Chapter 5.1 - Newton's First Law

In the previous section we saw the 'property of inertia' which was put forward by Galileo. In this section, we will see how Sir Isaac Newton based his studies on that concept.

• The three laws put forward by Newton laid the foundation of Mechanics. 
    ♦ The law of inertia put forward by Galileo was the starting point of newton's works.
■ Newton's first law states that:
Every body continues to be in it's state of rest or of uniform motion in a straight line unless compelled by an external force to act otherwise.
• The law is written in one sentence. But it contains many points. 
• We can elaborate it by the following 4 steps: 
1. A body can be in any one of the following two states:
    ♦ The body can be at the state of rest
    ♦ The body can be at the state of uniform motion in a straight line
2. Which ever be the state, it will continue to be in that state
3. If a 'change in it's state' is required, we must apply an external force on that body 
4. As we have seen in the previous section,
    ♦ If the external forces acting on the body cancel each other, the 'change in state' will not be achieved' 
    ♦ To achieve the 'change in state', there must be a net external force.

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■ Now, if a body is at rest, it means that, it has no acceleration
■ If the body is in uniform motion, then also it means that, it has no acceleration
We can all the above information symbolically:
    ♦ No net force ⇒ The body remains at rest ⇒ No acceleration
    ♦ No net force ⇒ The body remains at uniform motion ⇒ No acceleration
[The symbol '⇒' stands for 'implies']
• We can connect the first and the third. We get:
    ♦ No net force ⇒ No acceleration
■ That is., if no net force act on a body, that body will have zero acceleration
• We can write the converse: 
If we want a body to have no acceleration, we must not apply a net force on it

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From the above discussion, the following points become clear:
■ If we apply a net force on a body, it will experience acceleration 
We can write the converse of this also:
■ If a body experiences an acceleration, then a net force is acting on the body
Let us see some practical applications of the above information:
Example 1:
1. A child pulls a non-electric toy car by a string 
See fig.5.3(a) below:

Fig.5.3

• Let the toy car move with uniform velocity
• Then the car experience zero acceleration
2. Based on Newton's First law, we can write:
■ No net force is acting on the car
3. But we must analyse the situation and write the reason for 'no net force and hence no acceleration '
• We see that the child is applying a force through the string. Even then we say that, net force is zero. 
• So what happened to the force applied through the string?
• The answer is that, the 'force applied through the string' is canceled by 'another force in the opposite direction'.
4. This 'another force' is the frictional force which is acting at the interface between the tyres and the floor
• The child tries to pull the car forward
• The frictional force oppose this forward motion
5. The two forces obviously have opposite directions
• So if the two forces are equal in magnitude, they will cancel each other
• If the child's pull is greater in magnitude, then the car will move with acceleration.
■ We can give the inference by writing the three statements below:
1. The car is observed to be moving with uniform velocity. 
2. So the net external force on it must be zero
3. By Newton's first law, we conclude that, the following two  forces are equal in magnitude but opposite in direction:
(a) Force applied by the child
(b) Frictional force at the interface between the wheels and the floor

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■ We must not write the inference in this way:
1. The following two  forces are equal in magnitude but opposite in direction:
(a) Force applied by the child
(b) Frictional force at the interface between the wheels and the floor
2. The two forces cancel each other 
3. So we observe the car to be moving with uniform velocity
■ Why are we not able to write the inference by the above three statements?
Ans: When beginning to solve a problem, we may not be knowing all of the following items:
(i) The forces which are acting
(ii) The magnitudes of the forces
(iii) The directions of the forces
• It is by the 'application of Newton's first law', that we make an inference:
The frictional force is equal in magnitude but opposite in direction to the applied force

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Example 2:
1. A book rests on a table. See fig.5.3(b) above
• It has no motion at all
• Then the book has zero acceleration
2. Based on Newton's First law, we can write:
■ No net force is acting on the book
3. But we must analyse the situation and write the reason for 'no acceleration and hence no net force'
• In example 1 above, we see a 'visible force' which is applied through a sting. 
• We asked: What happened to that force?
• And we found the answer
• But here, we see no 'visible force'
• We are inclined to conclude that:
The case of 'no net force' need not be considered here. Because no force is acting
4. But on all objects on earth, the gravitational force is acting. 
• Because of this force, the book will be pulled down wards. 
• The magnitude of this force is equal to 'W', the weight of the book
5. So now we have to consider 'net force'
• We have a force 'W' acting on the book
• Yet we see no acceleration
• What happened to 'W'?
• The answer is that, 'W' is cancelled by 'another force in the opposite direction'.
6. This 'another force' is the reaction 'R' exerted by the table
• It is easy to see that, if the table is not present to provide 'R', the book will fall 
• The 'W' tries to pull the book downwards
• The 'R' opposes this downward motion
■ We can give the inference by writing the three statements below:
1. The book is observed to be at rest. 
2. So the net external force on it must be zero
3. By Newton's first law, we conclude that, the following two forces are equal in magnitude but opposite in direction:
(a) The weight W of the book
(b) The reaction R exerted by the table

