇒ v² = 0 + 2×10×4 ⇒ v² = 80 
• So the kinetic energy, E_k = ¹⁄₂ × mv² = ¹⁄₂ × 25 × v² = ¹⁄₂ × 25 × 80 = 1000 J
    ♦ So gain in kinetic energy = E_k at stage 4 - E_k at stage 0 = 1000 - 0 = 1000 J 
• Also we can say that, at stage 4, total energy, E_p + E_k = 500 + 1000 = 1500 J 

6. Consider the stage 5 when it travels 5 m from start . 
• That is., at stage 5, the object is at a height of 1 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 1 = 250 J
    ♦ So loss of potential energy = E_p at stage 0 - E_p at stage 5 = 1500 - 250 = 1250 J 
• At this stage, we want to know it’s kinetic energy. 
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v² = u² + 2as ⇒ v² = 0 + 2gs ⇒ v² = 0 + 2×10×5 ⇒ v² = 100 
• So the kinetic energy, E_k = ¹⁄₂ × mv² = ¹⁄₂ × 25 × v² = ¹⁄₂ × 25 × 100 = 1250 J
    ♦ So gain in kinetic energy = E_k at stage 5 - E_k at stage 0 = 1250 - 0 = 1250 J 
• Also we can say that, at stage 5, total energy, E_p + E_k = 250 + 1250 = 1500 J

7. Consider the stage 6 when it travels 6 m from start . 
• That is., at stage 6, the object is at a height of 0 m from the ground.
    ♦ That means the object has reached the ground. When it reaches the ground, it's velocity will be zero. So we will consider the instant when it just reaches the ground. At this stage, the velocity will be the maximum.
• At this stage, it’s potential energy = mgh = 25 × 10 × 0 = 0 J
    ♦ So loss of potential energy = E_p at stage 0 - E_p at stage 6 = 1500 - 0 = 1500 J 
• At this stage, we want to know it’s kinetic energy. 
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v² = u² + 2as ⇒ v² = 0 + 2gs ⇒ v² = 0 + 2×10×6 ⇒ v² = 120 
• So the kinetic energy, E_k = ¹⁄₂ × mv² = ¹⁄₂ × 25 × v² = ¹⁄₂ × 25 × 120 = 1500 J
    ♦ So gain in kinetic energy = E_k at stage 6 - E_k at stage 0 = 1500 - 0 = 1500 J 
• Also we can say that, at stage 6, total energy, E_p + E_k = 0 + 1500 = 1500 J

---

We can write the above results in a tabular form as shown below:

From the above table, we can note the following points:

• At any stage, the loss of potential energy is equal to the gain in kinetic energy

• That is., what ever potential energy is lost, is gained by the kinetic energy

• So, there is a continuous transformation of energy from one form (potential) to another (kinetic)

• The total energy, that is., the sum E_p + E_k is always a constant

■ So this is an excellent example to prove the Law of conservation of energy

---

Now we will see a solved example

Solved example 4.8

An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. Take g = 10 m s^-2

Solution:
Mass m = 40 kg, Height h = 5 m
Part (i): E_p = mgh = 40 × 10 × 5 = 2000 J
Part (ii): We know that, when an object is allowed to fall freely, it accelerates towards the earth, and so it’s velocity increases. Using the third equation of motion, we can calculate the velocity at any stage. We want the velocity when it is 2.5 m from the ground.
• v² = u² + 2as ⇒ v² = 0 + 2gs ⇒ v² = 0 + 2×10×2.5 ⇒ v² = 50 [u = 0 because, the object falls from rest] 
• So the kinetic energy, E_k = ¹⁄₂ × mv² = ¹⁄₂ × 40 × v² = ¹⁄₂ × 40 × 50 = 1000 J

---

In the next section, we will see Power. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 4.2 - Potential Energy

In the previous section we saw basic details about kinetic energy. In this section we will see potential energy.

