I. Very short answer type questions:
1. Write one difference between gravitation and gravity.
Ans. Gravitation is the force of attraction between any two bodies and gravity is the force of attraction between a body and the Earth.
2. How is the acceleration due to the gravity of the falling body related to its mass?
Ans. Acceleration due to the gravity of a falling body doesn't depend on the mass of the body.
3. Why G is called universal constant?
Ans. The value of G remains constant at all the places in the universe, so it a called universal constant.
4. Define the mass and weight of an object.
Ans. Mass is the quantity of matter contained in the body and the weight of the body is the force with which Earth attracts the body.
5. Can the mass of an object be zero?
Ans. No.
6. What do you understand by the gravitational force of earth?
Ans. Gravitational force is the force with which earth attracts the other body.
7. How does acceleration due to gravity change with the shape of the earth?
Ans. The earth is not a perfect sphere. As the radius of the earth increases from the pole to the equator, the value of g becomes smaller at the equator and greater at the pole.
Ans. Gravitation is the force of attraction between any two bodies and gravity is the force of attraction between a body and the Earth.
2. How is the acceleration due to the gravity of the falling body related to its mass?
Ans. Acceleration due to the gravity of a falling body doesn't depend on the mass of the body.
3. Why G is called universal constant?
Ans. The value of G remains constant at all the places in the universe, so it a called universal constant.
4. Define the mass and weight of an object.
Ans. Mass is the quantity of matter contained in the body and the weight of the body is the force with which Earth attracts the body.
5. Can the mass of an object be zero?
Ans. No.
6. What do you understand by the gravitational force of earth?
Ans. Gravitational force is the force with which earth attracts the other body.
7. How does acceleration due to gravity change with the shape of the earth?
Ans. The earth is not a perfect sphere. As the radius of the earth increases from the pole to the equator, the value of g becomes smaller at the equator and greater at the pole.
8. Is the value of G dependent on the medium present between the two bodies?
Ans. No.
9. At what place on the earth's surface is the weight of a body be maximum and minimum?
Ans. The weight of a body is maximum at the pole and minimum at the equator of the earth.
10. If the mass of a body is 8 Kg and weight is 12 N then what will be its mass and weight in the moon?
Ans. Mass of the body will be 8 kg and weight will be one-sixth of 12 means 2N (12/6). Because the mass of the body does not change but the weight of a body is different in different places.
11. State the factors on which acceleration due to gravity 'g' and 'G' depends.
Ans. Acceleration due to gravity g depends on the mass of the earth and G which is universal constant depends on the distance or radius and mass of the planet.
12. On what factors does the weight of a body depend?
Ans. Mass(m) and gravity (g) are the two factors on which weight depends.
As Weight = mg.
13. How does the weight of a rocket change as it moves from earth to moon?
Ans. The weight of a rocket decreases as it moves from earth to moon because weight is directly related to g and at the moon, weight becomes one-sixth.
14. What is the relation between the weight of an object on the moon and on the earth?
Ans. The weight of an object on the moon is one-sixth the weight of the object on the earth.
Ans. The weight of a rocket decreases as it moves from earth to moon because weight is directly related to g and at the moon, weight becomes one-sixth.
14. What is the relation between the weight of an object on the moon and on the earth?
Ans. The weight of an object on the moon is one-sixth the weight of the object on the earth.
II. Short answer type questions
1. Give reasons for the following:
(a) A sheet of paper falls slower than when it is crumpled into a ball.
(b) What is meant by the statement that acceleration due to gravity is 9.8 m/s2?
(a) A sheet of paper falls slower than when it is crumpled into a ball.
(b) What is meant by the statement that acceleration due to gravity is 9.8 m/s2?
Ans. (a)A sheet of paper falls slower than when it is crumpled into a ball because the resistance provided by the air is more for a sheet of paper than the crumpled ball of paper. In a vacuum, both will touch the ground together.
(b) Acceleration due to gravity is 9.8 m/s2b means, every second an object which is in free fall, the velocity of the object to increase 9.8 m/s due to gravity.
2. What is meant by free fall? A stone and a small coin are dropped from the same height at the same time, which of the two will touch the ground first?
Ans. When a body falling from a height towards the earth under the gravitational force of the earth and no other force acting on it is called free fall.
When stone and a small coin are dropped from the same height at the same time, both the stone and coin will touch the ground together because the acceleration due to the gravity of the falling body doesn't depend on the mass of the body.
3. (i) Write the units of the following:
(a) Gravitational constant G (b) Mass (c) Weight.
(ii) Why a body weighs more at the pole than at equator?
(a) Gravitational constant G (b) Mass (c) Weight.
(ii) Why a body weighs more at the pole than at equator?
Ans. (i) (a) Unit of gravitational constant G is Nm2kg-2
(b) The unit of mass is kg.
(c) The Unit of weight is N.
(ii) A body weighs more at the pole than at the equator because the radius of the earth is smaller at the pole than the equator and weight is inversely proportional to the radius of the earth.
4.Explain what happens to the force between two objects if:
(a) The mass of one object is doubled?
(b) The distance between the two objects is tripled?
Ans. (a) The forces between two objects become double
when the mass of one object is doubled. Force is directly proportional to mass.
