Chapter 3.2 - Mass and Weight of objects

In the previous section we saw the acceleration due to gravity 'g'. We obtained it's value as 9.8 m s^-2. In this section we will see some of it's applications. Also later in this section we will see thrust and pressure.

Mass of an object

• Mass of an object is the amount of matter in that object. For example, if we consider a wooden block, it’s mass is the amount of wood in that block. 
• The mass of a particular object is a constant. Consider a wooden block of weight 3 kg. If we put this wooden block on one side of a weighing balance, we will have to put an exact 3 kg standard weighing block on the other side of the balance. 
• The mass of that wooden block will be the same 3 kg, where ever we take it in the universe.
• Mass is used in another situation also. We have seen that it is a measure of inertia. That is., greater the mass, greater is the inertia of that object. So we can say this: Mass is the resistance shown by the body to change it's state of rest or state of uniform motion.

Weight of an object

• Consider a wooden block kept on the floor. The earth is attracting the block towards it’s centre.  
• If we want to lift it from the floor, we will have to apply some force. This force that we need to apply is equal and opposite to the force applied on the block by the earth. 
• We know that this force is mg. Where m is the mass of the block and g is the acceleration due to gravity.  
• If there are two blocks, one of 3 kg and the other of 5 kg, we would prefer to lift the 3 kg block. This is because lesser force is sufficient to lift it when compared to the 5 kg block. 
• So we need a method to specify the force exerted by the earth on an object. 
• To describe this force, we use the term ‘weight’. So we can write: Weight of an object is the force exerted by the earth on that object. 
• For a 3 kg wooden block, the weight will be 3 × 9.8 = 29.4. Now we need a unit to specify weight. 
• Since weight is the force, the unit newton that we use for force is used for weight also. 
• So we can write: The weight of a 3 kg wooden block is 29.4 N. 
• The symbol for weight is W. So we can write in this way also:
For a 3 kg wooden block, W = 29.4 N

---

Now we will see an interesting problem:
Solved example 3.6
Weight of an object on earth is 4 kg. How much would it weigh on the moon?
Given: Mass of the moon = 7.36 × 10²² kg, Radius of the moon = 1.74 × 10⁶ m
Solution:
• On the earth, the object is placed on one side of the balance. Then on the other side, we will have to place a 4 kg standard weighing block. Then only it will balance. 
• Now the same object is taken to the moon. On landing on the moon, we will notice that it is easier to lift the same object. That is., we do not have to use the same amount of force, that we needed to lift it on earth. 
What could be the reason? Also, we want to know it’s exact weight on the moon. We can use the following steps:
1. Mass of 4 kg has not changed when the object was taken to the moon
2. But the force with which the object is attracted to the centre of the moon has changed. 
3. Let us calculate the force F_m with which any object on the surface of the moon would be attracted to it's centre. For this calculation we will use the universal law of gravitation (Details here):
(i) Let m₁ be the mass of the moon (given as 7.36 × 10²² kg) and m₂ the mass of any object placed on the surface of the moon. 
(ii) Let R be the radius of the moon (given as 1.74 × 10⁶ m). 
(iii) Then we have: F_m = G(^m₁m₂⁄_R²). Substituting the given values we get:

So we have: Force of attraction on an object of mass m₂ on the surface of the moon = F_m = 1.63 m₂.
4. The mass of our object = 4 kg. So F_m = Weight of the object on the moon = 1.63 × 4 = 6.52 N

---

Note: The final equation in step 3(iii) is: F_m = 1.63 m₂.
• Compare this with the equation: Force = mass × acceleration  
• Comparing the right sides, it is obvious that 1.63 is the acceleration. That is., acceleration due to gravity on the moon. So we can write:
■ Acceleration due to gravity on the moon = 1.63 m s^-2.
• Now, acceleration due to gravity on the earth = 9.8 m s^-2. 
• Let us take the ratio. We get: ^9.8⁄_1.63 = 6.01 = 6 (approximately)
So we can write:
■ Acceleration due to gravity on the earth is 6 times that on the moon
■ Consequently, weight of any object on the moon will be one sixth of it's weight on the earth

---

We will now see some solved examples

Solved example 3.7
Mass of an object is 10 kg. What is it's weight on the earth?
Solution:
1. Weight of an object on the earth = The force exerted by earth on that object = m × g
Where m is the mass of the object (given as 10 kg) and g is the acceleration due to gravity (which is equal to 9.8 m s^-2)
2. So weight = 10 × 9.8 = 98 N  