■ Take a rubber band. Hold it's one end and pull the other end. 
• The band stretches. In this stretched position, the band will try to pull our fingers at the two ends. We can feel it because, we have to apply force to keep the band in a stretched position. So it is clear that the band is applying a pulling force. 
• From where did it get the energy to apply this force?
• We know that a rubber band which is not stretched, will not be able to apply any force.
• So it is clear that, the 'process of stretching' gave it some energy. The energy that we applied to stretch the rubber band, got stored in it. It will be stored in it until it regains it's original shape.
■ Similar is the case of bow and arrow. When the string of the arrow is tightened, the bow bends and is stretched. The energy used to stretch the bow, gets stored in it. 
• When the arrow is placed in the bow and pulled back wards, the bow bends further and more energy gets stored. 
• When the arrow is released, the bow regains it's original position. When this happens, the stored energy is released. This energy is converted into kinetic energy of the arrow and so the arrow moves forward.
■ Consider the stone raised to a height above ground. It possess some energy. It can now do some work, like driving a nail into a wooden piece. 
• It acquired energy because some work was done to raise it to higher position. 
• A stone at ground level will not be able to do work. 

---

■ In the above three cases, the objects rubber band, bow and stone acquired energy because some work was done on them. 
■ The work done got stored in them as potential energy. 

---

We have to note some peculiarities of potential energy:
• The 'work done on an object' is stored as potential energy.
• But we saw that kinetic energy is also equal to the 'work done on the object'. 
■ So what is the difference?
• In kinetic energy, the work done causes the object to move continuously with a certain velocity
• But in potential energy, there is no continuous movement for the object. There is only 'change in configuration' or 'change in position'
    ♦ In the case of rubber band, the band changed to a stretched configuration
    ♦ In the case of bow, it changed to a stretched configuration
    ♦ In the case of stone, it's position changed from ground level to a higher level
■ The potential energy possessed by the object is the energy present in it by virtue of its position or configuration.

Gravitational potential energy

We have seen that, when an object is raised to a height, potential energy gets stored in that object. Let us calculate how much joules of energy is stored in this way:
1. Consider an object of mass m. We have seen that it will have a weight W = mg. Where g is the acceleration due to gravity. (Details here)
The earth is pulling this object towards it's centre. The force of this pull is mg newton. 
2. So we have to apply an equal and opposite force to raise the object above the ground. That is., we must apply a force of mg in the upward direction. 
3. We know that work done = force × displacement. If the object is raised to a height h, the displacement is h. So work done on the object = mgh. 
4. This much work gets stored in the object as potential energy. It is called gravitational potential energy. 
5. So, the gravitational potential energy of an object of mass m situated at a height h is equal to mgh. We can write it in the form of an equation:
Eq.4.2:
Gravitational potential energy E_p = mgh

---

We have to note two important points while considering gravitational potential energy:
First point:
• Consider 'object A' with mass m, remaining at ground level in fig.4.3 below. It cannot do any work on another object which is at the same ground level.

Fig.4.3

• So, if we take the ground level as our 'zero level' or 'datum level':
    ♦ The object A at height 'h' above datum will have an Ep equal to mgh
    ♦ The object A at the ground level is 'useless' as far as potential energy is concerned
• But, if there is a basement level at a height h₁ below the ground level, the object A at ground level is 'not useless'. It can do a work equal to mgh₁ on an object at the basement level.
■ So it is important to specify a datum when we use potential energy
Second point:

• Consider fig.4.4 below. In the first case, the object is taken to a height h above the ground level, through a straight line path. This is indicated as 'path 1'.

Fig.4.4

• In the second case, the object is taken through 'path 2' which is a zigzag path. 
• The final position in both cases is 'at a height h above the ground level'. 
• So the final potential energy in both cases will be the same: mgh. 
■ So we can say that, the potential energy depends on the height from the datum level. It does not depend on the path taken to reach the height.

---

Now we will see some solved examples
Solved example 4.6
Find the energy possessed by an object of mass 12 kg when it is at a height of 4 m above the ground. Given, g = 9.8 m s^-2.
Solution:
• Mass of the object = 12 kg
• Height above the ground at which the object is situated = 4 m
• Given g =  9.8 m s^-2.
• We have: E_p = mgh = 12 × 9.8 × 4 = 470.4 J

Solved example 4.7
An object of mass 15 kg is at a certain height above the ground. If the potential energy of the object is 900 J, find the height at which the object is with respect to the ground. Given, g = 10 m s^-2.
Solution:
• Mass of the object = 15 kg
• Potential energy of the object = 900 J
• Given g =  10 m s^-2.
• We have: E_p = mgh ⇒ 900 = 15 × 10 × h ⇒900 = 150h ⇒h = 6 m

---

In the next section, we will see Law of Conservation of Energy. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 4 - Work, Power and Energy

