Frank ICSE Solutions for Class 9 Physics Chapter 3 – Laws of Motion Solutions
Frank ICSE Solutions for Class 9 Physics Chapter 3 – Laws of Motion Solutions
PAGE NO: 113.
Solution – 01.
The property by which a body neither changes its present state of rest or of uniform motion in a straight line nor tends to change the present state,is known as inertia.
Solution – 02.
A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.
Solution – 03.
The greater is the MASS , the greater is the inertia of the object.
Solution – 04.
An object possess two kind of inertia, inertia of rest and inertia of motion.A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.
Solution – 05.
1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it. 1 newton would produce acceleration of 1 ms-2 in a mass of 1 kg.
Solution – 06.
The acceleration produced bya force in an object is directly proportional to the applied FORCE and inversely proportional to the MASS of the object.
Solution – 07.
SI unit of force is Newton (N).
Solution – 08.
Acceleration is the physical quantity associated with N kg-1.
Solution – 09.
1 N = 105 Dyne.
Solution – 10.
As mass of loaded van is greater than sports car so it would require more force to stop.
Solution – 11.
We know force = mass X acceleration.
a= f/m = 12 N / 4 kg. = 3 ms-2
so acceleration of the body would be 3 ms-2.
Solution – 12.
SI unit of force is Newton whereas CGS unit of force is dyne.
1 newton / 1dyne = 105.
Solution – 13.
SI unit of momentum is kgms-1.
Solution – 14.
Momentum is defined as the amount of motion contained in the body. It is given by the product of the mass of the body and its velocity.
Solution – 15.
Momentum is the physical quantity associated with the motion of the body.
Solution – 16.
Momentum is possessed by bodies in MOTION.
Solution – 17.
A fast pitched soft ball has more momentum.
Solution – 18.
SI unit of momentum is kgms-1 and CGS unit of momentum is g cms-1.
And their ratio is = 1000 X 100 g ms-1= 1:10.
Solution – 19.
A body at rest has zero momentum as its velocity is zero.
Solution – 20.
According to Newton’s third law, for every action there is always an equal and opposite reaction.
Solution – 21
When a force acts on a body then this is called an action.
Solution – 22.
No, action and reaction never act on a same body they always act simultaneously on two different bodies.
Solution – 23.
2nd law of motion gives the definition of force.
Solution – 24.
Newton’s third law explains this statement.
Solution – 25.
Force is a vector quantity.
Solution – 26.
This means these forces are balanced forces.
Solution – 27.
Passengers tend to fall sideways when the bus takes a sharp turn due to the inertiaof direction.
Solution – 28.
Passengers are thrown in the forward direction as the running bus stops suddenly because due to their inertia of motion, their upper body continues to be in the state of motion even though the lowerbody comes to rest when the bus stops
Solution – 29.
Passengers tends to fall in backward direction when bus starts suddenly because due to their inertia of rest, as soon as the bus starts, their lower body comes in motion but the upper body continues to be in the state of rest.
Solution – 30.
No, internal forces cannot change the velocity of a body.
Solution – 31.
When a hanging carpet is beaten using a stick, the dust particles will start coming out of the carpet because the part of the carpet where the stick strikes, immediately comes in motion while the dust particle sticking to the carpet remains at rest . Hence a part of the carpet moves ahead alongwith the stick, and the dust particles fall down due to the earth’s pull.
Solution – 32.
When we shake the branches of a tree, the fruits and leaves remain in state of rest while branches comes in rest so fruits and leaves are detached from the tree.
Solution – 33.
We know force = mass X acceleration
F1 = 10 X 5 = 50 dyne.
F2 = 20 X 2 = 40 dyne.
So first body require more force
Solution – 34.
PAGE NO: 114.
Solution – 35.
initial velocity of the object = 0 ms-1
Acceleration of the object = 8 ms-2.
Time = 5 s.
Distance covered would be S = ut + 1/2 at2.
S = 1/2 X 8 X 5 X5 = 100 m.
Solution – 36.
Initial velocity of the truck = 0 ms-1
Distance covered by truck = 100 m
Time taken to cover this distance = 10 s.
We know Distance covered would be S = ut + 1/2 at2.
100 =1/2 Xa X100
a= 2 ms-2.
Mass of truck = 5 metric tons = 5000 kg.
Force acted on truck = mass X acceleration
Force = 5000 X 2 = 10000 N.