(b) The forces between two objects become one-ninth when the distance between the two objects is tripled.
F1 = m1m2/r2 and when r is triple,new force
F2 = m1m2/(3r)2
= m1m2/9r2
F2 = F1/9
(c) The Unit of weight is N.
(ii) A body weighs more at the pole than at the equator because the radius of the earth is smaller at the pole than the equator and weight is inversely proportional to the radius of the earth.
4.Explain what happens to the force between two objects if:
(a) The mass of one object is doubled?
(b) The distance between the two objects is tripled?
Ans. (a) The forces between two objects become double
when the mass of one object is doubled. Force is directly proportional to mass.
(b) The forces between two objects become one-ninth when the distance between the two objects is tripled.
F1 = m1m2/r2 and when r is triple,new force
F2 = m1m2/(3r)2
= m1m2/9r2
F2 = F1/9
5. What is the importance of the universal law of Gravitation? Or State any four natural phenomena explained by the universal law of Gravitation.
Ans. The universal law of gravitation successfully explained several phenomena which are:
(i) the force that binds us to the earth;
(ii) the motion of the moon around the
earth;
(iii) the motion of planets around the Sun;
and
(iv) the tides due to the moon and the Sun.
Ans. The universal law of gravitation successfully explained several phenomena which are:
(i) the force that binds us to the earth;
(ii) the motion of the moon around the
earth;
(iii) the motion of planets around the Sun;
and
(iv) the tides due to the moon and the Sun.
6. How does the force of gravity between two objects change when the distance between them is reduced to half?
Ans. The force of gravity between two objects becomes 4 times when the distance between them is reduced to half.
The force of gravity:
F1 = m1m2/r2
And when the distance between them is reduced to half the force of gravity:
F2 = m1m2/(r/2)2
= 4 m1m2/r2 .
Ans. The force of gravity between two objects becomes 4 times when the distance between them is reduced to half.
The force of gravity:
F1 = m1m2/r2
And when the distance between them is reduced to half the force of gravity:
F2 = m1m2/(r/2)2
= 4 m1m2/r2 .
7. Stone and the earth attract each other with an equal and opposite force. Why then we see only the stone falling towards the earth but not the earth rising towards the stone?
Ans . According to the third law of motion, stone and the earth attract each other with an equal and opposite force. But according to the second law of motion, for a given force, acceleration is inversely proportional to the mass of an object. The mass of stone is negligibly small compared to that of the earth. So, we see only the stone falling towards the earth but not the earth rising towards the stone.
Ans . According to the third law of motion, stone and the earth attract each other with an equal and opposite force. But according to the second law of motion, for a given force, acceleration is inversely proportional to the mass of an object. The mass of stone is negligibly small compared to that of the earth. So, we see only the stone falling towards the earth but not the earth rising towards the stone.
8. Write three differences between 'g'(gravitational acceleration) and 'G'(gravitational constant).
Ans. g (gravitational acceleration)
1. g is the acceleration due to gravity acquired by the body due to the earth gravitational pull.
2. g is a vector quantity.
3. It is different in different places on the earth's surface.
G (gravitational constant)
1. G is numerically equal to the force of attraction between two masses of 1kg which are separated by a distance 1m.
2. G is a scaler quantity.
3. Its value is the same everywhere in the universe. earth surface.
9. Give the relation between mass and weight. What kind of quantity is the weight?
Ans. Mass is the quantity of matter contained in the body and the weight of the body is the force with which Earth attracts the body.
The relation is: W, is the product of the mass m of the object and the magnitude of the gravitational acceleration g
= weight/acceleration (W/g)
Weight is a vector quantity.
10. A man of mass 60 Kg is standing on the floor holding a stone weighing 40N. What is the force with which the floor is pushing him up?
Ans. The mass of the man is 60kg.
Weight of man = mg= 60 x 9.8 =588N
Weight of stone = 40N
According to the newton's third law of motion total force exerted by the man to the ground = the force with which the floor is pushing him up=588 +40 = 628N
The force with which the floor is him up= 628N.
11. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:
Ans. g (gravitational acceleration)
1. g is the acceleration due to gravity acquired by the body due to the earth gravitational pull.
2. g is a vector quantity.
3. It is different in different places on the earth's surface.
G (gravitational constant)
1. G is numerically equal to the force of attraction between two masses of 1kg which are separated by a distance 1m.
2. G is a scaler quantity.
3. Its value is the same everywhere in the universe. earth surface.
9. Give the relation between mass and weight. What kind of quantity is the weight?
Ans. Mass is the quantity of matter contained in the body and the weight of the body is the force with which Earth attracts the body.
The relation is: W, is the product of the mass m of the object and the magnitude of the gravitational acceleration g
= weight/acceleration (W/g)
Weight is a vector quantity.
10. A man of mass 60 Kg is standing on the floor holding a stone weighing 40N. What is the force with which the floor is pushing him up?
Ans. The mass of the man is 60kg.
Weight of man = mg= 60 x 9.8 =588N
Weight of stone = 40N
According to the newton's third law of motion total force exerted by the man to the ground = the force with which the floor is pushing him up=588 +40 = 628N
The force with which the floor is him up= 628N.
11. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:
(a) The maximum height to which it reaches.