Solved example 3.8
An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?
Solution:
1. We know that, weight of any object on the moon will be one sixth of it's weight on the earth. 
2. So we get: Weight of the given object on the moon =  ¹⁄₆ × 10 N = 1.67 N 

---

• So now we know what weight is. It is a force. Even an object which is simply placed on the floor is experiencing a force. This force is it’s weight W. 
• Now, according to Newton’s third law, every force has an equal and opposite reaction. So which is the reaction here?
• The object placed on the floor is trying to move downwards and reach the centre of the earth. It is exerting a force on the floor. This force is the action and is equal to it's weight W
• The floor obstructs the downward movement of the object, by applying an equal and opposite force. This is the reaction. 
    ♦ If the floor is well built, with materials like stone or concrete, it can resist the object easily. 
    ♦ But if the floor is of materials like loose sand, the object will penetrate some distance into the floor.
• Another example: A book rests on a table. The book tries to move downwards and reach the centre of the earth. It’s weight W is the action. The table provides an equal and opposite reaction. So the downward movement of the book is prevented. 

Thrust and Pressure

Consider the following situation: 
1. A person stands on loose sand, his feet would go deep into the sand. 
2. But if he lay down on the sand, his body will not go that deep. 
3. Note that the same downward force is acting in both cases. Because, in both cases, it is the weight W of the person. There is no other force. How is the sand able to provide better resistance when the person lay down? 
• The answer is that, the resisting capacity of the sand did not increase or decrease. It remains the same. In the latter case, a larger area of sand could take part in providing the resistance. It is like dividing a task among more number of workers. This can be explained as follows:
    ♦ Two workers are given the task of lifting a timber log
    ♦ The same log can be lifted more easily by 7 workers
• In the same way, more sand was able to take part in the task of 'resisting the weight of the body'.
• Now, how did ‘more sand’ come into the picture?
Ans: The contact area between the body and sand increased when the person lay down.
• So ‘contact area’ has an important role to play here.
• Let the same force F act on two areas A₁ and A₂.  Let A₂ be greater than A₁. Then the effect of the force F will be greater on the area A₁. 
■ So we need a new quantity to measure the effect of a force. This new quantity is called pressure. 
• Pressure is the ratio of force to the area on which that force acts. 
• So to obtain pressure, we will be dividing force by the area. When we do that division, what we get is the 'force per unit area'. 
■But we have to note one point: The force under consideration must be acting in an exact perpendicular direction to the area. This is shown in the fig.3.5 below:

Fig.3.5

• In the fig.3.5 above, 
    ♦ The green surface is perfectly horizontal
    ♦ The red surface is a sloping surface
    ♦ The blue arrow shows a perfectly vertical force
    ♦ The yellow arrow is a sloping force, but is perpendicular to the red surface
• Consider the green surface and the blue force. 
    ♦ We have a vertical force on a horizontal surface. That means the force is acting perpendicular to the area. So the blue force can be considered for calculating the pressure on the green surface
    ♦ Similarly, the yellow force can be considered for calculating the pressure on the red surface
• By the same argument: 
    ♦ The yellow force cannot be considered for calculating the pressure on the green surface. Because yellow force is not perpendicular to the green surface
   ♦ Similarly, the blue force cannot be considered for calculating the pressure on the red surface. Because blue force is not perpendicular to the red surface

---

• We give a special name for the 'force in the perpendicular direction'. It is the thrust. So thrust can be defined as follows:

■ Force acting on a body in a direction perpendicular to it’s surface is called thrust.
• So we cannot use ‘any force’ to calculate pressure. We must use only ‘thrust’.  
• Thus the definition of pressure becomes:
■ Pressure is the thrust per unit area.
Mathematically we write:
Eq.3.4:
• Pressure = ^Thrust⁄_Area 
• Substituting the units of thrust and area, we will get the unit of pressure. Thus:
• Unit of pressure = ^N⁄_m² = N m^-2
• This N m^-2 is given a special name ‘pascal’ in honour of the scientist Blaise Pascal. The short form is pa
• So we can use either N m^-2 or pa. Both are same.

