CBSE Tutorials https://cbse.myindialist.com/ One stop for all CBSE resource Thu, 21 Jul 2022 17:41:31 +0000 en-US hourly 1 https://wordpress.org/?v=6.9.4 23684851 noOne stop for all CBSE resource Floatation–Density and Relative Density | Class 9 https://cbse.myindialist.com/floatation-density-and-relative-density-class-9/ Thu, 21 Jul 2022 17:41:31 +0000 https://cbse.myindialist.com/?p=3905 Density We describe the lightness or heaviness of different substances by using the word density. The density of a substance is defined as mass of the substance per unit volume. That is The density of a given substance, under specified conditions remains the same. The therefore the density of a substance is one of the […]

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]]> Density

We describe the lightness or heaviness of different substances by using the word density.

The density of a substance is defined as mass of the substance per unit volume. That is

\displaystyle \mathbf{Distance}=\frac{\mathbf{mass}\,\,\mathbf{of}\,\,\mathbf{the}\,\,\mathbf{substance}}{\mathbf{vol}\mathbf{.}\,\,\mathbf{of}\,\,\mathbf{the}\,\,\mathbf{substance}}

\displaystyle \mathbf{8}\mathbf{.9}\,\mathbf{unit}\to \mathbf{kg/}{{\mathbf{m}}^{\mathbf{3}}}\,\,\,\mathbf{or}\,\,\,\,\mathbf{kg}\,{{\mathbf{m}}^{-\mathbf{3}}}

The density of a given substance, under specified conditions remains the same. The therefore the density of a substance is one of the characteristic property of a substance.

For example, density of gold is \displaystyle \mathbf{19300}\,\mathbf{kg/}{{\mathbf{m}}^{\mathbf{3}}} while that of water is \displaystyle 10\mathbf{00}\,\mathbf{kg/}{{\mathbf{m}}^{\mathbf{3}}}.

The density of a given sample of a substance can help us to determine its purity.

Relative Density

The relative density of a substance is the ratio of its density to that of water.

Relative density of a substance \displaystyle =\frac{\mathbf{Density}\,\,\mathbf{of}\,\,\mathbf{the}\,\,\mathbf{substance}}{\mathbf{Density}\,\,\mathbf{of}\,\,\mathbf{water}}

Since the relative density is a ratio of similar quantities, it has no unit.

The relative density of a substance expresses the heaviness of the substance in comparison to water. For example the relative density of iron is 7.8. This means that iron is 7.8 times as heavy as an equal vol. of water.

The relative density of water is 1. Now if the relative density of a substance is more than 1, then it will be heavier than water and hence it will sink in water. On the other hand if the relative density of a substance is less than 1, it will be lighter than water and hence float in water.

Question: The mass of 2 m3 of steel is 15600 kg. Calculate the density of steel is S.I. Units.

Solution: We know that density \displaystyle =\frac{\mathbf{mass}}{\mathbf{volume}}

\displaystyle =\frac{15600\,kg}{2{{m}^{2}}} \displaystyle =7800\,kg/{{m}^{3}}

Question: An object of mass 50 kg has a vol. of 20 m3. Calculate the density of the object. If the density of water be \displaystyle 1\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}. State whether the object will float or sink is water

Solution: Density \displaystyle =\frac{\mathbf{mass}}{\mathbf{volume}}

Mass of the object = 50 g

Volume of the object = \displaystyle 20\,\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}

Density \displaystyle =\frac{50\,g}{20\,c{{m}^{3}}}=2.5\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}

Question: The relative density of silver is 10.8. If the density of water be \displaystyle 1\times {{10}^{3}}\,\mathbf{kg/}{{\mathbf{m}}^{\mathbf{3}}}. Calculate the density of silver is S.I. Units.

Solution: Relative Density \displaystyle =\frac{\mathbf{Density}\,\,\mathbf{of}\,\,\mathbf{the}\,\,\mathbf{substance}}{\mathbf{Density}\,\mathbf{of}\,\mathbf{water}}

Relative density of silver = 10.8

Density of silver = ?