(b) The total time it takes to return to the earth's surface.
Ans. (a) Let the max. height to which it reaches is h.
We have v2-u2= 2gh
h = v2-u2/2g where v = 0 and u = - 49 m/s
h = -(49)2/2 x (-9.8) = 2401/19.6 = 122.5 m.
(b) Let the total time it takes to return to the earth's surface s t.
We have v = u +gt
and t = (- u) /-g
t = -49/-9.8 = 5 s.
12. A body of weight 600N rests on a lift. If the lift begins to fall freely under the gravity, what is the force with which the body presses on the floor?
Ans. When the lift is falling freely under the gravity then the body of weight 600N kept in it also falls freely under the action of gravity. In this case, the reaction force of the floor of the lift on the body is zero. So the action force of the body on the floor of the lift should also be zero.(by Newton's III Law of motion). Hence no force is exerted by the body on the floor of the lift when the lift is falling freely under the gravity. The body is weightless in such a situation.
13. A sphere of mass 40 kg is attracted by another sphere of mass 80 kg. by a force of 2.5 x10-6 N and when their centers are 300 mm apart Find the value of G.
G = (2.5 x 90000 x 10-12)/3200 = (25 x 90 x 10-12)/3200
= 2250 x 10-12/3200
= 22500 x 10-11/3200
G = 7.02135 x 10-11
14. A ball is thrown upwards from the ground at a tower with a speed of 20 meters per second. There is a window in the tower at a height of 15 m from the ground. How many times and when will the ball pass a window?
Ans. Initial velocity = u = 20 m/s.
The maximum height the ball will reach h =?
Using equation v2- u2= 2gh
At maximum height v = 0 m/s
h= (v2- u2)/2g = - u2/2g = - 202/2 x 10 = 400/20 = 20m.
h = 20m
The ball will reach the height of 20m.
The ball will reach the window two times when the ball goes upward and then comes back downward.
Now to calculate the time:
We have h = ut - 1/2gt2 15 = 20t - 0.5 x10t2
15 = 20t - 5t2
5t2- 20t + 15 = 0
t2 - 4t + 3 = 0
t2 - 3t - t + 3 = 0
t(t - 3) - 1(t -3) = 0
t = 1 , 3 s
Therefore ball pass the window at 1 sec. and 3 sec. respectively.
15. (a) Seema byes a few grams of gold at the pole as per the instruction of one of her friends. She hands over the same when she meets her friend at the equator. Will the friend agree with the weight of gold bought? If not why?
(b) If the moon attracts the earth, why does the earth not move towards the moon?
Ans. (a) No. The weight of the same amount of gold will be less on the equator because weight is the force of attraction with which the earth attracts the body and weight = mg. Therefore weight is directly proportional to g and the value of g is less at the equator than the pole.
(b) The moon attracts the earth but the earth does not move towards the moon because the mass of the moon is very less as compared to the earth.
16. Suppose a planet exists whose mass and radius both are half those of earth. Calculate the acceleration due to gravity on the surface of this planet.
Ans. If ge and gp are the acceleration due to gravity on the earth and planet surface then
ge= GM/R2 and gp= Gm/r2,
Where R = radius of the earth
r = radius of the planet.
m = mass of the planet and
M = mass of the earth.
According to question m = 1/2M and r = 1/2R
Therefore gp = Gm/r2
= G x M x 4
2 x R2
= 2ge
=2 x 9.8 = 19.6m/s2
III. Long answer type questions:.
(b) The total time it takes to return to the earth's surface.
Ans. (a) Let the max. height to which it reaches is h.
We have v2-u2= 2gh
h = v2-u2/2g where v = 0 and u = - 49 m/s
h = -(49)2/2 x (-9.8) = 2401/19.6 = 122.5 m.
(b) Let the total time it takes to return to the earth's surface s t.
We have v = u +gt
and t = (- u) /-g
t = -49/-9.8 = 5 s.
12. A body of weight 600N rests on a lift. If the lift begins to fall freely under the gravity, what is the force with which the body presses on the floor?
Ans. When the lift is falling freely under the gravity then the body of weight 600N kept in it also falls freely under the action of gravity. In this case, the reaction force of the floor of the lift on the body is zero. So the action force of the body on the floor of the lift should also be zero.(by Newton's III Law of motion). Hence no force is exerted by the body on the floor of the lift when the lift is falling freely under the gravity. The body is weightless in such a situation.
13. A sphere of mass 40 kg is attracted by another sphere of mass 80 kg. by a force of 2.5 x10-6 N and when their centers are 300 mm apart Find the value of G.
Ans. The force of gravitation exerted by Earth or any other planet on the objects on its surface is called gravity. Every object attracts every other object due to the force of gravitation. So gravity is a special case of Gravitation.
F = GMm/r2
2.5 x10-6 = G (80 x 40)/(300 x 10-3)2
2.5 x 10-12 = G (3200)/90000F = GMm/r2
2.5 x10-6 = G (80 x 40)/(300 x 10-3)2
G = (2.5 x 90000 x 10-12)/3200 = (25 x 90 x 10-12)/3200
= 2250 x 10-12/3200
= 22500 x 10-11/3200
G = 7.02135 x 10-11
14. A ball is thrown upwards from the ground at a tower with a speed of 20 meters per second. There is a window in the tower at a height of 15 m from the ground. How many times and when will the ball pass a window?