---

Now we will see some solved examples

Solved example 3.6
A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.
Solution:
• Mass of the wooden block = 5 kg
• Force exerted on the table top = Weight of the block = mg = 5 × 9.8 = 49 N  
• The block is kept on the table top in two different ways as shown in the fig.3.6 below:

Fig.3.6

Case (i):
1. In fig.3.6(a), the block is resting on a 20 × 10 face
• So area = 20 × 10 = 200 cm² = ²⁰⁰⁄₁₀₀₀₀ m² = 0.02 m².
2. The force, which is the weight, is acting in a vertical direction. So it is perpendicular to the top surface of the table.
• So the force can be considered as the thrust
3. Thus we get: pressure = ^Thrust⁄_Area = ⁴⁹⁄_0.02 = 2450 Nm^-2.
Case (ii):
1. In fig.3.6(a), the block is resting on a 40 × 20 face
• So area = 40 × 20 = 800 cm² = ⁸⁰⁰⁄₁₀₀₀₀ m² = 0.08 m².
2. In this case also, the force, which is the weight, is acting in a vertical direction. So it is perpendicular to the top surface of the table.
• So the force can be considered as the thrust
3. Thus we get: pressure = ^Thrust⁄_Area = ⁴⁹⁄_0.08 = 612.5 Nm^-2.
■ Comparing the two cases, we can see that, when area increases pressure decreases 

---

In the next section, we will see Work, Power and Energy. 

PREVIOUS      CONTENTS       NEXT

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

Chapter 3.1 - Acceleration due to Gravity

In the previous section we saw the Universal  gravitation constant. In this section we will see acceleration due to Gravity.

1. Consider a stone thrown up from the ground. After reaching a height, it will fall down. 
2. While falling down, the only force acting on it is the ‘gravitational force of attraction’. 
■ That stone is said to be in ‘free fall’. 
3. When there is a force, there is an acceleration. We have seen that Force = mass × acceleration. Details here. 
4. So the stone which is in free fall is under acceleration. Because of the acceleration, the velocity of the stone increases. This acceleration is due to the gravitational force of the earth. 
■ So it is called ‘acceleration due to gravity’. It is denoted by the letter g. 
5. The unit of g is same as that of acceleration. That is., ms^-2

---

1. Let the mass of an object on the surface or near the surface of the earth be m₂. Then the force F exerted by the earth on that object is given by:

F = G(^m₁m₂⁄_d²) Details here.

• Where m₁ is the mass of the earth. 
• d is the distance between the object and the centre of the earth
• G is the universal gravitation constant.
2. Note that the object under consideration is 'on the surface or near the surface of the earth'. So the distance 'd' becomes 'R', the radius of the earth. So we can write:
F = G(^m₁m₂⁄_R²) 
3. We know that F = m₂a. Here, a = g, the acceleration due to gravity. So we can write: F = m₂g. That means, the force acting on the object is equal to the product of it’s mass and the acceleration due to gravity. 
4. No other force is acting on the object. So we can equate (2) and (3). We get:
m₂g = G(^m₁m₂⁄_R²) 
■ From this we get:
Eq.3.2:
g = G(^m₁⁄_R²)
• Where m₁ is the mass of the earth 

---

We know that earth is not a perfect sphere. The value of R is greater at the equator than at the poles. This is shown in the fig.3.3 below:

Fig.3.3

Let us see how this variation in radius affects the value of g:

At the equator, R is greater. In equation 3.2 above, R is in the denominator. So, the greater value of R will give a lesser value of g at the equator. That means, g is lesser at the equator than at the poles. The values are:
• Value of g at the polar regions = 9.83 m s^-2
• Value of g at the equator = 9.78 m s^-2
■ The average value of g on the surface of the earth is taken as 9.8 m s^-2 for solving numerical problems. 
• This is obtained when we use an average value of R as  6.4×10⁶ m. Let us do that calculation ourselves:
From Eq.3.2 above, we have:
g = G(^m₁⁄_R²)
• Where m₁ is the mass of the earth = 6×10²⁴ kg 
• R is the mass of the earth = 6.4×10⁶ m
• G is the universal constant of gravitation = 6.7×10^-11 m 
Substituting the values, we get:

---

We have seen 'motion of objects' in chapter 1. We derived the three equations of motion. Details here. Those equations can be used to calculate the following unknown quantities: 

• Distance travelled s

• Initial velocity u

• Final velocity v

• Time of travel t 

Those three equations are valid when an object moves with a constant acceleration 'a'.

