Selina Concise Class 9 Physics Solution Chapter No. 3- ‘Laws of Motion’ For ICSE Board Students.
Exercise -3A Solution
1.) Solution:
Ans:
a) Contact forces:
Contact forces are the forces which are acting on a body when they are in physical contact with each other.
Contact forces are produced only when one body comes in physical contact with another body.
For example:
Firctional force between two bodies
Normal reaction force
Force of tension exerted by string
Now, when two bodies comes closer physically, there will be a frictional force is generated which is called as contact force.
b) Non-contact force:
Non-contact forces are produced even when no two bodies are in physical contact. Hence, non-contact forces are also called as forces at a distance.
For example:
Gravitational force, electrostatic force, magnetic force.
For example:
In universe, every body attracts another body due to their mass with a force called as gravitational force. And this force is exerted even no two bodies are in physical contact. Hence, gravitational force is the non-contact force.
2.) Solution:
Ans:
Frictional force, normal reaction force, force of tension in a string are the contact forces.
And gravitational force, electrostatic force, magnetic force are the non-contact forces.
3.) Solution:
Ans:
a) the force of contact:
Frictional force is the best example of force of contact as it is produced due to the physical contact of the two bodies.
When one body is rubbed or pushed over the other body then there will be frictional force which resist the motion of body.
When we place a book on table and we push the book by applying force on it. Then it is seen that there will be frictional force acting on the book which resist it’s motion also. Suppose if we applied force in right direction then frictional force will be act in its opposite direction i.e. in left direction and hence its motion is opposed. Thus, we can say that frictional force is the contact force generated due to physical contact of two bodies.
b) force at a distance:
Force at a distance is the non-contact force in which force is produced even when no two bodies are in physical contact with each other.
Gravitational force is the best example of non contact force.
For example, when we thrown a stone in upward direction after sometime it returns to down towards earth’s surface because of the acceleration due to gravity which acts on stone as it has mass. This is the gravitational force of attraction between two bodies, which are not in contact also. Thus, we can say that gravitational force is the non contact force.
4.) a) a ball is hanging by a string from a ceiling of the roof. Draw a neat labeled diagram showing the forces acting on the ball and the string.
b) a string is compressed against the rigid wall. Draw a neat labeled diagram showing the forces acting on the string.
c) a wooden block is placed on a table top. Name the forces acting on the block and draw a neat labeled diagram to show the point of application and direction of these forces.
5.) Solution:
Ans:
The magnitude of non-contact forces depends on the distance of separation between the two bodies. If the distance of separation is increased the non-contact force decreases. While if the distance of separation is decreased then non-contact force increases. Thus, magnitude of non-contact force is inversely proportional to the square of the distance of separation.
6.) Solution:
Ans:
We know that, magnitude of non-contact forces depends on the distance of separation.
And gravitational force is the non-contact force whose magnitude is inversely proportional to the square of the distance of separation.
If r is the distance of separation then,
F a 1/r2
And if the distance of separation is made half then r become r/2.
Hence, F a 1/(r/2)2
Hence, F a 4/r2
Thus, the magnitude of gravitational force is increased four times.
7.) Solution:
Ans:
a) a non-rigid body:
Non rigid body is the body in which the interspacing between any two particles will be changed if external force is applied.
Thus, when we apply force to such non rigid body, the interspacing between its particles get changed and hence the shape of body indirectly dimensions of the body changes. And as a result motion is produced in the non rigid body.
b) rigid body:
Rigid body is the body in which interspacing between any two particles will be remained fix although we apply an external force also.
Thus, when force is applied to such rigid body, its interparticle spacing does not get disturbed and hence its dimensions does not changes. But, the motion is produced in the rigid body also due to external force.
8.) Solution:
Ans:
a)
When a car is moving and suddenly a dog came in front of car then by applying the brakes of car the car suddenly get stopped.
b)
In case of bullock cart, the ox pushes the cart in forward direction by applying force on the cart.
c)
When we press any rubber material then it changes its size as it is elastic.
d)
When we pour a air inside the empty balloon its s get changed it becomes bigger as compared to initial size.