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■ We must not write the inference in this way:
1. The following two  forces are equal in magnitude but opposite in direction:
(a) The weight W of the book
(b) The reaction R exerted by the table
2. The two forces cancel each other.
3. So we observe the book to be at rest
■ Why are we not able to write the inference by the above three statements?
Ans: When beginning to solve a problem, we may not be knowing all of the following items:
(i) The forces which are acting
(ii) The magnitudes of the forces
(iii) The directions of the forces
• It is by the 'application of Newton's first law', that we make an inference:
The Reaction R is equal in magnitude but opposite in direction to the weight W

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■ We see that the property of inertia put forward by Galileo is contained in Newton's first law
• We experience inertia in many day to day situations
Let us see an example. We will write it in steps: 
1. Consider a person standing inside a bus
• Initially, the bus is at rest
• When the bus starts to move forward, the person tends to fall backwards
2. This can be explained as follows: 
• The feet are in contact with the floor. 
• When the floor moves forward the feet (due to inertia), would want to stay at rest.
• The feet would not want to move with the floor. 
3. But the friction (between the feet and the floor) will not allow the feet to stay at rest. 
• It will carry the feet forward. 
4. This motion of the feet should carry the entire body forward. 
• But the human body is somewhat flexible. It is not a rigid object. 
• So the upper parts will not experience the same motion of the feet.
5. The upper parts like to stay at rest, and somewhat succeeds in doing so. 
• But the feet is not present straight below to carry the upper part. It has moved forward. 
• So the person falls back
6. The opposite happens when the bus stops.
• The feet comes to stop due to friction. 
• The upper parts tend to continue being in the state of motion. 
• But the feet has stopped moving. There is no feet straight below to carry the upper part. 
• So the person falls forward

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In the next section, we will see how Newton's second law.

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Chapter 5 - Laws of Motion

In the previous chapter we completed a discussion on motion in two dimensions. In this chapter, we will see Laws of motion.

• In the preceding chapters we saw the details about velocity and acceleration of objects. 
• Based on the knowledge about velocity and acceleration, we were able to write the details about position and displacement of objects.  
• We saw that:
    ♦ If the velocity is constant, the body will not be having any acceleration 
    ♦ If the velocity changes, the body will be having an acceleration/retardation
• If we say that 'a body is in motion', it implies that it has a velocity 
■ But how did the body attain that velocity?
■ Further, 'if that velocity is not uniform', how did the body attain the acceleration/retardation?
In this chapter we try to answer those questions. Let us consider some simple cases:
1. We see a football in motion. We can certainly guess how is attained its velocity 
• Someone must have clicked it.
    ♦ That is., someone applied a force on it
2.  We see a stone rising in the air
• Clearly, someone must have thrown it
    ♦ That is., someone applied a force on it.  
3. We see the branches of a tree swinging
• The breeze must be causing them to swing
    ♦ That is., the breeze applied a force on it.  
■ We can easily conclude that a force is required to create motion

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■ It may be noted that in the examples above, the forces which caused the motion were in physical contact with the objects
1. To kick a ball, the player has to make physical contact with the ball
2. To throw a stone, the thrower has to make physical contact with the stone
3. To swing the branches, the air has to make physical contact with the branches 
■ But there are cases where such physical contact is not required. Let us see some examples:
1. Consider a stone dropped from the top of a building
• The person who was holding the stone just dropped it.
    ♦ He did not apply any force 
• Yet we see that stone is moving downwards 
• We are inclined to think that no force is required for producing motion in this case 
• But the truth is that, force is be applied in this case also.
    ♦ It is the force applied by Earth. It is called the gravitational force. We will learn about this force in the next chapter
• If gravitational force was absent, the stone would not fall.
• This will be clear if we consider dropping the stone in outer space. Let us see how:
• Consider a person holding the stone in outer space
• He just lets it go without applying any force
    ♦ He does not apply any force in the horizontal direction 
    ♦ He does not apply any force in the vertical direction (upwards or downwards)
    ♦ He does not apply any force in any inclined directions 
• He just let it go. Then the stone will not fall. In fact, the stone will not move in any direction.
• This is clear proof that, a force is required to cause motion.   
2. Consider an iron nail lying on the table.
• If we bring a bar magnet near the nail, then the nail will move towards the magnet
• Here no physical contact is made. Yet the magnetic force is able to give motion to the nail
■ So we can write: Gravitational and magnetic forces are capable of setting objects in motion, with out physical contact
■ The overall conclusion is: A force is required to start a motion