Solution – 37.
Momentum is used for quantifying the motion of body.
Solution – 38.
When we fire a gun, a force is exerted in the forward in the forward direction as the bullets comes out; in reaction to which an equal and opposite force is act in the backward direction and hence, we feel a backward jerk on the shoulder.
Solution – 39.
A person applies force on water in backward direction and water according to third law of motion water apply an equal and opposite force in forward direction which helps a person to swim.
Solution – 40.
Newton’s third law of motion is involved in the working of a jet plane.
Solution – 41.
Yes, a rocket can propel itself in a vacuum once it is given initial velocity.
Solution – 42.
Action is equal and opposite to reaction but they act on different bodies and object moves as movement requires an unbalanced force and these are provided once inertia is overcome.
PAGE NO: 125.
Solution – 01.
Sir Isaac Newton stated the law of gravitation.
Solution – 02.
Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
Solution – 03.
Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
Solution – 04.
Acceleration due to gravity is the acceleration experienced by a body during free fall.
Solution – 05.
g = GM/R2
Solution – 06.
We know that law of gravitation.
F = G ( m1 X m2)/R2.
Here G is universal constant and is called constant of gravitation. It doesnot depend upon on the value of m1,m2 or R.
Its value is same between any two objects in the universe.
Solution – 07.
SI unit of constant of gravitation is Nm2kg-2.
Solution – 08.
we know that law of gravitation.
F = G ( m1 X m2)/R2.
(a) If distance between them is halved then put R = R/2.
F = 4 X G( m1 X m2)/ R2.
F1 = 4 F.
(b) If distance between them is doubled then put R = 2R.
F = G( m1 X m2)/ 4R2.
F1 = F/4.
(c) If distance between them is made four times then put R = 4R.
F = G( m1 X m2)/16 R2.
F1 = F/16.
(d) If distance between them is infinite then put R = infinite.
F = G( m1 X m2)/ R2.
F1 = 0.
(e) If distance between them is almost zero then put R = 0.
F = G( m1 X m2)/ 0.
F1 = infinite.
Solution – 09.
All objects in the universe attract each other along the line joining their CENTRES.
Solution – 10.
The force of attraction between any two material objects is called FORCE OF GRAVITATION.
Solution – 11.
The gravitational force of the earth is called earth’s GRAVITY.
Solution – 12.
The Gravity is a particular case of GRAVITATIONAL FORCE OF EARTH.
Solution – 13.
The value of G is extremely SMALL.
Solution – 14.
Yes the law of gravitation is also applicable in case of the sun and moon.
Solution – 15.
we know that law of gravitation.
F = G ( m1Xm2)/R2.
Mass of earth = 6X1024kg.
Mass of the person = 100 kg.
G = 6.7 X10-11 Nm2kg-2.
Radius of earth = 6.4 X 1014.
F = (6.7 X10-11X 100 X 6 X1014 )/ (6.4 X6.4 X1012) = 981.4N
Force of gravity due to earth acting on a 100 kg person is 981.4 N.
Solution – 16.
Objects fall towards the earth due to force of gravitation.
Solution – 17.
Because the masses of persons are not large enough to overcome the value of small constant of gravitation so the force of gravitation is very small and negligible to feel.
Solution – 18.
Initial speed of ball is = 4.9 ms-1.
Acceleration due to gravity = -9.8 ms-2.
(a) We know v2 – u2 =2as
At highest point final velocity is zero so
0 – 4.9 X 4.9 = 2 X (-9.8) S
S = 1.125 m
(b) We know v = u + at
0 = 4.9 – 9.8 t
T = 0.5 sec.
(c) for highest point initial velocity is zero
Acceleration due to gravity is = 9.8 ms-2.
Final velocity at ground is v
V2 – 0 = 2 X9.8 X 1.125
V = 4.9 ms-1.
Time taken to reach ground from highest point
V = u + at
4.9 = 0 + 9.8 t
T = 4.9/9.8 = 0.5 sec.
So time of ascent is equal to time descent.
Solution – 19.
g = GM/R2.
Solution – 20.
Value of the g at the surface of the earth is 9.8ms-2
Solution – 21.
Mass of the body is constant at all positions so mass will not change. But weight will change as gravity on the surface of earth is almost 6 times than on the surface of the moon, so its weight will increase almost 6 times on the surface of earth.
Solution – 22.
We will weigh more on the surface of the earth.