Density or eater \displaystyle =1\times {{10}^{3}}\,\,\mathbf{kg/}\,{{\mathbf{m}}^{\mathbf{3}}}

\displaystyle 10.8=\frac{\mathbf{Density}\,\,\mathbf{of}\,\,\mathbf{silver}}{1\times {{10}^{3}}}

Density of silver \displaystyle =10.8\times {{10}^{3}}\,\,\mathbf{kg/}{{\mathbf{m}}^{\mathbf{3}}}

Question: The volume of a solid mass 500g is \displaystyle \mathbf{350}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}

(a) What will be the density of this solid?

(b) What will be the mass of water displaced by this solid?

(c) What will be the relative density of the solid?

(d) Will it float or sink in water?

Solution: (a) Density \displaystyle =\frac{\mathbf{Mass}}{\mathbf{Volume}} \displaystyle =\frac{500}{350}\,\,\,\,=1.42\,\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}

(b) The solid will displace water equal to its an volume. Since the volume of solid is \displaystyle \mathbf{350}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}} so it will displace \displaystyle \mathbf{350}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}of water. Now vol. of water displaced is \displaystyle \mathbf{350}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}} and the density of water is \displaystyle \mathbf{1}\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}.

Density of water \displaystyle =\frac{mass\,of\,water}{volume\,of\,water}

\displaystyle \mathbf{1}\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}=\frac{\mathbf{mass}\,\mathbf{of}\,\mathbf{water}}{\mathbf{350}\,\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}}

Mass of water \displaystyle =\mathbf{1}\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}\times \mathbf{350}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}} \displaystyle =350\,\mathbf{g}

(c) \displaystyle \mathbf{R}\mathbf{.D}\mathbf{.}=\frac{\mathbf{Density}\,\,\mathbf{of}\,\,\mathbf{solid}}{\mathbf{Density}\,\,\mathbf{of}\,\,\mathbf{water}}=\frac{\mathbf{1}\mathbf{.42}\,\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}}{\mathbf{1}\,\,\mathbf{g/c}{{\mathbf{m}}^{\mathbf{3}}}}=\mathbf{1}\mathbf{.42}

Since the R.D. of this solid is greater than R. D. of water i.e., 1 therefore this solid is heavier than water and hence it will sink in water.

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]]> 3905 Floatation | Archimedes’ Principle & Buoyancy | Class 9 https://cbse.myindialist.com/floatation-archimedes-principle-buoyancy-class-9/ Thu, 21 Jul 2022 17:39:24 +0000 https://cbse.myindialist.com/?p=3902 Pressure and Fluids All liquids and gases are called as fluids. A solid exerts pressure on a surface due to its weight. Similarly, fluids have weight, and they also exert pressure on the base and walls of the container in which they are enclosed. A fluid exerts pressure in all directions even upwards. Buoyancy To […]

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]]> Pressure and Fluids

All liquids and gases are called as fluids. A solid exerts pressure on a surface due to its weight.

Similarly, fluids have weight, and they also exert pressure on the base and walls of the container in which they are enclosed. A fluid exerts pressure in all directions even upwards.

Buoyancy

To understand the term Buoyancy let us perform and activity.

Take an empty plastic bottle. Close the mouth of the bottle with an air tight stopper. Put it is a bucket filled with water.

· We will see that the bottle floats.

· Now push the bottle into the water. We will feel an upward push. If we push it further down, we will find it difficult to push deeper and deeper.

· This indicates that water exerts a force on the bottle in the upward direction.

· This upward force exerted by the water goes on increasing as the bottle is pushed deeper till it is completely immersed.

· If we release the bottle, it bounces back to the surface.

Explanation

The force due to the gravitational attraction of the earth acts on the bottle in the downward direction. So the bottle is pulled downwards. But the water exerts an upward force on the bottle. Thus, the bottle is pushed upwards.