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• Thus we know how an object begins to move. 
■ Our next aim is to get the 'details about an object which is already in motion'. 
• We see a body in motion. 
    ♦ When will it come to rest? 
    ♦ How will it come to rest?
Let us see an example. We will write it in steps: 
1. Consider a ball rolling on the floor. 
•  The ball will come to a stop when the 'energy available for the motion' is completely used up
2. What if there is no agent to use that energy?
• In other words, What if the ball experience no resistance against it's motion?
• Then it would mean that, the ball has no work to do. 
■ All it's available energy will remain as such and it will continue to move for ever
• In such a situation is encountered, we will have to apply an external force to stop the ball
3. This will be clear if we consider a scenario in outer space.
• Let a ball be kicked in outer space. 
• It will continue it's motion for eternity. 
• It will never come to a stop unless acted upon by an external stopping force. 
4. But on earth, we see that the ball comes to a stop after some time, even if we see no external stopping force. 
• It is not true that there is no stopping force acting on the ball. 
• There is indeed a stopping force. We just do not notice it. 
5. It is the frictional force. It is a force acting at the interface between the ball and the ground. 
• We will learn more about frictional force later in this section. 
• At present it is sufficient to know that, the ball comes to a stop because of frictional force. The energy of the moving ball is gradually used up to do work against friction. 
• If this frictional force was completely absent, the ball would roll on forever just like it would, in outer space
6. Let us see a real life example of the scenario in (3):
• The voyager 1 space probe launched by NASA in 1977 is still travelling at a very high speed of 15 km/s. (Details here) 
• It is not using any propulsion for the travel. 
■ It is still moving because, there is no external force available to stop it. 
• Also note that, it is not moving in any random direction. It was set to move in a definite direction by giving it an initial velocity in that direction. 
• The following two points are worth noting:
(i) Voyager has not deviated from it's intended path
(ii) It is continuing to travel at a uniform velocity away from the earth
• The  'continuity of motion' and 'consistency in direction' are accomplished without any help from it's rocket engines
• If a change in direction becomes necessary, then those engines will have to be turned on

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■ Based on the above 6 steps, we can write: An external force is required to stop a body in motion
■ Earlier, we saw that, an external force is required to set a body in motion 
Now let us consider each of the above two points again. We must add one more information.
1. The first point:
• An external force is required to set a body in motion.
This is illustrated in fig.5.1(a) below:

Fig.5.1

• We see a wooden block at rest. It will continue to be at rest because no forces are acting on it.
    ♦ That is., no force is available to set the block in motion
• Now, we see another wooden block at rest. Based on what we have learned, we are inclined to think that no force is acting on the block. But it need not be true.
• Consider the wooden block in fig.5.1 (b) above. It is acted upon by several forces. But the block is at rest. How is that possible?
• The answer is that, the forces on the left cancel the effects of the forces on the right. 
• In other words, the net force on the block is zero. 
• That is., in effect, no force is acting on the block. That is why the block is at rest. 
• So we must analyze each situation carefully. An object being at rest does not necessarily mean that, no force is acting on it. But we can say no net force is acting on it.
• We must clearly distinguish between no force and no net force
2. The second point:
• An external force is required to stop a body in motion
This is illustrated in fig.5.2(a) below:

Fig.5.2

• We see a wooden block in motion in space. It will continue to be in motion because no forces are acting on it. 
    ♦ That is., no force is available to stop the block
• Now, we see another wooden block in motion in space. Based on what we have learned above, we are inclined to think that, no force is acting on the block. But it need not be true.
• Consider the wooden block fig.5.2(b) above. It is acted upon by several forces. But the block is in motion. How is that possible?
• The answer is that, the forces on the left cancel the effects of the forces on the right. 
• In other words, the net force on the block is zero. 
• That is., in effect, no force is acting on the block. That is why the block is in motion. 
• So we must analyze each situation carefully. An object being in uniform motion does not necessarily mean that, no force is acting on it. But we can say no net force is acting on it.
• We must clearly distinguish between no force and no net force.

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So we can write a summary:

■ If the net external force is zero, then:
(i) A body at rest continues to remain at rest
(ii) A body in motion continues to move with uniform velocity
■ This property of the body is called inertia.
• We can write: Inertia is a property of a body by which it resists the following two items:
(i) Any change in it's state of rest 
(ii) Any change in it's state of uniform motion   
• Because of this property, we encounter the following two situations:
(i) We have to apply a net external force to make a body (at rest) to move
(ii) We have to apply a net external force to make a body (in motion) to stop

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• It was the studies made by the Italian scientist Galileo Galilei that brought these phenomena to the notice of the world for the first time

• Later, the studies made by the English scientist Sir Isaac Newton gave more insight about them

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In the next section, we will see how Sir Isaac Newton developed those phenomena.