Solution – 23.
Beam balance is used to measure the mass of a body.
Solution – 24.
Spring scale is used to measure the weight of a body.
Solution – 25.
The weight is greater at the poles than the equator.
Solution – 26.
Newton 1N = 9.8 kgwt.
Solution – 27.
We will weigh more on earth surface as value of g is greater on earth surface.
Solution – 28.
No, the force of gravitation between two objects does not depend on the medium between them.
Solution – 29.
we know that law of gravitation.
F = G ( m1Xm2)/R2.
Now m1 = 2 m1
m2 = 2 m2
R = 2 R
F1 = G ( 2m1 X2 m2)/4R2.
F1 = F
So force between them remains same.
PAGE NO: 126.
Solution – 30.
Yes, in absence of gravity all freely falling body have same force acting on them.
Solution – 31.
g= GM/R2
it means acceleration due to gravity is directly proportional to the mass of body and inversely proportional to the square of distance between earth and object.
Solution – 32.
Yes a body falling freely near the earth surface has a constant acceleration.
Solution – 33.
As we know
g= 1/R2
so value of g is more at poles than equator so value of g is maximum near a camp site in Antarctica as this lie on the pole.
Solution – 34.
PAGE NO: 128.
Solution – 01.
Force is that external agency which tends to change the state of rest or the state of motion of a body.
Solution – 02.
1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.
Solution – 03.
Newton is the SI unit of force whereas dyne is the CGS unit of force.
1 N = 105 dyne.
Solution – 04.
No, force is a vector quantity.
Solution – 05.
A force can produce MOTION in an objectat rest. It can ACCELERATE an object and can change its DIRECTION of motion.
Solution – 06.
(a) force changes the shape of skin.
(b) force produces stretching in the rubber.
(c) force provides retardation to the car and finally stops the car.
(d) force decreases the momentum of ball and finally stops the ball.
Solution – 07.
No, every force does not produce motion in every type of body.
Solution – 08.
The amount of inertia of a body depends on its MASS.
Solution – 09.
You can change the direction in which an object is moving by APPLYING FORCE ON IT.
Solution – 10.
A man riding on a car has INERTIA of motion.
Solution – 11.
When a body is at rest , it will continue to remain at rest unless some external force is applied to change its state of rest. This property of body is called inertia of rest.
Solution – 12.
(i) Weight of the book is action and normal force applied by table on book is reaction.
(ii) Force applied by man on ground is action and force of friction is the reaction.
(iii) Force applied by hammer on nail is action and normal force applied by nail on hammer is reaction to this force.
(iv) Firing of bullet is the action and recoiling of gun is the reaction.
(v) Force applied by us on wall is action and opposite force applied by wall on us or we can say that resistance of wall to our force is reaction.
Solution – 13.
A book lying on a table will remain placed on table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.This is an example of inertia of motion.
Solution – 14.
Unbalance external force causes motion in the body.
Solution – 15.
Linear Momentum is defined as the amount of motion contained in a body. It is given by the product of the mass of the body and its velocity.
Solution – 16.
SI unit of momentum is kgms-1.
Solution – 17.
According to Newton’s first law force is that external agency which tends to change the state of rest or the state of motion of a body.
Solution – 18.
According to Newton’s first law, everybody continues in its state of rest or in uniform motion in a straight line unless compelled by some external force to act otherwise.
PAGE NO: 129.
Solution – 19.
Out of all these, as mass of truck is greatest and mass is measure of inertia so a truck has maximum inertia.
Solution – 20.
It is advantageous to run before taking a long jump because after running we get motion of inertia which helps in long jumping.
Solution – 21.
Ball moving on a table top stops eventually due to force of friction between the ball surface and table surface.
Solution – 22.
Force is equal to the rate of change of linear momentum.
Solution – 23.
According to newton second law of motion, when a force acts on a body, the rate of change in momentum of a body is equal to the product of mass of the body and acceleration produced in it.
Yes, Newton’s first law is contained in the second law as if force is zero then acceleration would be zero which means body would remain in its state of rest or in state of constant motion.
Solution – 24.
1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.
1 newton / 1dyne = 105.
Solution – 25.
1 newton = 1 kg X 1 ms-1 = 1000 g X 100 cms-1 = 105 cms-1.
1 dyne 1 g X 1 cms-1 = 1cms-1.
So 1 newton = 105 dyne
Solution – 26.