The weight of an object is the force due to gravitational attraction of the earth. When the bottle is immersed, the upward force exerted y the water on the bottle is greater than its weight therefore it rises up when released.

To keep the bottle completely immerse the upward force on the bottle due to water must be balanced. This can be achieved by an externally applied force acting downwards. This force must at least be equal to the difference the upwards force and the weight of the bottle.

The upward force acting on an object immersed in a liquid Buoyant force or upthrust.

Factors affecting Buoyant force

The magnitude of buoyant force acting on an object immersed in a liquid depends on two factors.

(i) volume of object immersed in the liquid.

(ii) density of the liquid.

The buoyant force exerted by a liquid depends on the vol. Of the solid object immersed in the liquid. As the volume of solid object immersed inside the liquid increases, the upward buoyant force also increases. And when the object is completely immersed in the liquid, the buoyant force becomes the maximum and remain constant. Also, the magnitude of buoyant force acting in a solid object does not depend on the nature of the solid object.

It depends only on its volume. For e.g. If two balls made of different metals having different weights but equal volumes are fully immersed is a liquid, they will experience an equal upward ‘buoyant force’ i.e. equal loss in weight.

The buoyant force exerted by a liquid depends on the density of the liquid in which the object is immersed. As the density of liquid increases, the buoyant force exerted by it also increases, for e.g. Sea water has higher density then fresh water therefore sea water will exert more buoyant force on an object immersed in it than the fresh water. Therefore it is easier to swim is sea water because sea water exerts a greater buoyant force on the swimmer due to its higher density.

Even a very heavy material like an iron block floats in mercury because mercury exerts a very high buoyant force on iron block due it its very high density.

Archimede’s principle

Introduction:

Archimedes was a greek scientist. He discovered the principle, subsequently named after him, after noticing that the water in a both tub overflowed when he stepped into it. He raw through the streets shoaling “Eureka”! which means “I have got it”. This knowledge helped him to determine the purity of the gold in the crown made for the king. This work in the field of Geometry and Mechanics made him famous. His understanding of levers, pulleys, wheels – axle helped the greek army is its war with roman army.

“Archimedes’ principle states that when a body is partially or wholly immersed is a liquid, it experiences an upward force that is equal to the weight of the fluid displaced by it.

Activity of understand Archimede’s Principle

(a) Take a piece of stone and tie it to one end of a rubber string or a spring balance.

(b) Suspend the stone by holding the balance or the string as shown in figure.

(c) Note the elongation of the string or the reading on the spring balance due to the weight of the stone.

(d) Now slowly dip the stone in the water is a container. You will find that the elongation of the string or the reading of the balance decreases as the stone is gradually lowered is the water.

(e) However no further change is observed once the stone gets fully immersed is the water.

Explanation:

The elongation is produced is the string or the spring balance due to the weight of the stone. Since the extension decreases once the stone is lowered is water, it weans that same force acts on the stone in upward direction. As a result, the net force on the string decreases and hence the elongation also decreases. This upward force exerted by water is known as the force of buoyancy.

Add figure 10.6 (N.C.E.R.T)

Applications of Archimede’s Principle

(a) Archemede’s principle is used in determing the relative density of a substance.

(b) The hydrometers used for determing the density of liquids are baded on Archimede’s principle.

(c) The lactometers used for determing the purity of milk are based on Archimede’s principle.

(d) Archimede’s principle is used is designing ships and submarines.

Why objects float or sink is a liquid

Then an object is put is a liquid, then two forces act on it.

Weight of the object acting downwards due to the gravitational pull of the earth on the object.

Buoyant force acting upwards which tends to push up the object.

Sinking of an object in Water

If we place an Iron nail on the surface of water in a beaker then the nail sinks. The force due to the gravitational attraction of the earth on the iron nail pulls it downwards. There is an upthrust of water on the nail, which pushes it upwards. But the downward force acting on the nail is greater than the upthrest of water on the nail, so it sinks.

An object will sink is a liquid if its density is more than that of the liquid.