Ans. Initial velocity = u = 20 m/s.
The maximum height the ball will reach h =?
Using equation v2- u2= 2gh
At maximum height v = 0 m/s
h= (v2- u2)/2g = - u2/2g = - 202/2 x 10 = 400/20 = 20m.
h = 20m
The ball will reach the height of 20m.
The ball will reach the window two times when the ball goes upward and then comes back downward.
Now to calculate the time:
We have h = ut - 1/2gt2 15 = 20t - 0.5 x10t2
15 = 20t - 5t2
5t2- 20t + 15 = 0
t2 - 4t + 3 = 0
t2 - 3t - t + 3 = 0
t(t - 3) - 1(t -3) = 0
t = 1 , 3 s
Therefore ball pass the window at 1 sec. and 3 sec. respectively.
15. (a) Seema byes a few grams of gold at the pole as per the instruction of one of her friends. She hands over the same when she meets her friend at the equator. Will the friend agree with the weight of gold bought? If not why?
(b) If the moon attracts the earth, why does the earth not move towards the moon?
Ans. (a) No. The weight of the same amount of gold will be less on the equator because weight is the force of attraction with which the earth attracts the body and weight = mg. Therefore weight is directly proportional to g and the value of g is less at the equator than the pole.
(b) The moon attracts the earth but the earth does not move towards the moon because the mass of the moon is very less as compared to the earth.
16. Suppose a planet exists whose mass and radius both are half those of earth. Calculate the acceleration due to gravity on the surface of this planet.
Ans. If ge and gp are the acceleration due to gravity on the earth and planet surface then
ge= GM/R2 and gp= Gm/r2,
Where R = radius of the earth
r = radius of the planet.
m = mass of the planet and
M = mass of the earth.
According to question m = 1/2M and r = 1/2R
Therefore gp = Gm/r2
= G x M x 4
2 x R2
= 2ge
=2 x 9.8 = 19.6m/s2
1. What happens to the magnitude of the force of gravitation between two objects if:
(a) Mass of one of the objects is tripled?
(b) The distance between the objects is doubled?
(c) The mass of both objects is doubled.
(b) The distance between the objects is doubled?
(c) The mass of both objects is doubled.
Ans. (a) The force of gravitation between two objects is given by 'Universal gravitational Law'. It is numerically stated as:
F = Gm1m2/r2
(a) Mass of one of the objects is tripled: F = G3m1m2/r2
F = 3 Gm1m2/r2
Force will be tripled.
(b) The distance between the objects is doubled:
F = Gm1m2/(2r)2
F = 1/4 Gm1m2/r2
Force will be one-fourth of its previous value.
(c) The mass of both objects is doubled:
F = G2m12m2/r2
F = 4Gm1m2/r2
Force will be four times its previous value.
3. (I) Write the name of the scientist:
(a) Who discovered that force is the cause of motion?
(b) Who formulated the universal law of gravitation?
(c) Who found out the value of gravitational constant 'G’.
(b) Who formulated the universal law of gravitation?
(c) Who found out the value of gravitational constant 'G’.
(II) (a) Explain why the value of 'g' is greater at poles than the equator.
(b) Why does a body reach the ground quicker at poles than equator when dropped from the same height?
Ans. (a) Galileo Galilei discovered that force is the cause of motion.
(b) Sir Issac Newton formulated the universal law of gravitation.
(c) Henry Cavendish found out the value of gravitational constant 'G’.
(II) (a) The value of g directly proportional to, the radius of the earth and radius of the earth is less at the pole than the equator, so the value of 'g' is greater at poles than the equator.
(b) A body reaches the ground quicker at poles than the equator when dropped from the same height because g is greater at the pole than the equator.
4. Derive an expression for acceleration due to gravity in terms of mass M of earth and radius R.
Ans. According to the Newtons third law of motion, the force of attraction between the earth and the earth and body is given by F = GMm/r2
Where m = mass of the object, M = mass of the earth, r = distance between earth and object and G =universal constant.
And according to Newton's second law:
F = mg. (where g is the acceleration due to gravity)
From eq. 1 and 2 we get
mg = GMm/r2 g = GM/r2.
If the body is located on the earth's surface of the earth then r = R.
Therefore g = GM/R2
This expression gives acceleration due to gravity at the surface of the earth where the value of g is 9.8 m/s2
Where m = mass of the object, M = mass of the earth, r = distance between earth and object and G =universal constant.
And according to Newton's second law:
F = mg. (where g is the acceleration due to gravity)
From eq. 1 and 2 we get
mg = GMm/r2 g = GM/r2.
If the body is located on the earth's surface of the earth then r = R.
Therefore g = GM/R2
This expression gives acceleration due to gravity at the surface of the earth where the value of g is 9.8 m/s2
5. A stone is dropped from the edge of the roof. It passes a window 2m high, in 0.1sec. How far is the roof above the window?
Ans. Given :
When a stone is dropped from the edge of the roof,
its initial velocity = 0 .
The stone passes the window of 2m high in 0.1s.
Let the distance between the roof and that window be 'h' m and the time taken to reach that "h"m is t.