No, the body will not move as the two forces are equal and opposite and they constitute balanced forces.
Solution – 27.
As these forces are balanced so they will not affect the motion and motion of the body will remain unaffected.
Solution – 28.
According to Newton’s third law, for every action there is always an equal and opposite reaction. Rocket works on the same principle. The exhaust gases produced as the result of the combustion of the fuel are forced out at one end of the rocket. As a reaction , the main rocket moves in the opposite direction.
Solution – 29.
According to Newton’s third law, every action has equal and opposite reaction so force exerted by the wall on the boy is 30 N.
Solution – 30.
Newton stated the law of inertia.
Solution – 31.
Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
Law of gravitation is called universal because it applies to all bodies of universe.
Solution – 32.
Gravity is the force of attraction betwen the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
Solution – 33.
Person will weigh more at Delhi as we know that gravity decreases with increase in height. Now as Shimla is at a height from Delhi so weight is less in Shimla and more in Delhi.
Solution – 34.
Spring scale is used to measure the weight of a body.
Solution – 35.
Gravity is another kind of FORCE. It exerts all through the UNIVERSE. The sun’s gravity keeps the PLANETS in their orbits. Gravity can only be felt with very large MASS.
Solution – 36.
(i) Objects fall on the earth due to gravitational force between the earth and object.
(ii) Atmosphere doesnot escape because molecules of atmosphere are attracted by earth due to gravitational force of earth.
(iii) A moon rocket needs to reach a certain velocity because during its motion earth attracts the rocket towards it by its gravitational force.
Solution – 37.
‘g’ is acceleration due to earth’s gravity and ‘G’ is universal gravitational constant.
Solution – 38.
Free fall means motion of a body under the gravity of earth only.
Solution – 39.
Yes, we have a gravitational force of attraction between us and a book. But our mass is very small so the force between us and book is very small almost negligible.
Solution – 40.
Yes, the force of gravitation of earth affects the motion of moon, because moon is revolving around earth and centripetal force for this revolution is provided by earth’s gravitation.
Solution – 41.
Inertial mass is measure of inertia of the object. According to second law of motion F = m X a
m= F/a and this mass is called as inertial mass.
Newton law of gravitation gives another definition of mass.
F = (G m1m2)/R2
Thus m2 is the mass of the body by which another body of mass m1 attracts it towards it by law of gravitation. This mass is called gravitational mass.
Solution – 42.
Newton law of gravitation is that Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
(i) Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
(ii) ‘g’ is acceleration due to earth’s gravity and ‘G’ is universal gravitational constant.
Solution – 43.
Yes, it is true that apple attracts the earth towards it with same force but the mass of earth is so huge that acceleration produced in it due to this force is very much small and negligible to notice.
Solution – 44.
we know that law of gravitation is
F = G ( m1Xm2)/R2
Here the F is force due to attraction and this force is equal to weight of the body m2g.
So m2g = G ( m1Xm2)/R2
g= (G Xm2)/R2.
Here R is the distance between earth centre and the object centre. Now if we go on higher altitude say ‘H’ then this R would increase to (R + H)
And value of gravity at height H becomes
g’= (G�m2)/( R +H)2.
As denominator increases so g’ would be less than g and hence we can say that gravity decreases on higher altitudes.
Solution – 45.
(i) The force exerted by the block on is the weight of box and that is equal to 20N.
(ii) The force exerted by string on block is equal to the tension in the string and this is also equal to the 20N.
Solution – 46.
we know F = m X a
m= F/a
so we can calculate mass of each body
Mass of body 1 m1 = 4/8 = 0.5 kg.
Mass of body 2 m2 = 4/20 = 0.2 kg.
Total mass when two masses are tied together M = 0.5 + 0.2 = 0.7 kg.
Now as force is acting on total mass so acceleration produced is
a= 4/0.7 = 5.71 ms-2.
PAGE NO: 130.
Solution – 47.
Solution – 48.
Initial speed of body = 5 ms-1
Final speed of body = 8 ms-1
Time taken to acquire this speed = 2 s.
Acceleration of body = ( v- u)/t
a= (8- 5)/2 = 1.5 ms-2.
Force applied on body = 0.9 N.
we know F = m X a.
m = f/a = 0.9/1.5 = 0.6 kg
mass of the body is 600 gm.
Solution – 49.
Solution – 50.