Multiple choice type:
1.) Which of the following is the contact force
Ans: c) frictional force
2.) The non contact force is
Ans: b) force due to gravity
Exercise-B Solution:
1.) Solution:
Ans:
The most required physical quantity for the motion of any body is the external force acting on it. Thus, force is the physical quantity required for motion of any body.
2.) Solution:
Ans:
If there is already a body is in motion then there is no requirement of force for its further motion.
3.) Solution:
Ans:
Initially when we give external force to the ball it will starts moving. While moving on table, the ball is in physical contact with the surface of table also, so table’s surface applies force of friction on the ball which causes to stop the motion of ball slowly. And hence, finally ball get stopped moving completely.
4.) Solution:
Ans:
A ball is moving already and on the perfectly smooth horizontal surface which never applied any frictional force on the ball. As there is no external resistive force on the ball the motion ball will be continues with the same speed. Thus, motion of ball remains unchanged i.e. its speed remains unchanged.
5.) Solution:
Ans:
According to Galileo’s law of inertia, any body continues in a state of its rest or in the state of its motion till when we apply external unbalanced force on it.
6.) Solution:
Ans:
According to Newton’s first law of motion, any body is in a state of rest or in a state of motion will remain moving in the same direction with the same speed unless we can apply an external unbalanced force on it.
7.) Solution:
Ans:
According to Newton’s first law of motion, any body is in a state of rest or in a state of motion will remain moving in the same direction with the same speed unless we can apply an external unbalanced force on it.
For example:
When a ball is kicked on the ground having rough surface it will stay in motion but the ground surface also exerts frictional force on it due to which after sometime it comes to rest. But, when a ball is kicked on smooth ground surface then it remains in its state of motion and it will be stopped only when we apply external force on it.
This is best example which explains the Newton’s first law of motion.
8.) Solution:
Ans:
Inertia is the property of an object due to which it remains in its state of rest or in its state of motion untill we apply external unbalanced force on it.
For example:
A book placed on a table remains as it is till when we apply force on it. When we displaced it then only it get displacement.
9.) Solution:
Ans:
From the Newton’s first law of motion, we define force as it is the quantity which can moves the object which is at rest or it can changes the motion of the body which is in motion already.
For example:
A book placed table get displaced only when an external force is applied on it.
10.) Solution:
Ans:
Inertia of the body is due to its mass only. Hence, inertia of the body depends on the mass of the body and it is directly proportional to the mass of the body.
Thus, if the mass of body is more its inertia will be more. And if mass of the body is less then it’s inertia will be less definitely.
11.) Solution:
Ans:
For example:
We know that, cricket ball is more massive than the tennis ball. Hence, more force is required to stop the cricket ball than the tennis ball which are in motion with the same speed.
This is only due to the inertia of the body, here mass of cricket ball is more hence its inertia will also be more.
Again, when we have to push a loaded trolley it requires more force while a empty trolley will be pushed easily. This is also due to the inertia of the trolley. The inertia of the loaded trolley is more as its mass is more as compared to the empty trolley.
12.) Solution:
Ans:
Take two trolleys, one is empty and the other is completely filled with objects. Then which trolley is easy to move? We will say that, the trolley with less mass is easy to move than completely filled trolley.
Because, the trolley filled with objects has more mass as compared to the trolley which is empty. Thus, the inertia of trolley filled with objects is more as compared to the empty trolley.
13.) Solution:
Ans:
There are two kinds of inertia.
a) inertia of rest and
b) inertia of motion
14.) Solution:
Ans:
a) inertia of rest:
A body in a state of rest remains in that state untill we apply external force on it. This property of a body is due to the inertia of rest.
For example:
When we apply jerk to the branches of tree the fruits will falls down. Because when we apply jerk to the branches of tree the comes in motion while the fruits which are more massive remains in state of rest. And after shaking for long time, the fruits which are massive and lightly attached to branches get detached from the branch and falls down due to gravitational pull acting on it.
b) inertia of motion:
A body is in motion remains in motion in the same direction with the same speed along the straight line unless an external unbalanced force acts on it.
For example:
Let us suppose a book is placed on table which is at rest. Since it remains as it is till we can apply a force on it.
Again, if a cycle in motion then it will be in motion unless we can apply the brakes on it.
