## Laws of Motion Class 9 Science Notes Maharashtra State Board

**Displacement and Distance**

‘Distance’ is the length of the actual path traveled by an object in motion while going from one point to another, whereas displacement is the minimum distance between the starting and finishing points. Even if the displacement of an object is zero, the actual distance traversed by it may not be zero.

**Speed and Velocity**

The units of speed and velocity are the same. In the SI system, the unit is m/s while in the CGS system, it is cm/s. Speed is related to distance while velocity is related to displacement. If the motion is along a straight line, the values of speed and velocity are the same, otherwise, they can be different. Velocity is the displacement that occurs in unit time.

The distance traveled in a particular direction by an object in a unit of time is called its velocity. Here, unit time can be one second, one minute, one hour, etc. If large units are used, one year can also be used as a unit of time. The displacement that occurs in unit time is called velocity.

**Effect of Speed and Direction on Velocity**

Sachin is traveling on a motorbike. Explain what will happen in the following events during Sachin’s ride (see figure).

- What will be the effect on the velocity of the motorcycle if its speed increases or decreases, but its direction remains unchanged?
- In case of a turning on the road, will the velocity and speed be the same? If Sachin changes the direction of the motorcycle, keeping its speed constant, what will be the effect on the velocity?
- If, on a turning, Sachin changes the direction as well as the speed of the motorcycle, what will be the effect on its velocity? It is clear from the above that velocity depends on speed as well as direction and that velocity changes by
- changing the speed while keeping the direction the same.
- changing the direction while keeping the speed the same.
- changing the speed as well as the direction.

The first scientist to measure speed as the distance /time was Galileo. The speed of sound in dry air is 343.2 m/s while the speed of light is about 3 × 10^{8} m/s. The speed of revolution of the earth around the sun is about 29770 m/s.

**Uniform and non-uniform linear motion**

If an object covers unequal distances in equal time intervals, it is said to be moving with non-uniform speed. For example, the motion of a vehicle being driven through heavy traffic. If an object covers equal distances in equal time intervals, it is said to be moving with uniform speed.

**Acceleration**

1. Cut a 1m long plastic tube lengthwise into two halves.

2. Take one of the channel-shaped pieces. Place one of its ends on the ground and hold the other at some height from the ground as shown in the figure.

3. Take a small ball and release it from the upper end of the channel.

4. Observe the ball’s velocity as it rolls down along the channel.

5. Is its velocity the same at all points?

6. Observe how the velocity changes as it moves from the top, through the middle, and to the bottom.

You must have all played on a slide in a park. You know that while sliding down, the velocity is less at the top, it increases in the middle and becomes zero towards the end. The rate of change of velocity is called acceleration.

Acceleration = \(\frac{Change in Velocity}{Time}\)

If the initial velocity is ‘u’ and in time ‘t’ it changes to the final velocity ‘v’,

Acceleration (a) = \(\frac{\text { Final velocity }- \text { Initial velocity }}{\text { Time }}\)

∴ a = \(\frac{(v-u)}{t}\)

If the velocity of an object changes during a certain period, then it is said to have accelerated motion. An object in motion can have two types of acceleration.

- If the velocity changes by equal amounts in equal time intervals, the object is said to be in uniform acceleration.
- If the velocity changes by unequal amounts in equal time intervals, the object is considered non-uniform acceleration.

**Positive, Negative, and Zero Acceleration**

An object can have positive or negative acceleration. When the velocity of an object increases, the acceleration is positive. In this case, the acceleration is in the direction of velocity. When the velocity of an object decreases with time, it has negative acceleration. Negative acceleration is also called deceleration. Its direction is opposite to the direction of velocity. If the velocity of the object does not change with time, it has zero acceleration.

**Distance-Time Graph for Uniform Motion**

The following table shows the distances covered by a car in fixed time intervals. Draw a graph of distance against time taking ‘time’ along the X-axis and ‘distance’ along the Y-axis in the figure.

An object in uniform motion covers equal distances in equal time intervals. Thus, the graph between distance and time is a straight line.

**Distance-Time Graph for Non-Uniform Motion**

The following table shows the distances covered by a bus in equal time intervals. Draw a graph of distance against time taking the time along the X-axis and distance along the Y-axis in figure. Does the graph show a direct proportionality between distance and time?

Here, the distance changes non-uniformly with time. Thus, the bus has non-uniform motion.

**Velocity-time graph for uniform velocity**

A train is moving with a uniform velocity of 60 km/hour for 5 hours. The velocity-time graph for this uniform motion is shown in the figure.

- With the help of the graph, how will you determine the distance covered by the train between 2 and 4 hours?
- Is there a relation between the distance covered by the train between 2 and 4 hours and the area of a particular quadrangle in the graph? What is the acceleration of the train?