Floating of an object in Water

If we place a piece of cork on the surface of water in a beaker than the cork floats. This happens because of the difference is their densities. The density of a cork is less than the density of water. This means that the upthrust of water on the cork is greater than the weight of the cork. So it floats.

Therefore objects of density less than that of a liquid float on the liquid. The objects of density greater than that of a liquid sink in the liquid.

Question: When an aluminim object is immersed in water it displaces 5 kg of water. How much is the buoyant force acting on the aluminim object is Newtons? \displaystyle (g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}})

Solution: According to Archimede’s principle the buoyant force acting on the aluminim object will be equal to the weight of water displaced by this aluminim object.

Now, \displaystyle W=m\times g

\displaystyle m=5\,kg

\displaystyle g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

\displaystyle W=5\times 10\displaystyle =50\,\mathbf{N}

The weight of water displaced by the aluminim object is 50 newtons, therefore the buoyant force acting on the aluminium object will also be 50 N.

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]]> 3902 Floatation | Thrust and Pressure | Class 9 https://cbse.myindialist.com/floatation-thrust-and-pressure-class-9/ Thu, 21 Jul 2022 17:35:30 +0000 https://cbse.myindialist.com/?p=3900 Thrust and Pressure The force acting on an object perpendicular to the surface is called thrust. That us understand the meaning of thrust and pressure practically. Situation-1: You fix a poster on a notice board and while doing so you need to press drawing pins with your thumb. So pressing drawing pins means applying force […]

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]]> Thrust and Pressure

The force acting on an object perpendicular to the surface is called thrust. That us understand the meaning of thrust and pressure practically.

Situation-1:

You fix a poster on a notice board and while doing so you need to press drawing pins with your thumb. So pressing drawing pins means applying force on the surface area of the head of the pin. This force is directed perpendicular to the surface area of the board.

Situation-2:

When you stand on loose sand your feet go deep into the sand, but when you lie down on the sand. You will find that your body will not go deep on the sand.

This is because when you stand on loose sand, the force that is, the weight of your body is acting on an area equal to area of your feet. When you lie down, the same force acts on an area equal to the contact area of your whole body, which is larger than the area of your feet.

Thus the effect of thrust on sand is larger while standing than while lying.

The thrust on unit area is called pressure. Thus

\displaystyle \mathbf{Pressure}=\frac{\mathbf{thrust}}{\mathbf{area}}.

5.9 unit of pressure is \displaystyle \mathbf{N/}{{\mathbf{m}}^{\mathbf{2}}} or \displaystyle \mathbf{N}{{\mathbf{m}}^{-\mathbf{2}}}.

In honour of scientist Blaise Pascal, the S.I. unit of pressure is called pascal, denoted as \displaystyle {{P}_{a}}.

Pressure depends on two factors.

(i) Force applied

(ii) Area over which force acts.

Question: Why school bags have wide straps?

Solution: A school bag has wide strap made of thick cloth so that the weight of bag may fall over a large area of the shoulder of the child producing less pressure on the shoulder and due to les pressure, it is more comfortable to carry the heavy school bag. On the other hand, if the school bag has a strap made of thin string, then the weight of school bag will fall over a small area of the shoulder. This will produce a large pressure on the shoulder of the child and it will become very painful to carry the heavy school bag.

Question: Why a sharp knife cuts better than a blunt knife?

Solution: A sharp knife has a very thin edge to its blade. A sharp knife cuts objects better because due to its very this edge, the force of out hand falls over a very small area of the object producing a large pressure. And this large pressure cuts the object easily. On the other hand, a blunt knife has a thicker edge. A blunt knife does not cut an object easily because due to its thicker edge, the force of our hand falls over a large area of the object and produces lesser pressure. This lesser pressure cuts the object with difficult.

Question: Why is the tip of a needle sharp?

Solution: The tip of a sewing needle is sharp so that due to its sharp tip, the needle may put the force on a very small area of the cloth, producing a large pressure sufficient to pierce the cloth being stitched.