In the previous section we saw mass, weight, thrust and pressure. In this section we will see work, power and energy.
■ Consider some of our day to day activities. We walk, run, play, read, think. We engage in many such activities. 
• For performing those activities, we need energy. Some activities like running and playing requires more energy. 
• We get the required energy from the food we eat.
■ Consider animals. They also perform various activities. Like running, hunting, hiding from enemies etc.,  Also we humans make them do activities like pulling vehicles, carrying loads etc., 
• Animals also get the required energy from the food that they eat
■ Consider machines. They perform activities like pumping water, pulling trains, lifting heavy loads etc., 
• They get their energy from the fuels like petrol, coal, diesel etc.,

To define energy from a scientific point of view, we must first understand about another concept 'work'. In day to day life we regularly come across the term work. Let us see some examples:
• Two laborers worked hard to load the bricks into the truck
• Much work is required to arrange all the newly arrived books in the library
• A person is holding a heavy luggage on his shoulders. He may have to stand in that position for a long time. He is applying upward force through his shoulders. He may even get exhausted. But we would not say that he is 'working' to keep the luggage on his shoulders.
• Talking with friends for a long time about some academic topics may be considered as 'work'. It involves expenditure of energy too.
• At the same time, taking with friends about a movie may not be considered as work
■ Thus, even in common day do day conversations, we give a difference between work and energy. 
• In day to day conversations, we would not mind if they are interchanged. 
• From a scientific point of view, such interchanging is not allowed.

---

■ In science, work is done only if two conditions are satisfied: 

• A force should act on an object
• The object must get displaced due to the force
Let us see some examples:
1. Push a pebble lying on the ground. The pebble moves through a distance
• A pushing force was applied on the object (the object here is pebble)
• The object got displaced
■ So work is done
2. A girl pulls a trolley. The trolley moves through a distance
• A pulling force was applied on the object (the object here is trolley)
• The object got displaced
■ So work is done
3. Lift a book through a height. For lifting the book, a force must be applied. The book rises up.
• A lifting force was applied on the object (the object here is book)
• The object got displaced
■ So work is done
4. Push a large rock on the ground. The rock does not move.
• A pushing force was applied on the object (the object here is rock)
• The object did not get displaced
■ So in this case, work is not done

Work done by constant force

We are now able to identify those situations in which a work is done. What we need next, is a method to calculate the amount of work done.
Let a constant force, F act on an object. Let the object be displaced through a distance, s in the direction of the force. This is shown in fig. 4.1 below:

Fig.4.1

Let W be the work done.
Then W = F × s
Thus, work done by a force is equal to the product of 
• The force and
• The displacement produced by the force
■ Next we need a unit for the force. 
• Unit for force is newton 
• Unit for distance is metre.
■ So unit for work is newton-metre. In short form, it is N m
• This 'N m' has another name: joule. In short form, it is J
• Let a force of 1 N move an object through a distance of 1 m. Then the amount of work done is obtained as follows:
W = Fs = 1 × 1 = 1N m = 1 joule
• Thus we can write:
1 N m or 1 joule of work is done on an object, if a force of 1 N moves it through a distance of 1 m along the line of action of the force.

---

The unit 'joule' for work is named after James Prescott Joule. He was a British scientist who made significant contributions in the field of electricity and thermodynamics. 

---

Now we will see a solved example:

Solved example 4.1

A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force. If the force acts on the object all through the displacement, calculate the work done.

Solution:

We have: W = F×s = 5 × 2 = 10 N m or 10 J

---

When we defined work, we specified an important point. That is:

■ The displacement should be along the line of action of the force. 

Now consider the situation in fig.4.2 below:
1. A force F is acting on a body. The body is being displaced. The displacement is along the line of the force.

Fig.4.2

2. But the displacement is in the opposite direction of the force.
■ Let us see an example for such a situation:
• A body is moving with a uniform velocity u
• A retarding force F acts on it in a direction opposite to the direction of motion
• As a result of this retarding force, the body comes to rest. That is., final velocity v = 0 
• In this situation, how will we define force?
• We can use the same principle. But note that the displacement is opposite to the direction of the force. We know that displacement has both magnitude and direction. So it's direction should be taken as negative. This is to indicate that, it is opposite to the direction of the force.
So we get: W = F × (-s) = -Fs
■ Thus, the work can be either positive or negative
Another example:
1. Lift an object upwards. When we lift an object we are applying a force on the object
• The direction of movement of the object is same as the direction of the force
• So the work done is positive
2. When the object is lifted up, another force is also acting on the object. It is the gravitational force
• This force acts in a direction opposite to the direction of movement of the object
• So work done by the gravitational force is negative