Therefore h = ut + 1/2gt² = 1/2gt²........eq (1)
A/Q Stone passes the window in 0.1 seconds.
So it travels h+2 m in t + 0.1seconds.
h + 2 = 1/2g(t+ 0.1)²
1/2gt² + 2 = 1/2g(t²+0.01 + 0.2t) From eq ..(1)
1/2gt² - 1/2gt² + 2 = 0.005 g + 0.1gt
2 = 0.005 x 10 + 0.1 x 10 x t
2= 0.05 + t
2 - 0.05 = t
t = 1.95 s.
Therefore, The stone takes 1.95 seconds to travel from roof to top of the window.
The distance between the roof and top of the window (h)
h = 1/2 x 10 x t²
h = 5t²
h= 5x 1.95²
= 19.0125 m.
Therefore, the height of the roof above the top of the window is 19.0125m
6. State the Universal law of Gravitation? Derive an expression for gravitational force between two bodies.
Ans. The universal law of gravitation states that the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Let the two bodies A and B be of masses m1 and m2 respectively.
According to universal law:
F ∝ m1m2........eq. (1)
F ∝ 1/r²............eq. (2)
From eq...1 & 2 we get
F ∝ m1m2/r2
F = G m1m2/r2
Where G is the universal constant.
7. A stone is dropped from a height of 50 m on earth. At the same time, another stone is thrown vertically upwards from the ground with a velocity of 50m/s. At what height from the ground will the two stones meet (g = -10m/s2)
Ans. Let the distance covered by stone A which is dropped from a height be h.
h = 1/2gt² ( u = 0 m/s).....1
Height of the Stone B which is thrown vertically upwards
= 50 - h
= ut - 1/2gt²
50 - h = 50 t - 1/2 g t²
50 - 1/2 gt² = 50 t - 1/2 g t² ( from eq. 1)
50 = 50t
t = 1s
Now distance from the ground = 50 - 1/2gt² = 50 - 1/2 x 10 x1 = 45 m
8. A ball is thrown vertically upwards. The speed of the ball was 10m/s when it had reached one half of its maximum height. How high does the ball rise?
Ans. (a) When a ball is thrown vertically upward, the speed of the ball at half of the max.height =v = 10 m/s.
Let the maximum height be H and the final velocity is zero at height this height.
Now distance = H.
We have v² - u² = 2x g x H
0² - u² = 2 x -10 x H
- u² = -20H
u² = 20H
When the ball reaches half the maximum height, the final velocity is 10 m/s, and the distance traveled is H/2.
We have v² - u² = -2gH/2
10² - 20H = 2 x -10 x H/2
100 - 20H = -10H
10H = 100
H = 10m
Therefore the maximum height is 10 m.
9. If the mass of one of the objects is doubled and the mass of others remains the same and if the distance between them is halved then how does the gravitational force change?
Ans. Let the masses of two objects are m1 and m2. and the distance between them = r
Therefore the gravitational force F1 = m1m2/r2.
Now, when the mass of one of the object is doubled and the mass of others remains the same and if the distance between them is halved then:
Force F2 = 2m1m2/(r/2)2.
F2 = 8 m1m2/r2.
F2 = 8F1
If the mass of one of the objects is doubled and the mass of others remains the same and if the distance between them is halved then the gravitational force becomes eight times.
10. The radius of the earth is about 6370 Km. An object of mass 30 Kg is taken to a height of 230 km above the surface of the earth.
(a) What is the mass of the body?
(b) What is the acceleration due to gravity at this height?
(c) What is the weight of the body at this height?
Ans. (a) The of the will not change because mass is constant in everywhere.
(b) We have g = g'(1+h/R)2
Where g = acceleration due to gravity.
h= height of the body aabove the surface of the earth =230km
R= radius of the earth =
g' = acceleration due to gravity at the height of 230km.
g = g'(1+ h/R)2
9.8 =g'(1 + 230/ 6370)2= g'(6370 + 230)/ 6370)2
9.8 = g'(6600/ 6370)2 = g'(1.04)2 = g'x1.08
g'= 9.8/1.08 = 9.07m/s2
(c) The weight of this body = mg = 30 x 9.07 = 272.1 kg.
11. What is the magnitude of the gravitational force exerted by a 15Kg mass on a 25 Kg mass separated by a distance of 25 cm? What is the acceleration produced on each mass?
Ans. Let the two masses m and n and the gravitational force between them is F then according to law
F= Gmn/r2, where r is the distance between them.
F = (6.674×10-11. x 15 x 25)/0.25 x 0.25
F = 40044x 10-11N = 4 x 10-7N
Acceleration produced in 15 kg mass a = F/m = 4 x 10-7N/15kg
a= 2.66 x10-8 m/s2
Acceleration produced in 25 kg mass a = F/a = 4 x 10-15N/25kg
a = 1.6 x 10-8 m/s2
12. A boy on a cliff 49 m high drops a stone, one second later he throws another stone vertically downwards. The two stones hit the ground at the same time. What was the velocity with which the second stone was thrown?