15.) Solution:
Ans:
The body will not moves. Because, the two equal and opposite forces are acting on the body, due to which the resultant force acting on the body is zero. Hence, it can’t moves anywhere.
16.) Solution:
Ans:
Here also, there is no effect on the motion of body. Because, the two equal and opposite forces are acting on the object which is in motion, due to which net resultant force acting on it zero. So it’s motion is not get affected.
17.) Solution:
Ans:
Since, the aeroplane is moving uniformly at a constant height. And the upward force (lift) is balanced by the downward force (weight), hence the resultant force acting on aeroplane is zero.
18.) Solution:
Ans:
When the person jumps from moving train he falls down. Because, person is in motion with the train. When he want to jump outside, first he put his legs on the ground.
When he jumps by putting legs on the ground his legs become in a state of rest but its upper body part still in motion in the direction of motion of train. Hence, he falls in the direction of motion of train due to property of inertia, when he jumps from the moving train.
19.) Solution:
Ans:
The coin drops into the tumbler because when we flick the card it get some momentary force due to which it moves away. But, coin is at rest and does not take any force at once and hence it remains in its state of rest due to inertia. And after flicking the card, due acceleration due to gravity the coin drops into the tumbler.
20.) Solution:
Ans:
Because, when ball is thrown upward in moving train, the throwers along with the ball also in motion with the train. And due to inertia when ball is in air for few seconds, the ball and person moves forward and finally again the ball come back to throwers hand.
21.) Solution:
Ans:
a)
Because, before starting the train the person is at rest, but when train start moving the lower part of person which is at rest is in close contact with the train and hence it get suddenly motion while the upper part will not get affected. Hence, upper part remains at that position while the lower get moved with the motion of train. And hence the passenger in the train falls backward.
b)
In case of corridor train, the frame of sliding doors is in close contact with the floor of the moving train. As train moves the frame of the sliding doors moves ahead with the train while the sliding doors due to inertia moves in opposite direction of the motion of train. And hence, sliding doors of compartment may open and closed sometimes.
c)
When the branches of tree are shaken the branches will comes in motion. While the fruits are more massive and due to their inertia remains in the state of rest. After shaking the branches sometime, the fruits which are massive and weakly attached get detached from the branches and falls down due acceleration due to gravity. Hence, People often shake branches of tree for getting down its fruits.
d)
Because, when people alight from the moving bus their lower part suddenly comes to rest while the upper part till in motion in the direction of motion of bus. And hence they falls down in the direction of motion of bus. To avoid this type of issues the people have to move in the direction of motion of bus for sometime.
e)
When we beat the carpet, it comes in motion while the dust particles on it remains in the state of rest at that position only. After beating the carpet it moves away from its initial position and the massive dust particles will falls down due to acceleration due to gravity.
f)
Because, after running before taking the long jump the body comes in motion which causes to take a long jump. Thus, when body is in motion we can take long jump.
Multiple choice type:
1.) The property of inertia is more in
Ans: b) a truck
2.) A tennis ball and a cricket ball, both are stationary, to start motion in them
Ans: b) a less force is required for the tennis ball than the cricket ball.
3.) A force is needed to
Ans: a) change the state of rest or state of motion of the body
Exercise – C Solution
1.) Solution:
Ans:
The following are the two factors on which the force needed to stop a moving body in a given time depends
Mass of body and velocity of the body.
2.) Solution:
Ans:
Linear momentum of the body is defined as, it is the product of mass of body and its velocity.
It is the vector quantity and denoted by p.
The direction of momentum is in the direction of motion of body.
Thus, p= m*v
The SI unit of momentum of body is kg m/s.
3.) Solution:
Ans:
a) when v<<c
∆p = ∆(mv) = m∆v
b) when v~c
∆p= ∆(mv)
c) when v<<c but m does not remain constant.
∆p = ∆(mv)
4.) Solution:
Ans:
When we apply force on the moving body then it’s velocity changes and hence it’s momentum also changes.
Let us suppose a body is moving initially with velocity u, we apply force F on the body then it’s velocity changes to v and hence it’s momentum also changes.