Solved example 3.2
A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage. 
Solution:
1. Work done W = F×s
2. Here displacement s = 1.5 m
3. We need to find the force F. 
4. We know that the luggage which has a mass of 15 kg is acted upon by a force. This force is the 'Gravitational force' F_g exerted by the earth towards it's centre. 
5. We know that this force is equal to the weight mg. Where m is the mass and g the acceleration due to gravity.
So gravitational force of the earth = W = mg = 15 × 10 =150 N
6. To lift the object, we need to apply an equal and opposite force. That is., we need to apply 150 N in the upward direction. The displacement of 1.5 m is in the same direction of the lifting force. So we can take (+1.5) m
Thus work done = F×s = 150 ×1.5 = 225 J 

---

In the next section, we will see Kinetic energy. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 2.4 - Solved examples on Law of Conservation of Momentum

In the previous section we completed the discussion on Newton's Third law of motion. We also saw the Law of conservation of Momentum. In this section we will see some more solved examples.

Solved example 2.13
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s^-1 . Calculate the initial recoil velocity of the rifle. 
Solution:
1. Masses:
• Given: Mass of bullet = m_B = 50 g = 0.05 kg
• Given: Mass of rifle = m_R = 4 kg 
2. Initial velocities:
• The bullet accelerates from rest. So u_B = 0
• The rifle accelerates from rest. So u_R = 0
3. Final velocities:
• Given that velocity of bullet = v_B= 35 m s^-1.
• We have to find v_R.
4. Total momentum before the pistol is fired = (m_Bu_B +m_Ru_R) = (0.05 × 0 + 4 × 0) = 0 kg m s^-1  
• Total momentum after the pistol is fired = (m_Bv_B +m_Rv_R) = (0.05 × 35 + 4 × v_R) = (4v_R+ 1.75) kg m s^-1  
5. According to the law of conservation of momentum, Total momentum after the fire = Total momentum before the fire. So we can write:
4v_R+ 1.75 = 0 ⇒ v_R = - ^1.75⁄₄ = -0.44 m s^-1.
(The negative sign indicates that, the direction of motion of the rifle is opposite to the direction of motion of the bullet)

Solved example 2.14
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s^-1 and 1 m s^-1 respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s^-1. Determine the velocity of the second object.
Solution:
1. Masses:
• Given: Mass of first object = m_A = 100 g = 0.1 kg
• Given: Mass of second object = m_B = 200g = 0.2 kg 
2. Initial velocities:
• Given: Initial velocity of first object = u_A = 2 m s^-1
• Given: Initial velocity of second object = u_B = 1 m s^-1
Note that both velocities are taken as positive because, they travel in the same direction.
3. Final velocities:
• Given: Final velocity of first object = v_A = 1.67 m s^-1
• We have to find v_B.
4. Total momentum before the collision = (m_Au_A +m_Bu_B) = (0.1 × 2 + 0.2 × 1) = 0.4 kg m s^-1  
• Total momentum after the collision = (m_Av_A +m_Bv_B) = (0.1 × 1.67 + 0.2 × v_B) = (0.2v_B+ 0.167) kg m s^-1  
5. According to the law of conservation of momentum, Total momentum after the collision = Total momentum before the collision. So we can write:
0.4 = 0.2v_B+ 0.167 ⇒ 0.2v_B = 0.233  ⇒ v_B =   ^0.233⁄_0.2 = 1.165 m s^-1.
(The positive sign of v_B indicates that, the direction of motion of the second object is same as the initial direction of motion of the objects)

Solved example 2.15
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s^-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Solution:
In this problem, we have a special situation:
• The objects stick together after the collision. That means, after the collision, the objects move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of object A = m_A = 1.5 kg
• Given: Mass of object B = m_B = 1.5 kg
2. Final mass = m_f = m_A + m_B = 1.5 + 1.5 = 3 kg 
3. Initial velocities:
• Object A was moving with a velocity. Given u_A = 2.5 m s^-1
• Object B was moving with a velocity. Given u_B = 2.5 m s^-1
    ♦ But object B was moving in a direction opposite to that of object A. So the velocity of object B has to be taken as negative. Thus we can write: u_B = -2.5 m s^-1
4. Final velocity:
• After collision, the two object move together. We have to find the final velocity v_f with which they move together.
5. Total momentum before collision = (m_Au_A + m_Bu_B) = (1.5 × 2.5 + 1.5 × -2.5) = 3.75 - 3.75 = 0 kg m s^-1.
• Total momentum after collision = m_fv_f = 3 × v_f = 3v_f  kg m s^-1  
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
-3v_f = 0 ⇒ v_f = 0

