Class 9 FREE FALL AND MASS

### Topic Covered

color{red} ♦ FREE FALL
color{red} ♦ MASS

### FREE FALL

Let us try to understand the meaning of free fall by performing this activity.

Activity _____________10.2

♦ Take a stone.
♦ Throw it upwards.
♦ It reaches a certain height and then it starts falling down.

We have learnt that the earth attracts objects towards it. This is due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.

Is there any change in the velocity of falling objects? While falling, there is no change in the direction of motion of the objects.

But due to the earth’s attraction, there will be a change in the magnitude of the velocity. Any change in velocity involves acceleration.

Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force.

Therefore, this acceleration is called the acceleration due to the gravitational force of the earth (or acceleration due to gravity). It is denoted by g. The unit of g is the same as that of acceleration, that is, m s^(–2).

We know from the second law of motion that force is the product of mass and acceleration. Let the mass of the stone in activity 10.2 be m. We already know that there is acceleration involved in falling objects due to the gravitational force and is denoted by g.

Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is,

F = m g ....(10.6)
From Eqs. (10.4) and (10.6) we have

 mg = G ( M xx m)/d^2

or  g = G M/d^2(10.7)

where M is the mass of the earth, and d is the distance between the object and the earth.

Let an object be on or near the surface of the earth. The distance d in Eq. (10.7) will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth,

 mg = G (M xx m)/R^2  .......(10.8)

 g = G M/R^2 ......(10.9)

The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

For most calculations, we can take g to be more or less constant on or near the earth. But for objects far from the earth, the acceleration due to gravitational force of earth is given by Eq. (10.7).

### TO CALCULATE THE VALUE OF g

To calculate the value of g, we should put the values of G, M and R in Eq. (10.9), namely, universal gravitational constant, G = 6.7 xx 10^(–11) N m^2 kg^(-2), mass of the earth, M = 6 xx 10^(24) kg, and radius of the earth, R = 6.4 xx 10^6 m.

 g = G M/R^2

 = ( 6.7 xx 10^(-11) N m^2 kg^2 xx 6 xx 10^(24) kg)/( 6.4 xx 10^6 m)^2

 = 9.8 m s^(–2).

Thus, the value of acceleration due to gravity of the earth, g = 9.8 m s^(–2).

10.2.2 MOTION OF OBJECTS UNDER THE INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH

Let us do an activity to understand whether all objects hollow or solid, big or small, will fall from a height at the same rate.

Activity _____________ 10.3

♦ Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. Observe whether both of them reach the ground simultaneously.

♦ We see that paper reaches the ground little later than the stone. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects.

The resistance offered by air to the paper is more than the resistance offered to the stone. If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.

We know that an object experiences acceleration during free fall. From Eq. (10.9), this acceleration experienced by an object is independent of its mass.

This means that all objects hollow or solid, big or small, should fall at the same rate. According to a story, Galileo dropped different objects from the top of the Leaning Tower of Pisa in Italy to prove the same.

As g is constant near the earth, all the equations for the uniformly accelerated motion of objects become valid with acceleration a replaced by g (see section 8.5). The equations are:

v = u + a t ....(10.10)

s = u t + 1/2 a t^2 ....(10.11)

v^2 = u^2 + 2 a s .......(10.12)

where u and v are the initial and final velocities and s is the distance covered in time, t.

In applying these equations, we will take acceleration, a to be positive when it is in the direction of the velocity, that is, in the direction of motion. The acceleration, a will be taken as negative when it opposes the motion.
Q 3205101068

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 ?
Class 9 Chapter 10 Example 2
Solution:

Time, t = ½ second
Initial velocity, u = 0 m s^(–1)
Acceleration due to gravity, g = 10 m s^(–2)
Acceleration of the car, a = + 10 m s^(–2) (downward)

(i) speed v = a t
v = 10 m s^(–2) xx 0.5 s
= 5 m s^(–1)

(ii) average speed  = (u + v)/2
= (0 m s^(–1) + 5 m s^(–1) )//2
= 2.5 m s^(–1)
(iii) distance travelled, s = ½ a t^2
= ½ xx 10 m s^(–2) xx (0.5 s)^2
= ½ xx 10 m s^(–2) xx 0.25 s^2
= 1.25 m