Thus, we can write as
Initial momentum= m*u
Final momentum=m*v
Change in momentum= final momentum- initial momentum
Change in momentum= (mv – mu) = m*(v – u)
Hence, we can write rate of change of momentum as,
Rate of change of momentum= m*(v-u)/t
But, we know that
Acceleration= change in velocity/time= (v – u)/t
Thus, rate of change of momentum = m*a = mass*acceleration
And this relation is true only when mass of body remains constant.
5.) Solution:
Ans:
Let body A is moving with velocity v of mass m.
And body B is moving with velocity 2v of same mass m.
Mass m is the measure of inertia. Hence,
Inertia of body A/ Inertia of body B= m/m= 1
Thus, inertia of body A: inertia of body B = 1:1
Now, momentum of body A is given by,
p1= m*v
Similarly, momentum of body B is given by,
p2= m*2v
Thus, p1/p2= 1/2
Hence, we can write
p1:p2= 1:2
6.) Solution:
Ans:
Let balls A and B of masses m and 2m are in motion with velocities 2v and v respectively.
And we know that mass is the measure of inertia.
Then,
Inertia of ball A : inertia of ball B = m:2m
Thus,
Inertia of ball A: inertia of ball B = 1:2
Now, momentum of first ball A is given by,
p1= m*2v
And momentum of second ball B is given by,
p2= 2m*v
Thus,
Momentum of ball A/ momentum of ball B= m*2v/2m*v = 1
Hence, momentum of ball A: momentum of ball B = 1:1
We know that, according to Newton’s second law of motion, the rate of change of momentum of a body is directly proportional to the force applied on it.
Thus,
Force required to stop ball A: force required to stop ball B = 1:1
7.) Solution:
Ans:
According to Newton’s second law of motion, the rate of change of momentum of body is directly proportional to the force applied on the body and the direction of change in momentum is in the direction of force applied.
Thus, this law gives the force acting on body in terms of mass and acceleration.
8.) Solution:
Ans:
Newton’s first law of motion states that, the body remains in a state of rest or in a state of motion unless an external unbalanced force is acting on it.
While Newton’s second law of motion states that, the rate of change of momentum is directly proportional to the force acting on the body and its direction is in the direction of force.
Thus, from this two laws we can say that, first law gives the qualitative definition of force only while the Newton’s second law gives the quantitative value of force in terms of measuring quantities such as mass and acceleration.
9.) Solution:
Ans:
Newton’s second law of motion states that, the rate of change of momentum of body is directly proportional to the force applied on the body.
Hence, we can write it’s mathematical form as,
Rate of change of momentum = force= mass*acceleration
Thus, f= m*a
This relation is valid only when velocity of body is much smaller than the velocity of light.
And when the mass of body remains constant.
10.) Solution:
Ans:
Newton’s second law of motion states that, the rate of change of momentum of a body is directly proportional to the force applied on the body and takes place in the direction of motion of body.
Mathematically it can be represented as,
F=mass*acceleration
F=m*a
And this form is valid only when the velocity of the body is much smaller than the velocity of light.
And when the mass of the body remains constant.
11.) Solution:
Ans:
According to Newton’s second law of motion, we can write as
F=m*a
Thus, when F=0 then a=0
That means when the force acting on the body is zero then there is no acceleration of the body.
Hence, if the body is in the state of rest then it remains in the state of rest only. And if it is in motion then it remains in the state of motion. This is similar to the Newton’s first law of motion which states that, the body continues to be in a state of rest or in a state of motion unless it is acted upon by an external unbalanced force.
12.) Solution:
a) acceleration on force for a constant mass
b) force on mass for a constant acceleration
13.) Solution:
Ans:
If the force is acting on body of different mass then the acceleration produced by the body is inversely proportional to the masses of body. If we plot the graph of acceleration Vs mass then it will be hyperbola as shown in the following graph.
14.) Solution:
Ans:
The SI unit of force is newton.
1newton= 1kg*1m/s2
Thus, 1 newton is the force acting on a body of mass 1kg, which produces an acceleration of 1m/s2.
15.) Solution:
Ans:
The CGS unit of force is dyne.
1dyne = 1g*1cm/s2
Thus, 1 dyne is the force acting on a body of mass 1g , which produces an acceleration of 1cm/s2.