■ In our present discussion, we have seen that, the objects under free fall are subjected to a constant acceleration of g. 
• So can we use those three equations for freely falling objects, by replacing a with g? Let us find out:

1. When an object falls freely, the force acting on it is mg. So it is clear that, with increase in mass, the force will also increase. 
2. Consider two objects shown in the fig.3.4 below. One has a greater mass than the other.

Fig.3.4

• If we drop both of them from the same height h, which one will reach the ground first?  
3. At a first glance, we will be inclined to think that the object 2 which has heavier mass will reach the ground first. 
• But the fact is that, both will reach the ground together at the same instant. Let us see the reason:
4. When the two objects are released from rest, their initial velocity u will be zero. Both of them will then begin to gain velocity. That means, both of them will be accelerated. 
5. No force other than the gravitational force is acting on them. So the 'acceleration experienced' will be the same 'g' for both of them. It is the acceleration due to gravity. 
6. From eq.3.2, we see that, g is a constant. Because on the right side of eq.3.2, G, m₁ and R are all constants. 
7. Now consider the second equation of motion: s = ut + ¹⁄₂at² 
8. In our present case s = h, u = 0, and a = g. Let us substitute these known values and find the unknown 't'. We get:
h = 0×t + ¹⁄₂×g×t² ⇒ h =  ¹⁄₂×g×t² ⇒ 
Eq.3.3:
t = √(^2h⁄_g)
9. We will get this same 't' for both object 1 and object 2. Because, the two objects are distinguished from each other by only one property which is their mass. The final equation 3.3 which gives us the time required to reach the ground does not involve m. That means t is same for both the objects. That means both the objects will reach the ground at the same instant.

---

This was proved experimentally by Newton. 

1. He placed a feather and a coin at the top, inside a long glass jar. 

Why use a jar made of glass? 

Because we want to see the speed at which the objects fall from the top of the jar to the bottom.

2. When the objects were released, the coin reached the bottom first. The feather reached only a little later. 

3. He then repeated the experiment. This time, all the air inside the glass jar was sucked out. So there was a perfect vacuum. 

4. When the objects were released, both of them reached the bottom at the same instant. So what happened when there was no vacuum?

Ans: When there was no vacuum, the air offered resistance to the fall of the objects. Resistance experienced by the feather will be much greater than that experienced by the coin. So the feather will reach only after some time.
A video can be seen here. 
■ Several centuries earlier, Galileo had conducted similar experiments. He dropped different objects from the top of the leaning tower of Pisa. He then published his findings. He argued that the feather took more time because, it experienced greater resistance from air. But there were no facilities to create vacuum in a jar at that time.

---

So now we know that, for free fall, the final velocity and time taken are independent of mass. All the three equations of motion can be applied for objects in free fall. Let us see some solved examples:

Solved example 3.4

A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m s^-2 (for simplifying the calculations).

(i) What is its speed on striking the ground?

(ii) What is its average speed during the 0.5 s?

(iii) How high is the ledge from the ground?

Solution:

1. The car falls off the ledge. So initial velocity u = 0

2. It reaches the ground in time t = 0.5 s

3. We have to find the final velocity v. We can use the first equation of motion:

v = u + at ⇒ v = u + gt ⇒ v = 0 + 10 × 0.5 ⇒ v = 5 m s^-1. This is the answer for part (i)

4. Average speed = ^{(initial velocity + final velocity)}⁄₂ = ⁽⁰⁺⁵⁾⁄₂ = 2.5 m s^-1. This is the answer for part (ii)

5. Height of fall is the distance travelled s. We can use the second equation of motion:

s = ut + ¹⁄₂at² ⇒ h = 0×t + ¹⁄₂×g×t² ⇒ h =  ¹⁄₂×10×0.5² ⇒ h = 1.25 m. This is the answer for part (iii)

Solved example 3.5

An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

Solution:

1. The object is thrown upwards with an initial velocity u. We have to find this u

2. The distance travelled = s = 10 m

3. This distance travelled is the maximum height reached by the object. When it reaches this maximum height, it's velocity will be zero. So we get: Final velocity v = 0

4. The object is subjected to the 'acceleration due to gravity' g. But g acts down wards. The object is travelling upwards. So acceleration should be taken as negative, since it is opposite to the direction of motion. Thus g = -9.8 m s^-2.