**Velocity-Time Graph for Uniform Acceleration**

The changes in the velocity of a car in specific time intervals are given in the following table.

The velocity-time graph in the figure shows that,

- The velocity changes by equal amounts in equal time intervals. Thus, this is uniformly accelerated motion. How much does the velocity change every 5 minutes?
- For all uniformly accelerated motions, the velocity-time graph is a straight line.
- For non-uniformly accelerated motions, the velocity-time graph may have any shape depending on how the acceleration changes with time.

From the graph in the figure, we can determine the distance covered by the car between the 10th and the 20th seconds as we did in the case of the train in the previous example. The difference is that the velocity of the car is not constant (unlike that of the train) but is continuously changing because of uniform acceleration. In such a case, we have to use the average velocity of the car in the given time interval to determine the distance covered in that interval.

From the graph, the average velocity of the car = \(\frac{32+16}{2}\) = 24 m/s

Multiplying this by the time interval, i.e. 10 seconds gives us the distance covered by the car.

Distance covered = 24 m/s × 10 s = 240 m

Check that, similar to the example of the train, the distance covered is given by the area of quadrangle ABCD.

A(ABCD ) = A(AECD ) + A(ΔABE)

**Equations of Motion Using Graphical Method**

Newton studied the motion of an object and gave a set of three equations of motion. These relate to the displacement, velocity, acceleration, and time of an object moving along a straight line. Suppose an object is in motion along a straight line with initial velocity ‘u’. It attains a final velocity ‘v’ in time ‘t’ due to acceleration ‘a’ its displacement is ‘s’. The three equations of motion can be written as

v = u + at – This is the relation between velocity and time.

s = ut + \(\frac{1}{2}\) at^{2} – This is the relation between displacement and time.

v^{2} = u^{2} + 2as – This is the relation between displacement and velocity.

Let us try to obtain these equations by the graphical method.

**The equation describing the relation between Velocity and Time**

Figure shows the change in velocity with time of a uniformly accelerated object. The object starts from point D in the graph with velocity u. Its velocity keeps increasing and after time t, it reaches the point B on the graph.

The initial velocity of the object = u = OD

The final velocity of the object = v = OC

Time = t = OE

Acceleration (a) = \(\frac{Change in Velocity}{Time}\)

= \(\frac{\text { (Final velocity – Initial velocity) }}{\text { Time }}\)

= \(\frac{(\mathrm{OC}-\mathrm{OD})}{\mathrm{t}}\)

∴ CD = at ……..(i) (OC – OD = CD)

Draw a line parallel to the Y axis passing through B. This will cross the X-axis in E. Draw a line parallel to the X-axis passing through D. This will cross the line BE at A.

From the graph, BE = AB + AE

∴ v = CD + OD (AB = CD and AE = OD)

∴ v = at + u …………(from i)

∴ v = u + at

This is the first equation of motion.

**The equation describing the relation between displacement and time**

Let us suppose that an object in uniform acceleration ‘a’ and has covered the distance ‘s’ within time ‘t’. From the graph in the figure, the distance covered by the object during the time ‘t’ is given by the area of quadrangle DOEB.

∴ s = area of quadrangle DOEB

∴ s = area (rectangle DOEA) + area of triangle (DAB)

∴ s = (AE × OE ) + (\(\frac{1}{2}\) × [AB × DA])

But, AE = u, OE = t and (OE = DA = t)

AB = at …….( AB = CD ) …..from (i)

∴ s = u × t + \(\frac{1}{2}\) × at × t

∴ Newton’s second equation of motion is s = ut + \(\frac{1}{2}\) at^{2}

**The equation describing the relation between displacement and velocity**

We have seen that from the graph in the figure we can determine the distance covered by the object in time t from the area of the quadrangle DOEB. DOEB is a trapezium. So we can use the formula for its area.

∴ s = area of trapezium DOEB

∴ s = \(\frac{1}{2}\) × sum of lengths of parallel sides × distance between the parallel sides

∴ s = \(\frac{1}{2}\) × (OD + BE ) × OE

But, OD = u, BE = v and OE = t

∴ s = \(\frac{1}{2}\) × ( u + v) × t ………(ii)

But a = \(\frac{(v-u)}{t}\)

∴ t = \(\frac{(v-u)}{a}\) ……..(iii)

∴ s = \(\frac{1}{2} \times(u+v) \times \frac{(v-u)}{a}\)

∴ s = \(\frac{(\mathrm{v}+\mathrm{u})(\mathrm{v}-\mathrm{u})}{2 \mathrm{a}}\)

∴ 2as = (v + u) (v – u) = v^{2} – u^{2}

∴ v^{2} = u^{2} + 2as

this is Newton’s third equation of motion.