16.) Solution:
Ans:
The SI unit of force is newton.
The CGS unit of force is dyne.
The SI unit and CGS unit of force are related to each other by the following equation,
1newton = 105 dyne
17.) Solution:
Ans:
When a glass vessel falls on the hard floor its time period of coming in a state of rest will be very short due to which when it comes to rest suddenly on hard floor, the hard floor again apply a huge force on it. Due to that force the glass vessel breaks immediately.
While when the same glass vessel falls on the carpet it’s time period to come in a state of rest will be more. So it take more time to come in rest after falling on the carpet. Hence, carpet apply a less force on the glass vessel so that it does not break.
18.) Solution:
Ans:
- a) a cricketer pulls his hand back while catching a fast moving cricket ball because by doing that he increases the time period of catching the ball. And hence the time period for change in momentum also decreases. Due to decrease in the rate of change of momentum when ball comes in our hand it does not make any injury as it apply less force on our hands after catching it.
- b) When a athlete lands on hard floor after taking high jump, its time period of legs come to rest will be less and it comes to rest immediately. Due to which the hard floor exerts a huge force on his legs and it may get injured badly.
While when he lands on sand after taking a long jump, its legs may push the sand also for sometime and hence the time period of his legs comes to rest will be increased. Due to which the sand exerts a small force on his legs which doesn’t injured him.
Hence, athlete prefers to land on sand instead of hard floor.
Multiple choice type:
1.) The linear momentum of the body of mass m moving with velocity v is
Ans: c) mv
2.) The unit of linear momentum is
Ans: a) Ns
3.) The correct form of Newton’s second law is
Ans: F=∆p/∆t
4.) The acceleration produced in a body by a force of given magnitude depends on
Ans: c) mass of the body
Numericals:
1.) Solution:
Ans:
Given that, mass of body is m= 5kg
Velocity v =2m/s
Hence, momentum of the body is given by,
p=m*v = 5*2=10kgm/s
2.) Solution:
Ans:
Mass of the ball is m=50g= 0.05kg
Momentum of the ball is p= 0.5kgm/s
Thus, we know that,
p=m*v
0.5=0.05*v
Thus, v = 0.5/0.05= 10m/s
3.) Solution:
Ans:
Given that,
Force F= 15N
Mass m= 2kg
Thus, we know that,
F=m*a
15=2*a
Thus, a= 15/2= 7.5m/s2
4.) Solution:
Ans:
Given that,
Force F= 10N
Mass m= 5kg
We know that,
F=m*a
10=5*a
Thus, a=10/5=2m/s2
5.) Solution:
Ans:
Given that,
Mass m=0.5kg
Acceleration a= 5m/s2
Force is given by,
F=m*a
F=0.5*5
F=2.5N
6.) Solution:
Ans:
Given that,
F=10N
Mass m = 2kg
Time t=3s
Initial velocity u=0
We know that,
F=m*a
10= 2*a
Thus, a=10/2=5m/s2
But, we know that, acceleration is the change in velocity per unit time.
Hence, we can write as
a=(v-u)/t
Thus, a*t= v – u
Hence, v= u + a*t
Thus, v= 0 + 5*3 = 15m/s
Thus, the velocity acquired by the body will be 15m/s.
The change in momentum of the body is given by,
∆p= m*v= 2*15=30kgm/s
7.) Solution:
Ans:
Given that,
Time t= 10s
Mass m= 100kg
Distance s=100m
Initial velocity u= 0
As it covers distance 100m in 5s, hence it’s velocity is given by,
Velocity v= 100/5= 20m/s
Thus, the velocity acquired by the body will be 20m/s.
We know the second equation of motion which is given by,
v2 = u2 + 2a*s
Thus, 2a*s= v2 – u2 =400-0
2a*s= 400
a= 400/2*100= 2m/s2
Thus the acceleration produced by the body is 2m/s2.
And the magnitude of force is given by,
F=m*a = 100*2= 200N
Thus, the magnitude of the force will be 200N.
8.) Solution:
The figure shows the velocity-time graph whose slope gives the value of acceleration.