Solved example 2.16
A hockey ball of mass 200 g travelling at 10 m s^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s^-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution:
1. Mass of the hockey ball = 200 g = 0.2 kg
2. Initial velocity of the hockey ball = 10 m s^-1
3. So initial momentum before collision = 0.2 × 10 = 2 kg m s^-1
4. Final velocity of the hockey ball = -5 m s^-1
Note that the final velocity is given a negative sign because it returns along the same path. That is., it travels in the opposite direction after collision
5. Final momentum after collision =  0.2 × (-5) = -1 kg m s^-1 . 
6. So difference between the two momenta = 2 - (-1) = 2 + 1 = 3 kg m s^-1.

Solved example 2.17
A bullet of mass 10 g travelling horizontally with a velocity of 150 m s^-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet. 
Solution:
1. Initial velocity with which the bullet hits the wooden block = u = 150 m s^-1
2. Final velocity of the bullet in the block = v = 0 m s^-1
3. Time in which the bullet comes to rest = t = 0.03 s
4. So the velocity decreases from 150 to zero m s^-1. It is clear that there is deceleration (negative acceleration). To find this acceleration 'a', we can use the first equation of motion.
5. We have: v = u + at ⇒ 0 = 150 + a × 0.03 ⇒ a = ^-150⁄_0.03 = -5000 m s^-2.
6. We now have the initial velocity u, final velocity v, acceleration a and time of travel t. We want the distance travelled. We can use the third equation of motion. Because it connects all these quantities
7. So we can write:
v² = u² + 2as ⇒ 0² = 150² + 2 × -5000 × s ⇒ 0 = 22500  + -10000s ⇒ s  = 2.25 m.
8. The negative acceleration was due to the resistance of the wooden block. That means, the wooden block applied a resisting force F on the bullet. We want the value of this F
9. We know that force = mass × acceleration ⇒ F = ma = 0.01× -5000 = -50 N (∵ mass = 10 gram = 0.01 kg)
10. So we can write:
• The distance of penetration = 2.25 m
• The force exerted by the wooden block on the bullet = 50 N

Solved example 2.18
An object of mass 1 kg travelling in a straight line with a velocity of 10 m s^-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Solution:
In this problem, we have a special situation:
• The objects stick together after the collision. That means, after the collision, the objects move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of object A = m_A = 1.0 kg
• Given: Mass of wooden block = m_B = 5.0 kg
2. Final mass = m_f = m_A + m_B = 1 + 5 = 6 kg 
3. Initial velocities:
• Object A was moving with a velocity. Given u_A = 10 m s^-1
• Wooden block was stationary. So u_B = 0 m s^-1
4. Final velocity:
• After collision, the two object move together. We have to find the final velocity v_f with which they move together.
5. Total momentum before collision = (m_Au_A + m_Bu_B) = (1 × 10 + 5 × 0) = 10  + 0 = 10 kg m s^-1.
• Total momentum after collision = m_fv_f = 6 × v_f = 6v_f  kg m s^-1  
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
6v_f = 10 ⇒ v_f = 1.66 m s^-1.
7. So the total momentum before the impact = 10 kg m s^-1
• The total momentum after the impact = 6v_f = 6 × 1.66 = 9.96 kg m s^-1.
• Velocity of the combined object = v_f = 1.66 m s^-1.

Solved example 2.19
An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s^-1 to 8 m s^-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Solution:
1. Mass of the object = 100 kg
2. Initial velocity of the object = 5 m s^-1
3. So initial momentum  = 100 × 5 = 500 kg m s^-1
4. Final velocity of the object = 8 m s^-1
5. So final momentum = 100 × 8 = 800 kg m s^-1 
6. Time in which the change in velocity took place = t = 6 s
7. So the velocity increases from 5 to 8 m s^-1. It is clear that there is acceleration. To find this acceleration 'a', we can use the first equation of motion.
8. We have: v = u + at ⇒ 8 = 5 + a × 6 ⇒ a = ³⁄₆ = 0.5 m s^-2.
9. We know that force = mass × acceleration ⇒ F = ma = 100 × 0.5 = 50 N