The force that acts on a body for a very short time but produces a large change in its momentum, is known as impulsive force.
Solution – 51.
initial velocity of body = 0 ms-1.
Final velocity of body = 100 ms-1.
Mass of body = 20 kg.
Force applied = 100N.
We know that
F X t = m (v – u)
100 t = 20 (100 -0)
T = 2000/100 = 20 s.
Solution – 52.
SI unit of retardation is ms-2.
Solution – 53.
Force applied is equal to the product of mass and acceleration produced in the body.
F = mass X acceleration.
Solution – 54.
According to Newton’s second law of motion, when a force acts on a body, the rate of change in momentum of a body equals the product of mass of the body and acceleration produced in it due to that force, provided the mass remains constant.
Mass of body = 400 g = 0.4kg
Force applied on body = 0.02 N
Acceleration = force/mass = 0.02/0.4 = 0.05 ms-2.
Solution – 55.
Linear Momentum is defined as the physical quantity which is associated with bodies in linear motion. It is given by the product of the mass of the body and its velocity.
Mass of body = 1 kg
Acceleration produced = 10 ms-2.
Force applied would be = 1 X 10 N = 10 N.
Mass of second body = 4 kg.
As same force has to be applied on second body so force = 10N.
Acceleration produced is = F/M =10/4 = 2.5 ms-2.
Solution – 56.
Mass of P is m1= m.
Velocity of P is v1 =2 v
Mass of Q is m2 = 2m
Velocity of Q is v2 = v.
(i) inertia of P/inertia of Q = m1/m2 = 1/2.
So ratio of inertia of two bodies is 1:2.
(ii) Momentum of P/momentum of Q = m1v1/m2v2 = 1
So ratio of momentum of two bodies is 1:1.
(iii) As force required to stop them is equal to change in their momentum from moving to rest.
So ratio would be same as the ratio of their momentum i.e 1: 1.
Solution – 57.
According to newton second law
F = m X a
a= (v – u)/t.
F = m(v -u)/t
F = (mv – mu)/t
As F= m X a
ma = (mv – mu)/t
so rate of change of momentum = mass X acceleration.
This relation holds good when mass remains constant during motion.
Solution – 58.
Solution – 59.
According to newton third law, for every action there is always an equal and opposite reaction.
To demonstrate newton third law blow a balloon and hold its neck tightly facing downwards. When we release the balloon, the balloon will moves up instead of falling to the ground. As air is releasing from bottom of balloon and this air apply equal and opposite force to the balloon and this force helps balloon to move upwards.
Solution – 60.
time for which force is applied = 0.1 s.
Mass of body = 2 kg
Initial velocity of body = 0 ms-1
Final velocity of body = 2 ms-1.
We know FX t = m (v – u)
F X 0.1 = 2 (2 – 0)
F = 4 /0.1 = 40 N.
Solution – 61.
mass of ball = 500g = 0.5 kg.
Initial speed of the ball = 30 ms-1
Final speed of ball = 0 ms-1
Time taken by player to stop the ball = 0.03 s.
We know FX t = m (v – u)
F X 0.03 = 0.5 (0 – 30)
F = – 1.5 / 0.03 = – 500 N
(- ) sign shows that player has to apply force in opposite direction of the motion of the ball.
Solution – 62.
Time for which force is applied =0.1 s.
Mass of the body = 3.2 kg.
Initial speed of body = 0 ms-1
After removal of forces body covers a distance of 3m in 1 second so final speed of body = 3/1 = 3ms-1.
We know FX t = m (v – u)
F X 0.1 = 3.2 (3 -0)
F = 9.6/0.1 = 96 N.
So applied force is 96 N.
Solution – 63.
PAGE NO: 131.
Solution – 64.
Time for which force is applied =3 s.
Mass of the body = 2 kg.
Initial speed of body = 0 ms-1
Force applied = 10 N.
(i) We know F X t = m (v – u)
10 X3 = 2 (v- 0)
v = 15 ms-1.
Final velocity is 15 ms-1.
(ii) As m(v – u) is change in momentum and this is equal to the F X t so change in momentum is equal to the 30 kgms-1.
Solution – 65.
(i) We always prefer to land on sand instead of hard floor while taking a high jump because sand increases the time of contact.As FX t = m ( v – u ) and our change in momentum is constant so if time increases then force experienced would decrease.
(ii) Again while catching a fast moving ball, we always pull our hands backwards to increase reaction time so force experienced would decrease.