Ans. For the first stone:
h= 49 m, u = 0 m/s, g = 9.8 m/s2
Then v2 - u2 = 2gh
v2= 2gh + u2 = 2 x 9.8 x 49 + 0 = 960.4m/s
v = 30.99 or 31m/s
time t = (v - u)/g = (31 - 0)/9.8 = 3.16s
Now for second stone :
h = 49 m, g = 9.8 m/s2, time t = 3.16 -1= 2.16s
We have to find u.
h = ut + 1/2 at2
49 =2.16u + 1/2 x 9.8 x 2.16 x 2.16 = 2.16u + 22.86
2.16u = 49 - 22.86 = 26.14
u = 26.14/2.16=12.10m/s
The velocity with which the second stone was thrown is 12.10m/s
13. A ball is thrown vertically upward and it returns after 6 sec. Find:
(a) The velocity with which it was thrown up.
(b) The max.height it reaches.
(c) Its position after 4 sec.
Ans. (a) Let the time for aximum height be t = 6/2 = 3 s.
At the maximum height v = 0
g = - 9.8 m/s2
(a) We have v = u + gt = u - 9.8 × 3
0 = u - 29.4
u = 29.4 m/s
The velocity with which it was thrown up = 29.4m/s2
(b) We have 2aS = v² - u²
h = v²- u²/2g
= 0 - 29.4 × 29.4/- 2× 9.8
= 44.1 m
Thus, Maximum height it reaches = 44.1 m.
14.[Do yourself 👉 (a) A ball is dropped from the roof of a building, it takes 10 seconds to reach the ground to find the height of the building.
(b) A stone is released from the top of a tower of height 19.6 m. Calculate its velocity just before touching the ground]
15. (a) Determine the magnitude of the gravitational force between a planet of mass 6 x 1024.kg and a 1 kg object on its surface. Let the radius of the planet be 6 x 106m G = 6.67 x 10-11Nm2kg.
(b) To estimate the height of the bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2 sec. to reach the water surface in the river. Calculate the height of the bridge from the water level.
Ans. (a) According to the law of gravitation F = GMm/R2
Where R = 6 x 106m
F= (6.67 x 10-11 x 6 x 1024 x 1 )/(6 x 106)2 = 40.02 x 1013 /36 x 1012
= 1.11 x 10 N
F = 11.1N.
(b) Let the height be h.
g = 9.8m/s2
u = 0 m/s
h= ut + 1/2gt2
h = 1/2 x 9.8 x 2 x 2 = 19.6m
16. An astronaut carried a pot containing soil waiting for 60 Newton from the earth to the surface of the motion moon he kept it there and just before returning journey from Moon to Earth he waited for the soil there on the surface of the moon and found that it was only 10 Newton where did the rest of soil go and how much mass of soil has lost?
Ans. We know that wt of the soil on the moon = 1/6th of the weight of the soil on the earth and it is due to, g of moon = 1/6th g of earth.
Given that weight of the soil on the earth surface = 60 N
Therefore the weight of the soil on the moon surface = 1/6 x 60 = 10 N.
As mass always remains constant therefore there is no loss in mass of soil on the moon surface and a decrease in weight is due to the difference in gravity.
17. [Do yourself 👉 (a) Find the weight of an 80 kg man on the surface of the moon? What should be his mass on the earth and on the Moon?
(b) An object of mass 8 units weight 300 Newton on the surface of earth what would be its mass and weight on the surface of the moon?]
18. There are 2 planets A and B masses m1 and m2 respectively. They revolve around the sun in a circular orbit. The orbital radius of A from the sun is twice that of orbital radius of B. The mass of A is twice that of the mass of B.Compare the forces of gravitation between A to Sun and B to Sun.
Ans. Given: Mass of the planet A = 2x Mass of the planet B and
the orbital radius of A from the sun = 2 x The orbital radius of B from the sun.
Therefor m1= 2m2 ....... eq (1)
and r1 = 2r2 ........ eq (2)
According to the law of gravitation :
F1 = GMm1/r12 and F2 = GMm2/r22
F1 = GM 2m2/(2r2)2. (m1= 2m2)
=GM2m2/4(r2)2. ( r1 = 2r2 )
= GMm2/2(r2)2
F1 = 1/2 F2
The force of gravitation between April to Sun is half of the force of gravitation between B to Sun.
19. (a) Calculate the force of gravitation between two objects of mass 50 kg and 120 kg., Kept at a distance of 10 m from one another.
(b) A stone is dropped from a height of 10 m on an unknown planet having g = 20 m/s2. Calculate the speed of the stone when it hits the surface of the planet. Also, calculate the time it takes to fall through this height.
Ans. (a) Given m1=50kg, m2= 120kg and r = 10 m.
G = 6.67 x 10-11 Nm2kg.h
The force of gravitation between two objects F = Gm1m2/r2
F = (6.67 x 10-11 x 50 x 120)/102
F = 400.20 x 10-11N
or F = 4 x 10-9 N
(b) Given height h= 10m, g = 20 m/s2 and u = 0 m/s
We have v2 - u2 = 2gh
v2 = 2 x 20 x 10 = 400. ( u = 0 )
v = 20 m/s
To find t:
We have v = u + gt
20 = 0 + 20 x t =20t
t = 20/20 = 1 s
t = 1 s .
Ans. Given :
When a stone is dropped from the edge of the roof,
its initial velocity = 0 .