The velocity of an accelerated object changes with time. Change in the velocity can be due to a change in direction or magnitude of the velocity or both.

**Uniform Circular Motion**

The speed of the tip is constant, but the direction of its displacement and therefore, its velocity is constantly changing. As the tip is moving along a circular path, its motion is called uniform circular motion.

- Draw a rectangular path as shown figure.
- Place the tip of your pencil on the middle of any side of the square path and trace the path.
- Note how many times you have to change the direction while tracing the complete path.
- Now repeat this action for a pentagonal, hexagonal, or octagonal path, and note the number of times you have to change direction.
- If you increase the number of sides of the polygon and make it infinite, how many times will you have to change the direction? What will be the shape of the path? This shows that as we increase the number of sides, we have to keep changing direction more and more times. And when we increase the number of sides to infinity, the polygon becomes a circle.

When an object is moving at a constant speed along a circular path, the change in velocity is only due to the change in direction. Hence, it is accelerated motion. When an object moves with constant speed along a circular path, the motion is called uniform circular motion, e.g. the motion of a stone in a sling or that of any point on a bicycle wheel when they are in uniform motion.

If an object, moving along a circular path of radius ‘r’, takes time ‘t’ to come back to its starting position, its speed can be determined using the formula given below:

Speed = \(\frac{\text { Circumference }}{\text { Time }}\)

v = \(\frac{2 \pi r}{t}\), r = radius of the circle

**Determining the direction of velocity in uniform circular motion**

Take a circular disc and put a five rupee coin at a point along its edge. Make it move around its axis by putting a pin through it. When the disc is moved at a higher speed, the coin will be thrown off as shown in the figure. Note the direction in which it is thrown off. Repeat the action of placing the coin at different points along the edge of the circle and observe the direction in which the coin is thrown off.

The coin will be thrown off in the direction of the tangent which is perpendicular to the radius of the disc. Thus, the direction in which it gets thrown off depends on its position at the moment of getting thrown off. It means that, as the coin moves along a circular path the direction of its motion changes at every point.

### Newton’s Laws of Motion

What could be the reason for the following?

- A static object does not move without the application of a force.
- The force that is sufficient to lift a book from a table is not sufficient to lift the table.
- Fruits on a tree fall when its branches are shaken.
- An electric fan keeps on rotating for some time even after it is switched off.

If we look for reasons for the above, we realize that objects have some inertia. We have learned that inertia is related to the mass of the object. Newton’s first law of motion describes this very property and is therefore also called the law of inertia. All instances of inertia are examples of Newton’s first law of Motion.

**Newton’s First Law of Motion**

Balanced and unbalanced force: You must have played tug-of-war. So long as the forces applied by both sides are equal, i.e. balanced, the center of the rope is static despite the applied forces. On the other hand, when the applied forces become unequal, i.e. unbalanced, a net force gets applied in the direction of the greater force, and the center of the rope shifts in that direction.

‘An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.’ When an object is at rest or in uniform motion along a straight line, it does not mean that no force is acting on it. Several forces are acting on it, but they cancel one another so that the net force is zero. Newton’s first law explains the phenomenon of inertia, i.e. the inability of an object to change its state of motion on its own. It also explains the unbalanced forces that cause a change in the state of an object at rest or in uniform motion.

**Newton’s Second Law of Motion**

The effect of one object striking another object depends both on the mass of the former object and its velocity. This means that the effect of the force depends on a property related to both the mass and velocity of the striking object. This property was termed ‘momentum’ by Newton.

Momentum has magnitude as well as direction. Its direction is the same as that of velocity. In the SI system, the unit of momentum is kg m/s, while in the CGS system, it is g cm/s. If an unbalanced force applied on an object causes a change in the velocity of the object, then it also causes a change in its momentum. The force necessary to cause a change in the momentum of an object depends upon the rate of change of momentum.

Momentum (P): Momentum is the product of the mass and velocity of an object.

P = mv

Momentum is a vector quantity.

‘The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.’

Suppose an object of mass m has an initial velocity u. When a force F is applied in the direction of its velocity for time t, its velocity becomes v.

∴ The initial momentum of the object = mu,

Its final momentum after time t = mv

∴ Rate of change of momentum = \(\frac{\text { Change in momentum }}{\text { Time }}\)

∴ Rate of change of momentum = \(\frac{\mathrm{mv}-\mathrm{mu}}{\mathrm{t}}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\) = ma

According to Newton’s second law of motion, the rate of change of momentum is proportional to the applied force.