Question: Why a nail has a painted tip?

Solution: A nail has a painted tip, so that when it is hammered the force of hammer falls on a very small area of wood or wall creating a large pressure which pushes the nail into wood or wall.

Question: Why buildings have wide foundations?

Solution: The foundation of buildings and dams are laid on a larger area of ground so that the weight of the building or dam produces less pressure on ground and they may not sink into the ground.

Question: Why the tractors have broad tyres?

Solution: The tractors have broad tyres so that there is less pressure on the ground and the tyres do not sink into comparatively soft ground in the fields.

Question: A block of wood is kept on a table top. The man 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

(b) 40 cm × 20 cm

 

Solution: The mass of the wooden block = 5 kg

The dimension = 40 cm × 20 cm × 10 cm

Here the weight of the wooden block applies a thrust on the table top.

That is,

Thrust, \displaystyle F=m\times g

\displaystyle =5\,kg\times 9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

\displaystyle =49\,\mathbf{N}

Area of a side \displaystyle =l\times b

\displaystyle =20\,cm\times \,10\,cm

\displaystyle =200\,c{{m}^{2}}=0.02\,{{m}^{2}}

Pressure \displaystyle =\frac{F}{A}=\frac{49}{0.02}=2450\,\mathbf{N/}{{\mathbf{m}}^{\mathbf{2}}}

When the block lies on its side of dimensions 40 cm × 20 cm, it exerts the same thrust.

Area \displaystyle =l\times b

\displaystyle =40\times 20

\displaystyle =800\,c{{m}^{2}}=0.08\,c{{m}^{2}}

Pressure \displaystyle =\frac{F}{A}=\frac{49}{0.08}=612.5\,\mathbf{N/}{{\mathbf{m}}^{\mathbf{2}}}

Question: A force of 100 N is applied to an object of area \displaystyle 2{{m}^{2}}. Calculate the pressure.

Solution: Pressure \displaystyle =\frac{Force}{Area}

Pressure \displaystyle =\frac{100N}{2\,{{m}^{2}}} \displaystyle =50\,\mathbf{N/}{{\mathbf{m}}^{\mathbf{2}}} \displaystyle =50\,{{P}_{a}}

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]]> 3900 Gravitation | Mass, Weight and Weightlessness | Class 9 https://cbse.myindialist.com/gravitation-mass-weight-and-weightlessness-class-9/ Wed, 20 Jul 2022 19:03:43 +0000 https://cbse.myindialist.com/?p=3891 MASS The mass of a body is the quantity of matter contained in it. Mass is a scalar quantity. The unit of mass is kilogram. A body contains the same quantity of matter whether it be on the earth, moon or even in outer space. Thus, the mass of a body is constant and does […]

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]]> MASS

The mass of a body is the quantity of matter contained in it.

Mass is a scalar quantity.

The unit of mass is kilogram.

A body contains the same quantity of matter whether it be on the earth, moon or even in outer space.

Thus, the mass of a body is constant and does not change from place to place.

Mass of a body is usually denoted by the small ‘m’.

Mass of a body is a measure of inertia of the body and hence it is also known as inertial mass.

The mass of a body cannot be zero.

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 gravity (g).

The weight of an object is the force with which it is attracted towards the earth.

We know that \displaystyle F=m\times a

That is \displaystyle F=m\times g

The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W.

So we have, \displaystyle W=m\times g

As the weight of an object is the force with which it is attracted towards the earth, the S.I. unit of weight is the same as that of force i.e. Newton (N).

The weight is a force acting vertically downwards; it has both magnitude and direction, so it is a vector quantity.

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, \displaystyle W\,\propto \,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 earths and on any planet whereas its weight depends on its location.

Weight of a freely falling Body

Let us suppose a body is placed on a lift, the weighing machine will show the weight of the body n its scale. If now the lift is made to fall freely due to gravity, both the weighing machine as well as the body will fall with same acceleration i.e., with g in the downward direction. The body will, therefore, not press the weighing machine with any force and hence show zero weight. Thus a body is weightless during free fall.