5. We can use the third equation of motion:

v² = u²+ 2as ⇒ v² = u²+ 2gs ⇒ 0² = u²+ 2 × (-9.8) × 10 ⇒ 0² = u²+ 2 × (-9.8) × 10 

⇒ 0 = u² - 196 ⇒ u² = 196 ⇒ u = √196 = 14 m s^-1. This is the answer for part (i)

6. To find the time, we can use the first equation of motion:

v = u + at ⇒ v = u + gt ⇒ 0 = 14 + (-9.8) × t ⇒ -14 = -9.8t 

⇒ 14 = 9.8t ⇒ t = 1.43 s. This is the answer for part (ii)

---

In the next section, we will see the relation between Mass and Weight. 

PREVIOUS      CONTENTS       NEXT

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

Chapter 3 - Gravitation

In the previous section we completed the discussion on Force and laws of motion. In this section we will see Gravitation.

■ Lift a small stone to a certain height and let go. 
• We can see that it falls down.
■ Consider a small stone thrown vertically into the air. 
• We can see that it’s velocity goes on decreasing. 
• After a while, at a point, the velocity finally becomes zero.  Then it starts to fall back. 
• While falling back, it’s velocity goes on increasing. 
• It’s velocity will be maximum at the time when it hits the ground. 

Let us analyse the above.
• In the first case, when the stone was let go, it immediately fell to the ground. Instead of floating around in the air or moving upwards. 
• In the second case, the stone went on decreasing it’s speed while ascending.
And while descending, went on increasing it’s speed. As if to reach the ground as early as possible. 
• In both the above cases, it looks as if the stone does not want to leave the surface of the earth.

In fact, it is not the wish of the stone, that is being fulfilled here. It is the wish of the earth. The earth attracts every object towards it’s centre. So the stone is not able to float away or rise high up. Note that the force of attraction is towards the centre of the earth. This is shown in the fig.3.1 below:

Fig.3.1

So, if the falling stone does not meet the ground, it will continue it’s fall till it reaches the centre of the earth.

---

• Now we consider another aspect of this force of attraction. The earth attracts the stone. Does the stone attract the earth?

• Indeed it does. By the newton's third law, there will be equal and opposite forces. 
• So why doesn’t the earth move towards the stone?
The following steps give the answer:
1. Let F_ES be the force with which the earth attracts the stone. We have seen that Force = mass acceleration. Details here. 
• Then F_ES = m_sa_s (Where m_s is the mass of stone and a_s is the acceleration with which the stone moves towards the earth)
2. Let F_SE be the force with which the stone attracts the earth
• Then F_SE = m_ea_e (Where m_e is the mass of earth and a_e is the acceleration with which the earth moves towards the stone)
3. These two forces are equal and opposite. So we write:
m_sa_s = -m_ea_e
• Here mass of earth m_e is enormous, while mass of stone m_s is negligibly small.
• So, to maintain the equality, a_s will be large, while a_e will be negligibly small.

---

• Thus it is clear that there is indeed a Force. A force of attraction. 

• When there is a force, there will be acceleration. That is why, the stone ‘accelerates’ towards the earth. That is., it increases it’s velocity continuously.

■ Now, if the earth attracts every object towards it’s centre, what about the moon? Does the earth attract the moon towards it’s centre?
• The answer is yes. The earth attracts the moon too. And the moon in turn attracts the earth.
• This force of attraction causes the sea level to rise up. We call it: the ‘high tide’. 
• But we do not see the moon falling onto the earth. We will see the reason when we learn about circular motion.

■ The sun attracts all the planets which orbits around it.
■ In fact, there is a force of attraction between every object in the universe. 
■ There is a force of attraction even between two small wooden blocks placed on top of a table. It is a universal phenomenon. 