The stone passes the window of 2m high in 0.1s.
Let the distance between the roof and that window be 'h' m and the time taken to reach that "h"m is t.
Therefore h = ut + 1/2gt² = 1/2gt²........eq (1)
A/Q Stone passes the window in 0.1 seconds.
So it travels h+2 m in t + 0.1seconds.
h + 2 = 1/2g(t+ 0.1)²
1/2gt² + 2 = 1/2g(t²+0.01 + 0.2t) From eq ..(1)
1/2gt² - 1/2gt² + 2 = 0.005 g + 0.1gt
2 = 0.005 x 10 + 0.1 x 10 x t
2= 0.05 + t
2 - 0.05 = t
t = 1.95 s.
Therefore, The stone takes 1.95 seconds to travel from roof to top of the window.
The distance between the roof and top of the window (h)
h = 1/2 x 10 x t²
h = 5t²
h= 5x 1.95²
= 19.0125 m.
Therefore, the height of the roof above the top of the window is 19.0125m
6. State the Universal law of Gravitation? Derive an expression for gravitational force between two bodies.
Ans. The universal law of gravitation states that the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Let the two bodies A and B be of masses m1 and m2 respectively.
According to universal law:
F ∝ m1m2........eq. (1)
F ∝ 1/r²............eq. (2)
From eq...1 & 2 we get
F ∝ m1m2/r2
F = G m1m2/r2
Where G is the universal constant.
7. A stone is dropped from a height of 50 m on earth. At the same time, another stone is thrown vertically upwards from the ground with a velocity of 50m/s. At what height from the ground will the two stones meet (g = -10m/s2)
Ans. Let the distance covered by stone A which is dropped from a height be h.
h = 1/2gt² ( u = 0 m/s).....1
Height of the Stone B which is thrown vertically upwards
= 50 - h
= ut - 1/2gt²
50 - h = 50 t - 1/2 g t²
50 - 1/2 gt² = 50 t - 1/2 g t² ( from eq. 1)
50 = 50t
t = 1s
Now distance from the ground = 50 - 1/2gt² = 50 - 1/2 x 10 x1 = 45 m
8. A ball is thrown vertically upwards. The speed of the ball was 10m/s when it had reached one half of its maximum height. How high does the ball rise?
Ans. (a) When a ball is thrown vertically upward, the speed of the ball at half of the max.height =v = 10 m/s.
Let the maximum height be H and the final velocity is zero at height this height.
Now distance = H.
We have v² - u² = 2x g x H
0² - u² = 2 x -10 x H
- u² = -20H
u² = 20H
When the ball reaches half the maximum height, the final velocity is 10 m/s, and the distance traveled is H/2.
We have v² - u² = -2gH/2
10² - 20H = 2 x -10 x H/2
100 - 20H = -10H
10H = 100
H = 10m
Therefore the maximum height is 10 m.
9. If the mass of one of the objects is doubled and the mass of others remains the same and if the distance between them is halved then how does the gravitational force change?
Ans. Let the masses of two objects are m1 and m2. and the distance between them = r
Therefore the gravitational force F1 = m1m2/r2.
Now, when the mass of one of the object is doubled and the mass of others remains the same and if the distance between them is halved then:
Force F2 = 2m1m2/(r/2)2.
F2 = 8 m1m2/r2.
F2 = 8F1
If the mass of one of the objects is doubled and the mass of others remains the same and if the distance between them is halved then the gravitational force becomes eight times.
10. The radius of the earth is about 6370 Km. An object of mass 30 Kg is taken to a height of 230 km above the surface of the earth.
(a) What is the mass of the body?
(b) What is the acceleration due to gravity at this height?
(c) What is the weight of the body at this height?
Ans. (a) The of the will not change because mass is constant in everywhere.
(b) We have g = g'(1+h/R)2
Where g = acceleration due to gravity.
h= height of the body aabove the surface of the earth =230km
R= radius of the earth =
g' = acceleration due to gravity at the height of 230km.
g = g'(1+ h/R)2
9.8 =g'(1 + 230/ 6370)2= g'(6370 + 230)/ 6370)2
9.8 = g'(6600/ 6370)2 = g'(1.04)2 = g'x1.08
g'= 9.8/1.08 = 9.07m/s2
(c) The weight of this body = mg = 30 x 9.07 = 272.1 kg.
11. What is the magnitude of the gravitational force exerted by a 15Kg mass on a 25 Kg mass separated by a distance of 25 cm? What is the acceleration produced on each mass?
Ans. Let the two masses m and n and the gravitational force between them is F then according to law
F= Gmn/r2, where r is the distance between them.
F = (6.674×10-11. x 15 x 25)/0.25 x 0.25
F = 40044x 10-11N = 4 x 10-7N
Acceleration produced in 15 kg mass a = F/m = 4 x 10-7N/15kg
a= 2.66 x10-8 m/s2
Acceleration produced in 25 kg mass a = F/a = 4 x 10-15N/25kg
a = 1.6 x 10-8 m/s2
12. A boy on a cliff 49 m high drops a stone, one second later he throws another stone vertically downwards. The two stones hit the ground at the same time. What was the velocity with which the second stone was thrown?