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Chapter 4.17 - River crossing Problems

In our present chapter, we are discussing about motions in two dimensions. In the previous section we completed a discussion on relative motion in two dimensions. In this section, we will see problems in river crossing.

Case 1:
1. Consider a man swimming across a river
• The velocity of the man is $\mathbf\small{\vec v_M}$
• The velocity of the river current is $\mathbf\small{\vec v_R}$
2. The $\mathbf\small{\vec v_R}$ is in a direction perpendicular to $\mathbf\small{\vec v_M}$
• So the man will not be able to swim straight ahead. He will be deviated. 
• This is shown in fig.4.45(a) below:

Fig.4.45

3. The new direction will be the direction of $\mathbf\small{\vec v}$, which is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$.
5. Let us apply the above results to an actual case. It is shown in fig.4.45(b)
• The yellow lines are the river banks. The width of the river is w
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
• But due to the current, he will be travelling along $\mathbf\small{\vec v}$.
• Because of this change in direction, he will reach B'
■ The distance BB' is called drift. It is denoted by 'x'
7. If we know the details about $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$, we can easily calculate the details of $\mathbf\small{\vec v}$
• That is., we can find the magnitude $\mathbf\small{|\vec v|}$ and direction θ. We have
(i) $\mathbf\small{|\vec v|=\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}$
(ii) $\mathbf\small{\theta =\tan^{-1}\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
8. Once we find θ, we will be able to find some more useful details. Let us see what they are:
(a) The drift x
(i) Consider fig.b. A perpendicular B'C is dropped from B' to the other bank
(ii) In the right triangle ACB', we have:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}}$
(iii) So we get Eq.4.34: $\mathbf\small{x=\frac{w}{\tan \theta}}$
(iv) But tan θ is equal to $\mathbf\small{\frac{|\vec{v_M}|}{|\vec{v_R}|}}$ also. So we can write:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}=\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
From this we get Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
(b) The actual distance travelled:
(i) This is equal to AB'
(ii) Clearly, it is given by: $\mathbf\small{AB'=\sqrt{x^2+w^2}}$
(c) Time of travel T
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$ 
(ii) The distance covered is AB'
(iii) So time of travel $\mathbf\small{T=\frac{AB'}{|\vec{v}|}}$
(iv) Thus we get Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
(d) Another formula for 'x':
(i) Consider the horizontal travel alone
• The horizontal travel is with a uniform speed of $\mathbf\small{|\vec v_R|}$ 
• This travel has a duration of T
(ii) So we can write an equation for horizontal distance travelled as: 
Eq.4.37: x = $\mathbf\small{|\vec v_R| \times T}$

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Now let us see another case:
Case 2:
1. Consider a man swimming across a river
• The velocity of the river current is $\mathbf\small{\vec v_R}$
• The velocity of the man is $\mathbf\small{\vec v_M}$
2. This time, the man aims to a point on the upstream.
• That is., $\mathbf\small{\vec v_R}$ is not directed towards the exact opposite point. It is directed towards a point on the upstream
• Due to the river current $\mathbf\small{\vec v_R}$, he will be deviated from his intended path
■ If the direction of $\mathbf\small{\vec v_M}$ is adjusted carefully, he can achieve an interesting result:
• The resulting $\mathbf\small{\vec v}$ will be directed towards the exact opposite point on the other bank
    ♦ Thus he will reach the exact opposite point on the other bank
This is shown in fig.4.46(a) below:

Fig.4.46

3. The new direction will be the direction of $\mathbf\small{\vec v}$, which is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$
5. Let us apply the above results to an actual case. It is shown in fig.4.46(b)
• The yellow lines are the river banks. The width of the river is w.
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
■ If he swim directely towards B, he will be carried away frm B due to the current
• So he aims for a point on the upstream
• We want to know the angle θ which will enable him to reach B
7. For obtaining θ, we can use the following steps:
$\mathbf\small{\vec v}$ is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
Let us write the steps for obtaining $\mathbf\small{\vec v}$:
(i) Considering horizontal components:
$\mathbf\small{\vec v_x=[\vec v_{Mx}+\vec v_{Rx}]=[-(|\vec v_{M}|\cos \theta )\hat{i}+|\vec v_{R}|\hat{i}]=[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}}$
• The negative sign is due to the fact that, the horizontal component of $\mathbf\small{\vec v_M}$ is directed towards the left
(ii) Considering vertical components:
$\mathbf\small{\vec v_y=[\vec v_{My}+\vec v_{Ry}]=[(|\vec v_{M}|\sin \theta )\hat{j}+0]=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
• The zero value comes in because, $\mathbf\small{\vec v_R}$ has no vertical component
8. Now, $\mathbf\small{\vec v=0}$ does not have horizontal component. That means: $\mathbf\small{\vec v_x=0}$
So we can equate 7(i) to zero. We get:
$\mathbf\small{[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}=0}$
$\mathbf\small{\Rightarrow -(|\vec v_{M}|\cos \theta )+|\vec v_{R}|=0}$
$\mathbf\small{\Rightarrow |\vec v_{R}|=|\vec v_{M}|\cos \theta}$
$\mathbf\small{\Rightarrow \cos \theta=\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
So we get Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
■ Thus we successfully calculated θ
9. Let us continue and find the magnitude of $\mathbf\small{\vec v}$ also:
(i) The [vertical component of $\mathbf\small{\vec v}$] is $\mathbf\small{\vec v}$ itself. That means: $\mathbf\small{\vec v_y=\vec v}$
(ii) So we can equate 7(ii) to $\mathbf\small{\vec v}$. We get:
$\mathbf\small{\vec v=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec v|=|\vec v_{M}|\sin \theta}$
10. So, if we have 'sin θ', we can multiply it with $\mathbf\small{|\vec v_M|}$ to obtain $\mathbf\small{|\vec v|}$
(i) We have already obtained 'cos θ' in step 8. From that, we can easily calculate 'sin θ'
(ii) From math classes we know that $\mathbf\small{\sin \theta = \sqrt{1-\cos^2\theta }}$
So we get: $\mathbf\small{\sin \theta = \sqrt{1-\left( \frac{|\vec v_{R}|}{|\vec v_{M}|} \right )^2}}$
$\mathbf\small{\Rightarrow \sin \theta = \frac{\sqrt{(|\vec{v_M}|^2-|\vec{v_R}|^2})}{|\vec{v_M}|}}$
(iii) Thus, from the result in (9), we get:
$\mathbf\small{|\vec{v}| = |\vec{v_M}|\times \frac{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}{|\vec{v_M}|}}$
■ So we can write Eq.4.39: $\mathbf\small{|\vec{v}| =\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}$
11. Once we obtain $\mathbf\small{|\vec v|}$, we can calculate the time of travel
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$ 
(ii) The distance covered is AB = w
(iii) So time of travel $\mathbf\small{T=\frac{AB}{|\vec{v}|}}$
(iv) Thus we get Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$

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Now we will see some solved examples
Solved example 4.19
A man can row a boat at a speed of 4 km/h in still water. He is crossing a river where the speed of current is 2 km/h
(a) In what direction should he be headed if he wants to reach a point directly opposite to his starting point?
(b) If the width of the river is 4 km, how long will it take to reach the opposite bank if he heads in the direction derived in (a)?
(c) In what direction should he be headed if he wants to reach the opposite bank in the shortest possible time? How much is this shortest time?
Solution:
Part (a)
1. This comes under case 2 that we saw above
• We can use Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
2. Substituting the values, we get: $\mathbf\small{\theta=\cos^{-1}\frac{2}{4}}$ 
Thus θ = cos^-1 (0.5) = 60°.
■ So the man should row the boat in such a way that, his direction makes an angle of 60° with the bank on his left side
Part (b)
1. We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
2. Substituting the values, we get:  $\mathbf\small{T=\frac{4}{\sqrt{4^2-2^2}}=\frac{4}{\sqrt{12}}=\frac{2}{\sqrt{3}}=1.155\: \text{h}}$
Part (c)
1. The shortest possible time is achieved when the direction is headed exactly to the opposite point   
• So this is case 1
2. We can use Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
3. But, first we have to calculate the drift 'x'. We can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
• Substituting the values, we get: $\mathbf\small{x = \frac{4 \times 2}{4}=2\: \text{km}}$
4. Substituting in (2), we get: $\mathbf\small{T=\frac{\sqrt{2^2+4^2}}{\sqrt{4^2+2^2}}=\frac{\sqrt{20}}{\sqrt{20}}=1\: \text{h}}$
■ Note: To achieve the least possible time, the direction should be headed to the exact opposite point. Any other direction will take up a longer time.