■ Sir Isaac Newton studied about this phenomenon and published his findings in the form of a law. It is the Universal Law of Gravitation. Let us see the details about this law:
• All bodies in the universe attract each other
• The force of this attraction between two bodies is
    ♦ Directly proportional to the product of their masses
    ♦ Inversely proportional to the square of the distance between them

Let us put it into a mathematical form:
In the fig.3.2 below, masses of the two bodies are m₁ and m₂. The distance between their centres is d

Fig.3.2

Based on this fig., we can derive an equation for the force of attraction F between any two bodies:

Unit of G
Eq. 3.1 above can be written as:

Now we can obtain the unit of 'G' as follows:
• Unit of force is newton N
• Unit of distance is m
• Unit of mass is kg
• So the unit of G will be:

■ The value of G was calculated by the British scientist Henry Cavendish. The accepted value of G is 6.673 × 10^-11 N m² kg^-2

---

We will now see some solved examples

Solved example 3.1
A child of mass 40 kg is sitting at a distance of 1 m from another child of mass 50 kg. Calculate the gravitational force of attraction between them
Solution:
1. m₁ = 40 kg • m₂ = 50 kg • d = 1 m • G = 6.7 × 10^-11 N m² kg^-2.
2. We have: F = G(^m₁m₂⁄_d²) = 6.7 × 10^-11 × (^40×50⁄_1²) = 13400 × 10^-11 = 1.34 × 10^-7 = 0. 000000134 N.
3. We can see that it is a very small force. Such a small force cannot overcome the frictional resistance and other forces. So two children sitting close to each other do not come closer to each other due to the attractive force between them. 

Solved example 3.2
A body of mass 50 kg and another body of mass 60 kg are separated by a distance of 2 m. What is the force of attraction between them?
Solution:
1. m₁ = 50 kg • m₂ =650 kg • d = 2 m • G = 6.7 × 10^-11 N m² kg^-2.
2. We have: F = G(^m₁m₂⁄_d²) = 6.7 × 10^-11 × (^50×60⁄_2²) = 5025 × 10^-11 = 0.5025 × 10^-7 = 0. 00000005025 N.

Solved example 3.3
The mass of the earth is 6 × 10²⁴ kg and that of the moon is 7.4 × 10²² kg. If the distance between the earth and the moon is 3.84 × 10⁵ km, calculate the force exerted by the earth on the moon. G = 6.7 × 10^-11.
Solution:
1. Mass of the earth, m₁ = 6 × 10²⁴ kg
2. Mass of the moon, m₂ = 7.4 × 10²² kg 
3. Distance between earth and the moon, d = 3.84 × 10⁵ km = 3.84 × 10⁵ ×1000 m = 3.84 × 10⁸ m
4. The force can be calculated as follows: 

---

Importance of the Universal law of gravitation
The universal law of gravitation successfully explained several phenomena that we see in the universe:
• The force that binds us to the earth. If there was no gravitational force, we would not be able to walk on the surface of the earth. We will all be floating.
• The motion of the moon around the earth.  If there was no gravitational force, the moon would not be orbiting around the earth in a definite orbit. It would travel away in a straight line 
• The motion of the planets around the sun.  If there was no gravitational force, the planets also would not be orbiting around the sun.
• The tides due to the moon and the sun.  If there was no gravitational force, there would not be any high tides or low tides

---

In the next section, we will see Acceleration due to Gravity. 

PREVIOUS      CONTENTS       NEXT

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

Chapter 1.9 - Solved examples on Motion of Objects

In the previous section, we completed the discussion on the motion of objects. In this section we will see some solved examples.

Solved example 1.16
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
1. Diameter of the circular track = d = 200 m. So radius = r = 100 m
2. Perimeter of the track = 2πr = πd = 200π m
3. Time for completing one round = t = 40 s
4. Speed of the athlete = ^distance⁄_time  = ^πd⁄_t  = ^200π⁄₄₀ = 5π m s^-1.
5. So the athlete covers a distance of 5π m in 1 s. We have to find the distance that the athlete covers in 2 mins and 20 s
6. 2 mins 20 s = 140 s
7. So the distance covered in this time = 140 × 5π = 700π m
8. Now we have to find the displacement at the end of 140 s
9. When the athlete covers one perimeter 200π, from the starting point, he reaches back to his starting point. At that time, his displacement is zero
10. In this way, when he travels 700π, he reaches back the starting point 3 times. After the third time he travels a 'certain distance' to make the 700π. This 'certain distance is the displacement'. It can be calculated as follows:
11. 700π = 3 × 200π + 100π.
So the 'certain distance' = displacement = 100π.