Solution – 66.
Height of cliff = 98 m.
Initial velocity of stone = 0 ms-1.
Acceleration due to gravity = 9.8 ms-2.
(i) We know H = ut + 1/2 gt2.
98 = 1/2 X 9.8X t2.
t2 = 98X2/9.8 = 20
t= 4.47 sec.
(ii) Final velocity when it strikes the ground
V2 – u2 = 2 g H
V2 = 2X9.8X98
V2= 1920
V= 44.6 ms-1.
Solution – 67.
Initial speed of ball is = 9.8 ms-1.
Acceleration due to gravity = -9.8 ms-2.
Final speed at maximum height = 0 ms-1.
We know v = u + at
0 = 9.8 – 9.8 t
T = 1 sec.
We know v2 – u2 =2as
At highest point final velocity is zero so
0 – 9.8 X 9.8 = 2 X (-9.8) S
S = 4.9 m.
for highest point initial velocity is zero
Acceleration due to gravity is = 9.8 ms-2.
Final velocity at ground is v
V2 – 0 = 2 X9.8 X 4.9
V = 9.8 ms-1.
Time taken to reach ground from highest point
V = u + at
9.8 = 0 + 9.8 t
T = 9.8/9.8 = 1 sec.
Total time = time of ascent + time of descent.
Total of flight = 1+ 1 = 2 seconds.
Solution – 68.
Initial speed of ball = 10 ms-1.
Acceleration due to gravity on ball = – 9.8 ms-2
We know that from first equation of motion
v= u + gt.
After 1 sec
v= 10 – 9.8 X1
v= 0.2 ms-1
so velocity after 1 sec would be 0.2 ms-1.
Velocity after 2 seconds
v= 10 – 9.8X2 = 10 – 19.6 = -9.6 ms-1.
Here negative sign shows that velocity is in downward direction and magnitude is 9.6 ms-1.
Solution – 69.
Maximum Height attained by ball = 19.6 m
Let initial speed of ball = u ms-1.
Acceleration applied on ball due to gravity = -9.8 ms-2.
Final speed of ball at maximum height = 0 ms-1.
We know that from second equation of motion
V2 – u2 = 2as
0 -u2 = 2 X(-9.8)X19.6
u2 = 19.6 X 19.6
u= 19.6 ms-1
so initial speed of ball to attain maximum height of 19.6 m should be 19.6ms-1.
Solution – 70.
Height of tower = 98 m
Acceleration due to gravity on stone = 9.8 ms-2.
Initial speed of ball= 0 ms-1.
Let initial speed of second stone is v ms-1.
We know from second equation of motion
S = ut + 1/2 a Xt2.
98 = 0 + 1/2 X9.8Xt2.
t2 = 20
t= 4.47 sec.
As second stone is thrown 1 sec later so time taken by second body to cover distance of 98 m is = 4.47 – 1 = 3.47sec.
So again put t= 3.47 sec and S = 98 m in second equation of motion we get
98 = vX3.47 + 1/2 X9.8X3.47X3.47.
98 = 3.47Xv + 59
3.47X v = 98 – 59
v= 39/3.47 = 11.23 ms-1.
Initial speed of second stone should be 11.23 ms-1.
Solution – 71.
Mass of object 1 m1 = 200 mg = 200 �10-6 kg = 2 �10-4 kg.
Mass of object 2 m2= 200 mg = 200 �10-6 kg = 2 �10-4 kg.
Distance between the two objects = 1 mm = 10-3 m
We know law of gravitation is
F = G ( m1�m2)/R2
F = (6.67 � 10-11�2�10-4�2�10-4)/(10-3�10-3)
F = 6.67 �2�2�10-11-4-4+3+3
F = 26.68 �10-13 N
So these two objects would experience a force of 26.68 �10-13 N.
Solution – 72.
Radius of earth = 6.38 �103 km = 6.38�106 m
G = 6.67 � 10-11
Acceleration due to gravity = 9.8 ms-2.
We know that
g= (G � M)/R2.
9.8 = (6.67 �10-11�M)/ ( 6.38�106�6.38 �106)
9.8 �6.38�6.38�1012 = 6.67 �10-11� M
398.9 �1012 = 6.67 �10-11� M
M = 398.9 �1012/6.67 �10-11
M = 59 � 1023 kg
So mass of earth is 59 �1023 kg.