Ans. For the first stone:
h= 49 m, u = 0 m/s, g = 9.8 m/s2
Then v2 - u2 = 2gh
v2= 2gh + u2 = 2 x 9.8 x 49 + 0 = 960.4m/s
v = 30.99 or 31m/s
time t = (v - u)/g = (31 - 0)/9.8 = 3.16s
Now for second stone :
h = 49 m, g = 9.8 m/s2, time t = 3.16 -1= 2.16s
We have to find u.
h = ut + 1/2 at2
49 =2.16u + 1/2 x 9.8 x 2.16 x 2.16 = 2.16u + 22.86
2.16u = 49 - 22.86 = 26.14
u = 26.14/2.16=12.10m/s
The velocity with which the second stone was thrown is 12.10m/s
13. A ball is thrown vertically upward and it returns after 6 sec. Find:
(a) The velocity with which it was thrown up.
(b) The max.height it reaches.
(c) Its position after 4 sec.
Ans. (a) Let the time for aximum height be t = 6/2 = 3 s.
At the maximum height v = 0
g = - 9.8 m/s2
(a) We have v = u + gt = u - 9.8 × 3
0 = u - 29.4
u = 29.4 m/s
The velocity with which it was thrown up = 29.4m/s2
(b) We have 2aS = v² - u²
h = v²- u²/2g
= 0 - 29.4 × 29.4/- 2× 9.8
= 44.1 m
Thus, Maximum height it reaches = 44.1 m.
14.[Do yourself 👉 (a) A ball is dropped from the roof of a building, it takes 10 seconds to reach the ground to find the height of the building.
(b) A stone is released from the top of a tower of height 19.6 m. Calculate its velocity just before touching the ground]
15. (a) Determine the magnitude of the gravitational force between a planet of mass 6 x 1024.kg and a 1 kg object on its surface. Let the radius of the planet be 6 x 106m G = 6.67 x 10-11Nm2kg.
(b) To estimate the height of the bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2 sec. to reach the water surface in the river. Calculate the height of the bridge from the water level.
Ans. (a) According to the law of gravitation F = GMm/R2
Where R = 6 x 106m
F= (6.67 x 10-11 x 6 x 1024 x 1 )/(6 x 106)2 = 40.02 x 1013 /36 x 1012
= 1.11 x 10 N
F = 11.1N.
(b) Let the height be h.
g = 9.8m/s2
u = 0 m/s
h= ut + 1/2gt2
h = 1/2 x 9.8 x 2 x 2 = 19.6m
16. An astronaut carried a pot containing soil waiting for 60 Newton from the earth to the surface of the motion moon he kept it there and just before returning journey from Moon to Earth he waited for the soil there on the surface of the moon and found that it was only 10 Newton where did the rest of soil go and how much mass of soil has lost?
Ans. We know that wt of the soil on the moon = 1/6th of the weight of the soil on the earth and it is due to, g of moon = 1/6th g of earth.
Given that weight of the soil on the earth surface = 60 N
Therefore the weight of the soil on the moon surface = 1/6 x 60 = 10 N.
As mass always remains constant therefore there is no loss in mass of soil on the moon surface and a decrease in weight is due to the difference in gravity.
17. [Do yourself 👉 (a) Find the weight of an 80 kg man on the surface of the moon? What should be his mass on the earth and on the Moon?
(b) An object of mass 8 units weight 300 Newton on the surface of earth what would be its mass and weight on the surface of the moon?]
18. There are 2 planets A and B masses m1 and m2 respectively. They revolve around the sun in a circular orbit. The orbital radius of A from the sun is twice that of orbital radius of B. The mass of A is twice that of the mass of B.Compare the forces of gravitation between A to Sun and B to Sun.
Ans. Given: Mass of the planet A = 2x Mass of the planet B and
the orbital radius of A from the sun = 2 x The orbital radius of B from the sun.
Therefor m1= 2m2 ....... eq (1)
and r1 = 2r2 ........ eq (2)
According to the law of gravitation :
F1 = GMm1/r12 and F2 = GMm2/r22
F1 = GM 2m2/(2r2)2. (m1= 2m2)
=GM2m2/4(r2)2. ( r1 = 2r2 )
= GMm2/2(r2)2
F1 = 1/2 F2
The force of gravitation between April to Sun is half of the force of gravitation between B to Sun.
19. (a) Calculate the force of gravitation between two objects of mass 50 kg and 120 kg., Kept at a distance of 10 m from one another.
(b) A stone is dropped from a height of 10 m on an unknown planet having g = 20 m/s2. Calculate the speed of the stone when it hits the surface of the planet. Also, calculate the time it takes to fall through this height.
Ans. (a) Given m1=50kg, m2= 120kg and r = 10 m.
G = 6.67 x 10-11 Nm2kg.h
The force of gravitation between two objects F = Gm1m2/r2
F = (6.67 x 10-11 x 50 x 120)/102
F = 400.20 x 10-11N
or F = 4 x 10-9 N
(b) Given height h= 10m, g = 20 m/s2 and u = 0 m/s
We have v2 - u2 = 2gh
v2 = 2 x 20 x 10 = 400. ( u = 0 )
v = 20 m/s
To find t:
We have v = u + gt
20 = 0 + 20 x t =20t
t = 20/20 = 1 s
t = 1 s .
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