Thus, slope = a =(20-0)/(5-0)= 20/5= 4m/s2
And hence force is given by, F=m*a = 0.1*4 = 0.4N
Thus, force acting on the particle is 0.4 N
9.) Solution:
Ans:
Given that,
A force F causes an acceleration of 10m/s2 in a body of mass 500g i.e. 0.5kg
Thus, F = m*a = 0.5*10= 5N
Now, we have to find the acceleration caused by the force of 5N in a body of mass 5kg.
Thus, we can write,
F=5*a
5=5*a
Hence, a= 1m/s2
Thus, the acceleration produced will be 1m/s2.
10.) Solution:
Ans:
Given that,
Mass of ball m= 150g = 0.15kg
Velocity u= 25m/s
Final velocity v= 0
Time required to stop the ball is 0.03s.
Thus, the average force applied is given by,
Average force= change in momentum/time = m(v-u)/t = 0.15*(0-25)/0.03= 3.75/0.03= -125N
Thus, the average force applied on the ball will be 125N.
And negative sign indicates that the force is acting on the opposite direction of the motion of the ball.
11.) Solution:
Ans:
Given that,
Mass m= 2kg
Initial velocity u=0
Velocity v = 2m/s
Time t= 0.1s
Acceleration is given by,
Acceleration a = change in velocity/time
a= (v-u)/t
a= (2-0)/0.1
a= 20 m/s2
Thus, magnitude of force is given by,
F=m*a= 2*20= 40N
12.) Solution:
Ans:
Given that,
Mass of body m= 500g = 0.5kg
Initial velocity u= 0
Distance s= 4m
Time t= 2s
Now, to find the force applied first we have to calculate the acceleration produced.
To find the acceleration produced we use the second equation of motion,
s= u*t + 1/2a*t2
4= 0 + 1/2*a*4
Thus, 2a= 4
Hence, a= 2m/s2
Thus, force applied is given by,
F=m*a = 0.5*2= 1N
13.) Solution:
Ans:
Given that,
Mass of car m= 480kg
Initial velocity u= 54km/h =54/3.6= 15m/s
Time t= 10s
Final velocity v=0
We know that, the acceleration is the rate of change of velocity.
Hence,
Acceleration a =(v-u)/time
a= (0-15)/10= -15/10= -1.5m/s2
The negative sign of acceleration indicates the decrease in acceleration i.e. retardation.
Hence, the force applied is given by,
F=m*a = 480*1.5= 720N
Thus, force applied on the car is 720N.
14.) Solution:
Ans:
Initial velocity u = 30m/s
Time t= 2s
F = 1500 N
Final velocity v=0
We know that, acceleration is the rate of change of velocity and it is given by,
Acceleration a= (v-u)/t = (0- 30)/2 = -15m/s2
Thus, the retardation produced in car is 15m/s2.
We know that, the force acting on the body is given by,
F=m*a
1500= m*15
Mass m= 1500/15= 100kg
Thus, the mass of the car is 100kg.
And the change in momentum is given by,
∆p = m*(v-u) = 100*(0-30)
Hence, ∆p= 3000 kgm/s
15.) Solution:
Ans:
Given that,
Mass of moving bullet m= 50g= 0.05kg
Initial velocity u= 100m/s
Final velocity v= 0
Distance s = 2cm = 0.2m
Initial momentum is given by,
p= m*u= 0.05*100= 5m/s
Final momentum is given by,
p= m*v = 0.05*0 = 0
Now, we can find the acceleration produced from the second equation of motion,
v2 = u2 + 2as
0 = 1000 + 2*a*0.02
0.04*a = -1000
Thus, a=-25,000m/s2
Here, negative sign indicates the decrease in acceleration called as retardation.
The resistive force exerted by the wooden block is given by,
F=m*a = 0.05*25,000= 1250N
Exercise – D Solution
1.) Solution:
Ans:
Newton’s third law of motion states that, to every action there is equal and opposite reaction. Hence, if any action is occurring in the universe then there is always reaction to it also.
2.) Solution:
Ans:
According to Newton’s third law of motion, to every action there is equal and opposite reaction.
For example: when we apply a force with our palm on wall then we feel the force experienced by the wall also.