Solved example 2.20
How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s^-1 .
Solution:
1. Mass of the dumb-bell = m = 10 kg
2. Height of fall = distance of travel = s = 80 cm = 0.8 m
3. Acceleration = a = 10 m s^-2.
4. The dumb-bell falls from rest. So it has an initial velocity u = 0
5. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
6. So we can write:
v² = u² + 2as ⇒ v² = 0² + 2 × 10 × 0.8 ⇒ v² = 16 ⇒ v  = 4 m s^-1.
7. Let us analyse the situation:
(i) The dumb-bell falls from a height of 80 cm
(ii) It's velocity goes on increasing because it is acted upon by an acceleration of 10 m s^-2.
(iii) When the distance of travel becomes 80 cm, the velocity becomes 4 m s^-1.
(iv) The velocity would have increased even more, if it was allowed to travel more distance.
(v) But the travel comes to an abrupt stop. Because it met the floor at 80 cm.
8. The momentum  of the dumb-bell at the time of hitting the floor = mv = 10 × 4 = 40 kg m s^-1.
9. Now we consider the law of conservation of momentum:
(i) The system consists of two objects: The dumb-bell and the floor
(ii) Momentum of the dumb-bell just before collision = 40 kg m s^-1.
(iii) Momentum of the floor just before collision = 0 kg m s^-1.
(iv) Total momentum just before collision = 40 + 0 = 40 kg m s^-1.
10. This must be the total momentum after collision also. So a momentum of 40 kg m s^-1 is transferred to the floor.

Solved example 2.21
Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s^-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Solution:
1. When two persons push the car, it moves with a uniform velocity. Uniform velocity means no acceleration. That means, no resultant external force is acting on the car. That means, the frictional resistance is just balanced by the effort of two persons.
2. Let the force provided by each person be F_P. Then the frictional force FF will be equal to 2F_P. So we can write: F_F = 2F_P
3. Now a third person also puts in the same effort F_P. The combined pushing force is greater than the frictional force. That is why, there is a resultant force acting on the car. Because of this resultant force, the car begins to move with acceleration 'a' which is given as 2 m s^-2.
4. The resultant force acting on the car will be equal to 3F_P - F_F.
This resultant force is equal to mass × acceleration = 1200 × 0.2 = 240 N
So we can write: 3F_P - F_F = 240 N
5. But from (2) we get F_F = 2F_P. Substituting this in (4) we get:  
3F_P - 2F_P = 240 N ⇒ F_P = 240 N 
So the force with which each person push the car is 240 N

Solved example 2.22
A hammer of mass 500 g, moving at 50 m s^-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Solution:
1. The hammer is moving with an initial velocity u of 50 m s^-1.
2. It is stopped by the nail. The time taken by the nail to stop the hammer is 0.01 s
3. The hammer is 'stopped' by the nail. That means, the final velocity v of the hammer is 0 m s^-1
4. Since there is a velocity change, there is acceleration. Here the velocity decreases from 50 m s^-1 to zero. So it is deceleration or negative acceleration. We have to find this acceleration first. We can use the first equation of motion:
5. v = u + at ⇒ 0 = 50 + a × 0.01 ⇒ a = ^-50⁄_0.01 = -5000 m s^-2.
6. The 500 g mass of the hammer is subjected to an acceleration of -5000 m s^-2.
7. So force =  mass × acceleration = 0.5 × -5000 = -2500 N
8. This much force is exerted by the hammer on the nail. By newtons third law, the nail exerts the same force in the opposite direction. 
9. So the force exerted by the nail on the hammer = + 2500 N  

Solved example 2.23
A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 kmph. Its velocity is slowed down to 18 kmph in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution:
1. Initial velocity u = 90 kmph = 90 × ¹⁰⁰⁰⁄₃₆₀₀ = 25 m s^-1.
2. Final velocity v = 18 kmph = 18 × ¹⁰⁰⁰⁄₃₆₀₀ = 5 m s^-1.
3. Time duration in which the velocity changed = 4 s
4. To find acceleration a, we can use the first equation of motion:
5. v = u + at ⇒ 5 = 25 + a × 4 ⇒ a = ^-20⁄₄ = -5 m s^-2.
6. Change in momentum = m(v-u) = 1200 × (5-25) = 1200 × -20 = -24000 kg m s^-1. 
7. Force = mass × acceleration = 1200 × -5 = -6000 N. The negative sign indicates that, the force is acting in the direction opposite to the direction of motion.

---

In the next chapter, we will see Gravitational force. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Physics lessons. blogspot.in - All Rights Reserved