Solved example 4.20
A man crosses a river in a boat. If he choose to cross within the least possible time, he can do so in 10 minutes. But there will be a drift of 120 m. If he choose to cross with the least possible distance, he can do so in 12.5 minutes. Find (a) Width of the river  (b) Speed of the boat (c) Speed of the river current
Solution:
1. Consider the 'crossing in least possible time'. This is case 1
• We can use Eq.4.37: drift = x = $\mathbf\small{|\vec v_R| \times T}$
2. Substituting the values, we get: 120 = $\mathbf\small{|\vec v_R| \times 10}$   
• So we get speed of the river current = $\mathbf\small{|\vec v_R|}$ = 12 m/min
3. Now we can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
Substituting the values, we get: $\mathbf\small{120 = \frac{w \times 12}{|\vec{v_M}|}}$ 
So we get: $\mathbf\small{\frac{w}{|\vec{v_M}|}=10}$
4. Now we consider 'crossing in least possible distance'. This is case 2
• We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
• Substituting the values, we get: $\mathbf\small{12.5=\frac{10\times|\vec{v_M}|}{\sqrt{|\vec{v_M}|^2-12^2}}}$
• Squaring both sides: $\mathbf\small{12.5^2 \times\left ( |\vec{v_M}|^2-12^2 \right )=10^2 \times |\vec{v_M}|^2}$
$\mathbf\small{\Rightarrow (12.5^2-10^2)\times |\vec{v_M}|^2=(12.5 \times 12)^2}$
• Solving this, we get $\mathbf\small{|\vec{v_M}|}$ = 20 m/min
5. Substituting this value in (3), we get: w = 200 m

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With this we complete our present discussion on two dimensional motion. In the next chapter, we will see laws of motion.

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Chapter 4.16 - Relative velocity in Two Dimensions

In our present chapter, we are discussing about motions in two dimensions. In the previous section we completed a discussion on uniform circular motion. In this section, we will see relative velocity in two dimensions.

We saw 'relative velocity in the case of motion in one dimension' in a previous chapter. Our present discussion is based on that.

1. Consider two objects A and B shown in fig.4.41(a) below:

Fig.4.41

• They are moving in two different directions and with two different magnitudes
    ♦ A is moving with a velocity of $\mathbf\small{\vec {v_A}}$
    ♦ B is moving with a velocity of $\mathbf\small{\vec {v_B}}$
2. We can write two cases:
(i) Case 1: When viewed from A, The other object B will be moving with a velocity of [$\mathbf\small{\vec{v_B}-\vec{v_A}}$]
• That is., Relative velocity of B with respect to A = $\mathbf\small{\vec{v_{BA}}=\vec{v_B}-\vec{v_A}}$  
(ii) Case 2: When viewed from B, The other object A will be moving with a velocity of [$\mathbf\small{\vec{v_A}-\vec{v_B}}$]
• That is., Relative velocity of A with respect to B = $\mathbf\small{\vec{v_{AB}}=\vec{v_A}-\vec{v_B}}$  
3. So we need to do vector subtraction to find relative velocities. We already know the methods to do such subtractions. We have both graphical and analytical methods.
■ Here we will use analytical method: 
• We use the horizontal and vertical components. Fig.4.41(b) shows those components. 
4. Consider the resultant $\mathbf\small{\vec{v_{BA}}}$ in case 1
    ♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Bx}}-\vec{v_{Ax}}}$
    ♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{By}}-\vec{v_{Ay}}}$
Once the components are obtained, we can easily calculate the resultant
• Consider the resultant $\mathbf\small{\vec{v_{AB}}}$ in case 2
    ♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Ax}}-\vec{v_{Bx}}}$
    ♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{Ay}}-\vec{v_{By}}}$
Once the components are obtained, we can easily calculate the original  

Solved example 4.16

Rain is falling vertically with a speed of 35 ms^-1. A woman rides a bicycle with a speed of 12 ms^-1 in east to west direction. What is the direction in which she should hold her umbrella ?  

Solution:

1. Let the rain fall be denoted by $\mathbf\small{\vec {v_R}}$ 

• Let the motion of bicycle be denoted by $\mathbf\small{\vec {v_B}}$ 

• The two vectors are shown in the fig.4.42(a) below:

Fig.4.42

2. We want to find 'how the velocity of the rain will appear, when viewed from the bicycle'

    ♦ That is., we want to find the velocity of the rain with respect to the bicycle. 