3.) Solution:
Ans:
According to Newton’s third law of motion, to every action there is equal and opposite reaction.
For example:
When we push the wall with our palm that is we have exerted a force on the wall. And at the same time we feel a reverse force which is acting on our palm from the wall. So this is an example of action and reaction force.
Also, when a bullet is fired from the gun then the bullet moves with the force in forward direction and there is also a reverse force acting on the gun which is called as recoiling of gun. This is the best example of action and reaction force.
4.) Solution:
Ans:
a) firing a bullet from gun:
In case of firing a bullet from gun, the force exerted on the gun due to which gun moves in forward direction is the action.
While the recoiling of gun is the reaction.
b) hammering a nail:
When we hit a nail on the wall, then nail goes inside the wall which is due to the force exerted by hammer. And this is the action.
At the same time the nail experiences a force on the hammer which is called as the reaction.
c) a book lying on a table:
When we place a book on table, then it remains on it as it is due to its weight acting vertically downwards. This is the action.
And the book remains in the same state because the table also exerts a force on it. Hence the resultant force on the book is zero due to which it remains as it is. Thus, the force exerted by the table on book is the reaction.
d) a moving rocket:
When the fuel is burnt inside the rocket due to which high temperature and pressure is developed and hence it releases the gases outside. At that time the rocket exerts the force on the gases to leave them out from the nozzle which is the action.
At the same time, the emitted gases exerts a force on the rocket due to which it moves in the forward direction which is called as the reaction.
e) a person walking on the floor:
When a person is walking on the floor it exerts a pressure on the floor which is the action.
And at the same time, the floor also exerts a force on the man which is the reaction.
f) a moving train colliding with the stationary train:
When a moving train collides with the stationary train then it exerts a force on the stationary train which is called as action.
While at the same time the stationary train also exerts some force on the moving train which is called as the reaction.
5.) Solution:
Ans:
When fuel inside the rocket burns it emits gases at high temperature and pressure through the nozzle. And the rocket exerts a force on these gases in order to emit the gases outside the rocket. At the same time, the emitted gases also exerts a equal and opposite force on the rocket due to which rocket moves in the forward direction.
In this way, the action and reaction is produced in the rocket which causes the motion of rocket.
6.) Solution:
Ans:
When a bullet is fired from the gun, the gun exerts force on the gun due to which it moves in forward direction called as action. While at the same time the bullet exerts equal and opposite force i.e. recoil on the gun called as reaction due to which the gun gets recoiled.
7.) Solution:
Ans:
When a man step a shore from stationary boat, he exerts a force on the boat so that he will step from the boat. And at the same time the the boat try to leave the shore due the force exerted by the man on it.
8.) Solution:
Ans:
Let us suppose two spring balances A and B are joined at their free ends as shown in figure.
Now we pull the spring balance B then we will observe that both the balances shows the same readings.
Because, here when we pull the spring balance B then it aging pulls the spring balance A due to which it shows some reading. The same reading will be observed in spring B because the force we applied to pull the spring balance B that same force is exerted on spring balance A also by the balance B. The pull on the spring B by the spring A is the action and the pull on the spring B by the spring A is the reaction. In this way, we can say that there is equal and opposite reaction to each force. Thus, due to equal and opposite forces the spring balances joined at their free ends will show the same readings.
9.) Solution:
Ans:
To move a boat ahead in water, the boatsman has to push the water backwards by his oar because when the boatsman pushes the water by applying a force in backwards with the help of oar is called as action. And at the same time the water exerts equal and opposite force on the boat called as reaction, due to which the boat moves in forward direction. And hence motion of boat is possible.
10.) Solution:
Ans:
When a person is exerting a force on the hard wall by his palm is called the action. While at the same time the hard wall exerts equal and opposite force on the palm of man is called as reaction. Due to which the man is liable to fall back.
11.) Solution:
Ans:
The above statement is true because the action and reaction are the equal and opposite forces acting simultaneously.
12.) Solution:
Ans:
The above mentioned statement is true. Because action and reaction are equal and opposite forces acting simultaneously. Thus, their magnitude will be same.