Weightlessness in space

Consider an astronaut in a space ship orbiting the earth about 1000 km above its surface. At that distance from the earth, the force of gravity of earth is still quite strong. Since the acceleration due to gravity is not zero, the weight of astronauts in the space ship certainly cannot be zero. But we all have seen them on T.V., floating is a space ship and believe that in this situation they are weightless. This can be explained as follows:

Then the astronaut in the space ship is orbiting the earths, then both, the astronaut and the spaceship are in a continuing state of free fall towards the earth with the same acceleration due to gravity. Since the downward acceleration of the astronaut is the same as that of the spaceship he does not exert any force on the sides of the space ship and a weighing machine kept is the space vehicle will show his weight to be zero. Though the free fall of a body produces a feeling of weightlessness but a true weightlessness can be experienced by a spaceship is a region of outer space where the acceleration due to gravity ‘g’ is zero.

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 Wm. Let the mass of the moon be Nm and its radius be Rm.

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

\displaystyle {{W}_{m}}=G\frac{{{M}_{m}}\times m}{R_{m}^{2}}

It the weight of the same object on the earth be \displaystyle {{W}_{e}}. The mass of the earth is M and its radius is R.

Now, \displaystyle {{W}_{e}}=G\frac{M\times m}{{{R}^{2}}}

Mass of earth \displaystyle =5.98\times {{10}^{24}}\,kg

Radius of earth \displaystyle =6.37\times {{10}^{6}}m

Mass of moon \displaystyle =7.36\times {{10}^{22}}kg

Radius of moon \displaystyle =1.74\times {{10}^{6}}m

Substituting the values:

\displaystyle {{W}_{m}}=G\frac{7.36\times {{10}^{22}}kg\times m}{{{(1.74\times {{10}^{6}}m)}^{2}}}

\displaystyle {{W}_{m}}=2.431\times {{10}^{10}}G\times m …(i)

And \displaystyle {{W}_{e}}=1.474\times {{10}^{11}}G\times m …(ii)

Diving equation (i) by (ii), we get

\displaystyle \frac{{{W}_{m}}}{{{W}_{e}}}=\frac{2.431\times {{10}^{10}}}{1.474\times {{10}^{11}}}

or \displaystyle \frac{{{W}_{m}}}{{{W}_{e}}}=0.165\,\,\approx \frac{1}{6}

\displaystyle \frac{Weight\,\,of\,\,the\,\,object\,\,on\,\,the\,\,moon}{Weight\,\,of\,\,the\,\,object\,\,on\,\,the\,\,earth}=\frac{1}{6}

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

Question: Mass of an object is 10 kg. What is its weight on the earth?

Solution: Mass, \displaystyle m=10\,\mathbf{kg}

Acceleration due to gravity, \displaystyle g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

\displaystyle W=m\times g

\displaystyle W=10\times 9.8=98N

Thus, the weight of the object is 98 N.

Question: An object weighs 10N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?

Solution: We know,

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

That is \displaystyle {{W}_{m}}=\frac{{{W}_{e}}}{6}=\frac{10}{6}N \displaystyle =1.67\,N

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

Question: A man weighs 600N on the earth, what is its mass? If it was taken to the moon, his weight would be 100 N. What is his mass on moon? What is his accelerations due to gravity on the moon.

Solution: (i) Let is be the mass of body on earth.

\displaystyle W=m\times g

\displaystyle 600=m\times 10

\displaystyle m=60\,kg

(ii) Weight on moon = mass × acceleration due to gravity on moon.

\displaystyle 100=60\times gm

\displaystyle \frac{100}{60}=gm

\displaystyle {{g}_{m}}=1.67\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

Question: What is the mass of object whose weight is 49 N?