    ♦ That is., we want to find $\mathbf\small{\vec {v_{RB}}}$

    ♦ That is., we want to find $\mathbf\small{\vec{v_R}-\vec{v_B}}$   

3. For that, we need to find the following two:

(i) x component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Rx}}-\vec{v_{Bx}}}$ 

(ii) y component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Ry}}-\vec{v_{By}}}$ 

4. Let us write the values:

• $\mathbf\small{\vec{v_{Rx}}}$ = 0 (∵ the rain is falling vertically and so has no horizontal component)

• $\mathbf\small{\vec{v_{Bx}}}$ = -12 $\mathbf\small{\hat{i}}$ (The travel is from left to right. So it is taken as negative)

• $\mathbf\small{\vec{v_{Ry}}}$ = -35 $\mathbf\small{\hat{j}}$ (The travel is from top to bottom. So it is taken as negative)

• $\mathbf\small{\vec{v_{By}}}$ = 0 (∵ the bicycle is travelling horizontally and so has no vertical component)

5. Substituting the values in 3(i), we get:

• x component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{RBx}}}$ = [0 - (-12 $\mathbf\small{\hat{i}}$)] = 12 $\mathbf\small{\hat{i}}$
    ♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
    ♦ This is shown in fig.b

Substituting the values in 3(ii), we get:
• y component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{RBy}}}$ = [(-35 $\mathbf\small{\hat{i}}$ - 0)] = -35 $\mathbf\small{\hat{j}}$
    ♦ We get a negative value. So this vector is directed towards the negative side of the y axis
    ♦ This is also shown in fig.b

6. Now we can find the magnitude:
Magnitude of $\mathbf\small{\vec{v_R}-\vec{v_B}}$ = $\mathbf\small{|\vec{v_R}-\vec{v_B}|}$ = $\mathbf\small{|\vec{v_{RB}}|}$ = $\mathbf\small{\sqrt{12^2+(-35)^2}}$ = 37 ms^-1.
7. Next we find the direction:
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v{RBy}}|}{|\vec{v{RBx}}|}=\frac{35}{12}}$
• Thus we get: θ = 71.076^o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
8. But what we want is 'φ' which the resultant makes with the vertical
• It can be easily obtained because, in fig.c, we can see that φ = (90-θ) 
• Thus we get: φ = (90-71.076) = 18.924^o.
9. So we can write the result:
• When viewed from the bicycle, rain falls towards the cyclist at an angle of 18.924^o with the vertical
• So the cyclist must hold the umbrella at an angle of 18.924^o with the vertical,
10. We must note the difference between this example and example 4.1.
• In that example, the boy was stationary. The rain was falling at an angle because of the wind
• But here, the rain is falling vertically. But to the cyclist, it appears to be falling at an angle 

Solved example 4.17
In a harbour, wind is blowing at a speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
1. The speed and direction of the wind is given:
• Speed is 72 km/h and direction is N-E
2. Let us set up the coordinate axes as follows:
    ♦ E-W direction is the x axis
    ♦ N-S direction is the y axis
• Then the N-E direction (in which the wind is blowing) will be inclined at 45^o with the x axis.
• This is shown in fig.4.43(a) below:

Fig.4.43

3. Now we can write the details about the velocity vector of wind
• Let us denote it as $\mathbf\small{\vec {v_W}}$
• We can write:
    ♦ $\mathbf\small{|\vec {v_W}|}$ = 72 km/h
    ♦ $\mathbf\small{\vec {v_W}}$ makes an angle 45^o with the x axis
• This is also shown in fig.a
4. This velocity of the wind will not change
• But when viewed from a moving boat, the wind will appear to have a different magnitude and different direction

• We want to find 'how the velocity of the wind will appear, when viewed from the moving boat'

    ♦ That is., we want to find the velocity of the wind with respect to the moving boat. 

    ♦ That is., we want to find $\mathbf\small{\vec {v_{WB}}}$

    ♦ That is., we want to find $\mathbf\small{\vec{v_W}-\vec{v_B}}$ 
Where $\mathbf\small{\vec {v_B}}$ is the velocity of the boat

5. For that, we need to find the following two:

(i) x component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Wx}}-\vec{v_{Bx}}}$ 

(ii) y component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Wy}}-\vec{v_{By}}}$ 

6. Let us write the values:

• $\mathbf\small{\vec{v_{Wx}}}$ = 72 cos 45 $\mathbf\small{\hat{i}}$ = 50.912 $\mathbf\small{\hat{i}}$ km/h

• $\mathbf\small{\vec{v_{Bx}}}$ = 0 (∵ the boat is travelling north and so has no component towards east)
• $\mathbf\small{\vec{v_{Wy}}}$ = 72 sin 45 $\mathbf\small{\hat{j}}$ = 50.912 $\mathbf\small{\hat{i}}$ km/h
• $\mathbf\small{\vec{v_{By}}}$ = 51 $\mathbf\small{\hat{j}}$

7. Substituting the values in 5(i), we get:

• x component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{WBx}}}$ = [50.912 $\mathbf\small{\hat{i}}$ - 0] = 50.912 $\mathbf\small{\hat{i}}$
    ♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
    ♦ This is shown in fig.b