Solved example 1.17
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Solution:
Case 1. From A to B:
1. Distance = 300 m
2. Time = 2 mins 30 s = 150 s
3. Speed = ^distance⁄_time = ³⁰⁰⁄₁₅₀ = 2 m s^-1.
4. Velocity = ^displacement⁄_time 
• Here displacement = 300 m
• So velocity = ^displacement⁄_time = ³⁰⁰⁄₁₅₀ = 2 m s^-1.
Case 2. From A to C
1. Distance = (300 m + 100 m) = 400 m
2. Time = (2 mins 30 s + 1 min) = 3 mins 30 s = 210 s
3. Speed = ^distance⁄_time = ⁴⁰⁰⁄₂₁₀ = 1.904 m s^-1.
4. Velocity = ^displacement⁄_time 
• Here displacement = 200 m
• So velocity = ^displacement⁄_time = ²⁰⁰⁄₂₁₀ = 0.952 m s^-1.

Solved example 1.18
Abdul, while driving to school, computes the average speed for his trip to be 20 kmph. On his return trip along the same route, there is less traffic and the average speed is 30 kmph. What is the average speed for Abdul’s trip?  
Solution:
1. Let the distance from home to school be 's'
2. Let the time required in the morning be t1
3. Then Abdul calculated his average speed in the morning to be 20 kmph in the following way:
Speed = ^distance⁄_time = ^s⁄_t1 = 20 kmph
4. From this we get: t1 = ^s⁄₂₀ .
5. In the evening he travels the same distance 's'
6. Let the time required in the evening be t2
7. Then he calculated his average speed in the evening to be 30 kmph in the following way:
Speed = ^distance⁄_time = ^s⁄_t2 = 30 kmph
8. From this we get: t2 = ^s⁄₃₀ .
9. So total time required for the travel = (^s⁄₂₀ + ^s⁄₃₀) = ^5s⁄₆₀.
10. Total distance = s + s = 2s
11. So average speed = ^{total distance}⁄_{total time} = 2s ÷ ^5s⁄₆₀ = 2s × ⁶⁰⁄_5s = 24 kmph

Solved example 1.19
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^-2 for 8.0 s. How far does the boat travel during this time? 
Solution:
1. The motorboat starts from rest. So initial velocity u = 0
• Time of travel = 8 s
• Acceleration = 3 m s^-2.
2. We want the distance s
The second equation of motion connects the above quantities. So we will use it:
3. s = ut + ¹⁄₂ at² ⇒ s = 0 × 300 + ¹⁄₂ × 3 × 8² ⇒ s = 96 m

Solved example 1.20
A driver of a car travelling at 52 kmph applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 30 kmph in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
The fig. shows the speed-time graph for the two cars:

1. The yellow line AB is the speed-time graph of the first car. The purple line PQ is the speed-time graph of the second car.
Note that, the speeds are converted from kmph to m s^-1. 
2. The area enclosed by a speed-time graph and the time axis will give the distance travelled by the object. We saw the details here. The areas are shown shaded in the above fig.
3. First car:
Area = Area of the triangle OAB = ¹⁄₂ × 5 × 14.44 = 36.1 m
[ ∵ Base = 5 units and altitude = 14.44 units]
4. Second car : 
Area = Area of the triangle OPQ = ¹⁄₂ × 10 × 8.33 = 41.65 m
[ ∵ Base = 10 units and altitude = 8.33 units]

Solved example 1.21
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Solution:
1. Height of fall = distance of travel = s = 20 m
2. Acceleration = a = 10 m s^-2.
3. The ball falls from rest. So it has an initial velocity u = 0
4. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
5. So we can write:
v² = u² + 2as ⇒ v² = 0² + 2 × 10 × 20 ⇒ v² = 400 ⇒ v  = 20 m s^-1.
6. To find the time, we can use the first equation of motion:
v = u + at ⇒ 20 = 0 + 10t ⇒ 20 = 10 t ⇒ t = 2 s

Solved example 1.22
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
1. Radius of the orbit = r = 42250 km
2. Perimeter of the orbit = 2πr = 2 × π × 42250 = 84500π km = 84500000 m
3. Time for completing one round revolution = t = 24 hours = 24 × 60 × 60 = 86400 s
4. Speed of the satellite = ^distance⁄_time  =  ^84500000π⁄₈₆₄₀₀ = 3070.95 m s^-1

---

So we have completed the discussion on the 'Motion of objects'. In the next chapter, we will see 'Forces acting on objects'. 

PREVIOUS      CONTENTS       NEXT

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