Solution: \displaystyle W=m\times g

\displaystyle 49=m\times 9.8

\displaystyle m=\frac{49}{9.8}=\frac{490}{98}=5kg

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]]> 3891 Gravitation | Equation of Motion of freely falling Bodies | Class 9 https://cbse.myindialist.com/gravitation-equation-of-motion-of-freely-falling-bodies-class-9/ Wed, 20 Jul 2022 19:01:35 +0000 https://cbse.myindialist.com/?p=3889 Equation of Motion of freely falling Bodies When the bodies are falling under influence of gravity, they experience acceleration g i.e. 9.8 . However, when these are going up against gravity they move with retardation of 9.8 . All the gravitation of motion already read by us are valid for freely falling body with the […]

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]]> Equation of Motion of freely falling Bodies

When the bodies are falling under influence of gravity, they experience acceleration g i.e. 9.8 \displaystyle \mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}. However, when these are going up against gravity they move with retardation of 9.8 \displaystyle \mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}.

All the gravitation of motion already read by us are valid for freely falling body with the different that a is replaced by g. For motions vertically upwards a is replaced by – g. Here ‘s’ is also replaced by h, the height.

\displaystyle v=u+at changes to \displaystyle v=u+gt

\displaystyle s=ut+\frac{1}{2}a{{t}^{2}} changes to \displaystyle h=ut+\frac{1}{2}g{{t}^{2}}

\displaystyle {{v}^{2}}={{u}^{2}}+2as changes to \displaystyle {{v}^{2}}={{u}^{2}}+2gh

Question: A car falls off a ledge and drops to the ground in 0.5s. Let \displaystyle g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}.

(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?

Solutions: Time, \displaystyle t=\frac{1}{2} second

Initial velocity, \displaystyle u=0\,\,\mathbf{m/s}

Acceleration due to gravity, \displaystyle g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

Acceleration of the car, \displaystyle a=+10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}} (downward)

(i) Speed \displaystyle v=u+gt

\displaystyle v=0+(10\times 0.5)

\displaystyle v=5\,\,\mathbf{m/s}

(ii) Average speed \displaystyle =\frac{u+v}{2}

\displaystyle =\frac{0+5}{2} \displaystyle =2.5\,\mathbf{m/s}

(iii) Distance traveled\displaystyle (S)=ut+\frac{1}{2}a{{t}^{2}}

\displaystyle =0\times 0.5+\frac{1}{2}\times 10\times {{(0.5)}^{2}}

\displaystyle =0+(5\times 0.25)\,\,\,=\,1.25\,\mathbf{m}

Question: An object is thrown vertically upwards and rises to a height of 10m. Calculate

(i) the velocity with which the object was thrown upwards and

(ii) the time taken by the object to reach the highest point.

Solutions: Distance traveled \displaystyle (h)=10\,\mathbf{m}

Final velocity \displaystyle (v)=0\,\mathbf{m/s}

Acceleration due to gravity, \displaystyle g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

Acceleration of the object, \displaystyle a=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

(i) \displaystyle {{v}^{2}}={{u}^{2}}+2gh

\displaystyle 0={{u}^{2}}+2(-9.8)\times 10

\displaystyle 0={{u}^{2}}-196

\displaystyle {{u}^{2}}=196

\displaystyle u=\sqrt{196}

\displaystyle u=14\,\mathbf{m/s}

(ii) \displaystyle v=u+gt

\displaystyle 0=14-9.8\times t

\displaystyle t=1.43\,s

Question: To estimate the height of a bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2 seconds to touch the water surface in the river. Calculate the height of the bridge from the water level \displaystyle (g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}).