Substituting the values in 5(ii), we get:
• y component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{WBy}}}$ = [(72 sin 45 $\mathbf\small{\hat{j}}$ - 51 $\mathbf\small{\hat{j}}$)] = -0.0883
    ♦ We get a negative value. So this vector is directed towards the negative side of the y axis
    ♦ This is also shown in fig.b
8. Now we can find the magnitude. But it is not asked in the question
9. But we do have to find the direction
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v_{WBy}}|}{|\vec{v_{WBx}}|}=\frac{0.0883}{50.912}}$
• Thus we get: θ = 0.0997^o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
10. We see that, the relative velocity vector falls between the east and south directions
• But the deviation from east is only 0.0997^o, which is a very small quantity
• So we can write:
■ The flag flutters in the east direction approximately

Solved example 4.18
When a man walks at a rate of 3 km/h, the rain appears to fall vertically. When he walks at the rate of 6 km/h, it appears to fall at an angle of 45° with the vertical. What is the original magnitude and direction of the rain?
Solution:
1. Consider the simple case when the rain falls in the exact vertical direction
■ For a man moving from right to left, the rain will obviously appear to be falling at a slope
• This slope will be towards the right.
• That is., $\mathbf\small{\vec{v_{RM}}}$ will slope towards the right
    ♦ Where $\mathbf\small{\vec{v_{RM}}}$ is the relative velocity of the rain with respect to the man
    ♦ This is shown in fig.4.44(a) below:

Fig.4.44

2. But in our present case, even when there is motion at 3 km/h, the rain appears to be exact vertical
■ So it is clear that, the original rain is sloping towards the left. This is shown in fig.b
• We are asked to find this original magnitude and direction
• That is., we need to find the details of $\mathbf\small{\vec{v_{R}}}$ in fig.b
3. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.b are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
4. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{Mx}}=3 \hat{i}}$
• Substituting in 4(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 4(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
5. Now consider the $\mathbf\small{\vec{v_{RM}}}$ in fig.c
• It is perfect vertical. That means, it has no horizontal component
• That means, $\mathbf\small{\vec{v_{RMx}}}$ = 0
• From 4(i)b, we can write: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$ = 0
■ From this we get: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
This is one of the two equations which we will need to find $\mathbf\small{|\vec{v_{R}}|}$ and θ.
6. To get the other equation, we consider the motion at 6 km/h
• This time, the $\mathbf\small{\vec{v_{RM}}}$ is indeed at a slope. This is shown in fig.d
    ♦ Given that, it makes an angle of 45° with the vertical
    ♦ So it will make the same angle of 45° with the horizontal also
Note the two points:
(i) $\mathbf\small{\vec{v_{R}}}$ in fig.d is same as that in fig.c. This is because, the original speed and direction of rain does not change 
(ii) $\mathbf\small{\vec{v_{M}}}$ in fig.d has the same direction as that in fig.c But the magnitude changed from 3 to 6 km/h
7. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.d are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
8. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{Mx}}=6 \hat{i}}$
• Substituting in 8(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-6 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 8(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
9. Using the 'formula for direction of the resultant', we can write:
$\mathbf\small{\tan 45=\frac{|\vec{v_{RMy}}|}{|\vec{v_{RMx}}|}=\frac{||\vec{v_R}|\sin \theta| }{|(|\vec{v_R}|\cos \theta)-6|}}$
10. But in (5), we obtained: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
Substituting this in (9), we get: $\mathbf\small{\tan 45=\frac{||\vec{v_R}|\sin \theta| }{|(3)-6|}=\frac{||\vec{v_R}|\sin \theta| }{|(-3)|}=\frac{|\vec{v_R}|\sin \theta }{3}}$
11. But tan 45 = $\mathbf\small{\frac{1}{\sqrt{2}}}$
Thus we get: $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
12.So we have two results:
(i) $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
(ii) $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
Dividing (ii) by (i), we get: $\mathbf\small{\tan \theta =\frac{1}{\sqrt{2}}}$
Thus we get θ = 45°.
13. Substituting this value of θ in 12(i), we get:
$\mathbf\small{|\vec{v_{R}}|\cos 45=3}$
$\mathbf\small{\Rightarrow |\vec{v_{R}}|=\frac{3}{\frac{1}{\sqrt{2}}}=3\sqrt{2}=4.243\: km/h}$
14. So we can write:
• The rain is actually falling at an angle of 45° towards the left
• It has a magnitude of 4.243 km/h
• But when the man walks ar the rate of 6 km/h, the rain appears to be falling at an angle of 45° towards the right

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In the next section, we will see a few more solved examples.

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