Thus,
(i) its speed on striking the ground = 5 m s^(–1)
(ii) its average speed during the 0.5 s = 2.5 m s^(–1)
(iii) height of the ledge from the ground = 1.25 m.
Q 3215101069

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.
Class 9 Chapter 10 Example 3
Solution:

Distance travelled, s = 10 m
Final velocity, v = 0 m s^(–1)
Acceleration due to gravity,  g = 9.8 m s^(–2)
Acceleration of the object, a = –9.8 m s^(–2) (upward motion)
(i) v^2 = u^2 + 2 a s
0 = u^2 + 2 xx (– 9.8 m s^(–2) ) xx 10 m
–u^2 = –2 xx 9.8 xx 10 m^2 s^(–2)
u = sqrt ( 196) m s^(-1)
u = 14 m s^(-1)
(ii) v = u + a t
0 = 14 m s^(–1) – 9.8 m s^(–2) xx t
t = 1.43 s.
Thus,
(i) Initial velocity, u = 14 m s^(–1), and
(ii) Time taken, t = 1.43 s.

### MASS

We have learnt in the previous chapter that the mass of an object is the measure of its inertia (section 9.3).

We have also learnt that greater the mass, the greater is the inertia. It remains the same whether the object is on the earth, the moon or even in outer space. Thus, the mass of an object is constant and does not change from place to place.

### Weight

We know that the earth attracts every object with a certain force and this force depends on the mass (m) of the object and the acceleration due to the gravity (g). The weight of an object is the force with which it is attracted towards the earth.
We know that

F = m xx a, .......(10.13)
that is,
F = m xx g. .......(10.14)
The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. Substituting the same in Eq. (10.14), we have

W = m xx g .......(10.15)

As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction.

We have learnt that the value of g is constant at a given place. Therefore at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, W ∝ m.

It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. The mass of an object remains the same everywhere, that is, on the earth and on any planet whereas its weight depends on its location.

ulbb"WEIGHT OF AN OBJECT ON THE MOON"

We have learnt that the weight of an object on the earth is the force with which the earth attracts the object. In the same way, the weight of an object on the moon is the force with which the moon attracts that object.

The mass of the moon is less than that of the earth. Due to this the moon exerts lesser force of attraction on objects.
Let the mass of an object be m. Let its weight on the moon be W_m. Let the mass of the moon be M_m and its radius be R_m.

By applying the universal law of gravitation, the weight of the object on the moon will be

 W_m = G (M_m xx m)/R_m^2 ....(10.16)

Let the weight of the same object on the earth be We. The mass of the earth is M and its radius is R.

From Eqs. (10.9) and (10.15) we have,

 W_e = G (M xx m)/R^2 .....(10.17)

Substituting the values from Table 10.1 in Eqs. (10.16) and (10.17), we get

 W_m = G ( 7.36 xx 10^(22) kg xx m)/( 1.74 xx 10^6 m)^2

 W_m = 2.431 xx 10^(10) G xx m ....(10.18a)

and = 1.474 xx 10^(11) G xx m .....(10.18b)

Dividing Eq. (10.18a) by Eq. (10.18b), we get

W_m/W_e = ( 2.431 xx 10^(10))/( 1.474 xx 10^(11))

or  W_m/W_e = 0.165 approx 1/6.......(10.19)

 text( Weight of the object on the moon)/text( Weight of the object on the earth) = 1/6

Weight of the object on the moon
= (1//6) xx its weight on the earth.

Q 3215201160

Mass of an object is 10 kg. What is its weight on the earth?
Class 9 Chapter 10 Example 4
Solution:

Mass, m = 10 kg
Acceleration due to gravity, g = 9.8 m s^(–2)
W = m xx g
W = 10 kg xx 9.8 m s^(-2) = 98 N
Thus, the weight of the object is 98 N.
Q 3225201161

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?
Class 9 Chapter 10 Example 5
Solution:

We know,
Weight of object on the moon = (1//6) xx its weight on the earth.
That is,

 W_m = W_e/6 = (10)/6 N.

 = 1.67 N.
Thus, the weight of object on the surface of the moon would be 1.67 N.