Solutions: Now, initial velocity of stone, \displaystyle u=0

Time taken, \displaystyle t=2s

Acceleration due to gravity, \displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

And, height of the bridge, \displaystyle h=? (To be calculated)

We know that for a freely falling body

Height, \displaystyle h=ut+\frac{1}{2}g{{t}^{2}}

Putting the above values in this formula, we get

\displaystyle h=0\times 2+\frac{1}{2}\times (-9.8)\times {{(2)}^{2}}

or \displaystyle h=\frac{1}{2}\times 9.8\times 4

or \displaystyle h=-19.6\,\mathbf{m}

Question: When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity, \displaystyle g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}})

Solutions: Here, Initial velocity of ball, \displaystyle u=? (To be calculated)

Final velocity of ball, \displaystyle v=0

Acceleration due to gravity, \displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

And height, \displaystyle h=19.6\,\mathbf{m}

Now, putting all these values in the formula:

\displaystyle {{v}^{2}}={{u}^{2}}+2gh

We get \displaystyle {{(0)}^{2}}={{u}^{2}}+2\times (-9.8)\times 19.6

\displaystyle 0={{u}^{2}}-19.6\times 19.6

\displaystyle {{u}^{2}}={{(19.6)}^{2}}

So, \displaystyle u=19.6\,\mathbf{m/s}

Here, Final velocity, \displaystyle v=0 (The ball stops)

Initial velocity, \displaystyle u=19.6\,\mathbf{m/s} (Calculated above)

Acceleration due to gravity, \displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

And, Time \displaystyle t=? (To be calculated)

So, putting these values in the above equation, we get

\displaystyle 0=19.6+(-9.8)\times t

\displaystyle 0=19.6-9.8t

\displaystyle 9.8t=19.6

\displaystyle t=\frac{19.6}{9.8} \displaystyle t=2s

Thus, the ball takes 2 seconds to reach the highest point of its upward journey. Please note that the ball will take an equal time that is 2 seconds to fall back to the ground. In other words, the ball will take a total of 2 + 2 = 4 seconds to reach back to the thrower.

Question: A cricket ball is dropped from a height of 20 metres.

(a) Calculate the speed of the ball when it hits the ground.

(b) Calculate the time it takes to fall through this height. \displaystyle (g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}})

Solution: (a) Here, initial speed, \displaystyle u=0

Final speed, \displaystyle v=? (To be calculated)

Acceleration due to gravity, \displaystyle g=-10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

And, Height, \displaystyle h=10\,\mathbf{m}

Now, we know that for a freely falling body,

\displaystyle {{v}^{2}}={{u}^{2}}+2gh

So, \displaystyle {{v}^{2}}={{(0)}^{2}}+2\times (-10)\times 20

\displaystyle {{v}^{2}}=-400

\displaystyle v=-\sqrt{400}

or, \displaystyle v=-20\,\mathbf{m/s}

Thus, the speed of cricket ball when it hits the ground will be 20 metres per second. The minus sign with speed (or velocity) shows that it is in the downward direction.

(b) Now, initial speed, \displaystyle u=0

Final speed. \displaystyle v=-20\,\mathbf{m/s} (Calculated above)

Acceleration due to gravity, \displaystyle g=-10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

And, Time, \displaystyle t=? (To be calculated)

Putting these values in the formula:

\displaystyle v=u+gt,

We get: \displaystyle -20=0+(-10)\times t

\displaystyle -20=-10t

\displaystyle 10t=20

\displaystyle t=\frac{20}{10}

\displaystyle t=2s

Thus, the ball takes 2 seconds to fall through a height of 20 metres.

Question: A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? \displaystyle (g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}).

Solution: Here, initial speed of ball, \displaystyle u=15\,\mathbf{m/s}

Final speed of ball, \displaystyle v=0 (The ball stops)

Acceleration due to gravity, \displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}

And, Height, \displaystyle h=? (To be calculated)

Now, putting all these values in the formula,

\displaystyle {{v}^{2}}={{u}^{2}}+2gh

We get, \displaystyle {{(0)}^{2}}={{(15)}^{2}}+2\times (-9.8)\times h

\displaystyle 0=225-19.6\,h

\displaystyle 19.6h=225

or \displaystyle h=\frac{225}{19.6}

\displaystyle h=11.4\,\mathbf{m}

Thus, the ball will go to a maximum height of 11.4 metres before it begins to fall.

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