∴ ma ∝ F

∴ F = k ma (k = Constant of proportionality and its value is 1)

F = m × a

Consider two objects having different masses which are initially at rest. The initial momentum for both is zero. Suppose a force ‘F’ acts for time ‘t’ on both objects. The lighter object starts moving faster than the heavier object. However, from the above formula, we know that the rate of change of momentum i.e. ‘F’ in both objects is the same and the total change in their momentum will also be the same i.e. ‘Ft’. Thus, if the same force is applied to different objects, the change in momentum is the same.

In the SI system, the unit of force is Newton.

Newton (N): The force necessary to cause an acceleration of 1 m/s^{2} in an object of mass 1 kg is called 1 Newton.

1 N = 1 kg × 1 m/s^{2}

In the CGS system, the unit of force is a dyne.

Dyne: The force necessary to cause an acceleration of 1 cm/s^{2} in an object of mass 1 gm is called 1 dyne.

1 dyne = 1 g × 1 cm/s^{2}

**Newton’s Third Law of Motion**

We have learned about force and its effect on an object through Newton’s first and second laws of motion. ‘However, in nature force cannot act alone.’ Force is a reciprocal action between two objects. Forces are always applied in pairs. When one object applies a force on another object, the latter object also simultaneously applies a force on the former object. The forces between two objects are always equal and opposite. This idea is expressed in Newton’s third law of motion. The force applied by the first object is called action force while the force applied by the second object on the first is called reaction force. ‘Every action force has an equal and opposite reaction force which acts simultaneously.’

Action and reaction are terms that express force. These forces act in pairs. One force cannot exist by itself. Action and reaction forces act simultaneously. Action and reaction forces act on different objects. They do not act on the same object and hence cannot cancel each other’s effect.

**Law of Conservation of Momentum**

Suppose an object A has mass m_{1} and its initial velocity is u_{1}. An object B has mass m_{2} and initial velocity u_{2}.

According to the formula for momentum, the initial momentum of A is m_{1}u_{1} and that of B is m_{2}u_{2}.

Suppose these two objects collide. Let the force on A due to B be F_{1}. This force will cause acceleration in A and its velocity will become v_{1}.

∴ The momentum of A after collision = m_{1}v_{1}

According to Newton’s third law of motion, A also exerts an equal force on B but in the opposite direction. This will cause a change in the momentum of B. If its velocity after the collision is v_{2},

The momentum of B after collision = m_{2}v_{2}

If F_{2} is the force that acts on object B,

F_{2} = -F_{1}

∴ m_{2}a_{2} = -m_{1}a_{1} (∵ F = ma)

∴ \(m_2 \times \frac{\left(v_2-u_2\right)}{t}=-m_1 \times \frac{\left(v_1-u_1\right)}{t}\) (∵ a = \(\frac{(v-u)}{t}\))

∴ m_{2} (v_{2} – u_{2}) = -m_{1} (v_{1} – u_{1})

∴ m_{2}v_{2} – m_{2}u_{2} = -m_{1}v_{1} + m_{1}u_{1}

∴ m_{2}v_{2} + m_{1}v_{1} = m_{1}u_{1} + m_{2}u_{2}

The magnitude of total final momentum = the magnitude of total initial momentum.

Thus, if no external force is acting on two objects, then their total initial momentum is equal to their total final momentum. This statement is true for any number of objects.

‘When no external force acts on two interacting objects, their total momentum remains constant. It does not change.’ This is a corollary to Newton’s third law of motion. The momentum is unchanged after the collision. The momentum gets redistributed between the colliding objects. The momentum of one of the objects decreases while that of the other increases. Thus, we can also state this corollary as follows.

‘When two objects collide, the total momentum before collision is equal to the total momentum after collision.’

To understand this principle, let us consider the example of a bullet fired from a gun. When a bullet of mass m_{1} is fired from a gun of mass m_{2}, its velocity becomes v_{1}, and its momentum becomes m_{1}v_{1}. Before firing the bullet, both the gun and the bullet are at rest and hence the total initial momentum is zero. According to the above law, the total final momentum also has to be zero. Thus, the forward-moving bullet causes the gun to move backward after firing. This backward motion of the gun is called its recoil. The velocity of recoil, v_{2}, is such that, m_{1}v_{1} + m_{2}v_{2} = 0 or v_{2} = \(-\frac{m_1}{m_2} \times v_1\)

As the mass of the gun is much higher than the mass of the bullet, the velocity of the gun is much smaller than the velocity of the bullet. The magnitude of the momentum of the bullet and that of the gun are equal and their directions are opposite. Thus, the total momentum is constant. Total momentum is also constant during the launch of a rocket.

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