13.) Solution:
Ans:
When a light ball falls on the ground then it exerts a force on the ground which is called action. At the same time the ground exerts equal and opposite force on the ball called as reaction. Due to which the the ball after striking the ground rises upwards.
14.) Solution:
Ans:
The given statement is wrong statement.
Because, according to Newton’s third law of motion, action and reaction are equal and opposite forces acting simultaneously on two different bodies at the same time. And hence, these forces do not cancel each other.
Hence, the sum of action and reaction is not zero.
Multiple choice type:
1.) Newton’s third law
Ans: c) explain the way of force acting on a body
2.) Action and reaction act on the
Ans: b) different bodies in opposite directions
Numerical:
1.) Solution:
Ans:
We know that, according to Newton’s third law of motion to every action there is equal and opposite reaction. Hence, when a boy pushes a wall with a force of 10N towards east. At the same time the wall exerts a force of 10N on the body in opposite direction i.e. in west direction.
2.) Solution:
Ans:
When a block of weight 15N is hanging from a rigid support by a string. Then
a) block exerts the force of 15N in the form of its weight on the string in the downward direction.
b) and the string exerts equal 15N and opposite force on the block in opposite direction i.e. in upward direction i.e. tension in the string.
The following figure shows the forces and their names.
Exercise-E Solution:
1.) Solution:
Ans:
According to Newton’s law of gravitation, the gravitational force of attraction between any two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
2.) Solution:
Ans:
The gravitational force between two masses is always attractive.
3.) Solution:
Ans:
According to Newton’s law of gravitation, the gravitational force of attraction between any two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. And this force of attraction is lie along the line joining the two bodies.
Let us suppose, two masses m1 and m2 are separated by distance r as shown in figure below.
Let F be gravitational force then according to Newton’s law of gravitation,
F a m1*m2
And F a 1/r2
Thus, F a m1*m2/r2
Hence, we can write as
F= G m1*m2/r2
Where G is the proportionality constant which is same everywhere and independent of nature of particle, temperature, medium etc. Hence, G is called as universal gravitational constant.
4.) Solution:
Ans:
According to Newton’s law of gravitation, the gravitational force of attraction between the two masses is inversely proportional to the square of the distance between them.
That means, if the distance between two bodies increases the gravitational force of attraction between them decreases.
5.) Solution:
Ans:
We know that, the gravitational force of attraction between the two masses is inversely proportional to the square of the distance between them.
If r be the distance between two masses then,
F a 1/r2
But, given that r = 2r
Thus, gravitational force of attraction becomes,
F a 1/4r2
Thus, the gravitational force of attraction between the two masses is reduced to one fourth.
6.) Solution:
Ans:
Gravitational constant is the numerical value which is equal to the magnitude of force of attraction between the two masses each having mass 1kg and separated by distance 1m.
G is the universal gravitational constant whose value is same at every place.
It’s value is G= 6.67*10-11Nm2Kg-2
7.) Solution:
Ans:
G is the universal gravitational constant whose value does not changes at any place.
It’s value is G = 6.67*10-11Nm2kg-2
Thus, it’s SI unit is Nm2kg-2
8.) Solution:
Ans:
The Newton’s law of gravitation is responsible for knowing about the weight of the body on which acceleration due to gravity is acting and hence we are on the ground.
It also explains the how the two objects affect each other in the presence of gravitational force.
The motion of planets around the sun is explained on the basis of this law only.
The motion of moon and satellite around the earth is also explained on the basis of Newton’s law of gravitation.
And it explains what is the motion of freely falling body.
In this way, Newton’s law of gravitation is most important in everyday life.
9.) Solution:
Ans:
The force of gravity is the force of attraction due to which the earth attracts each object towards its centre.
10.) Solution:
Ans:
Let m be the mass of body and M is the mass of earth. R is the radius of earth. Then the force due to gravity between the body of mass m and the earth is given by,
F=GMm/R2
Where,
G is the universal gravitational constant
And G = 6.67*10-11Nm2Kg-2
M is the mass of the earth
m is the mass of the body
R is the radius of the earth
And F is the force due to gravity.
11.) Solution:
Ans:
Acceleration due to gravity is defined as it is the acceleration produced in a freely falling body due to the gravitational force of attraction of the earth.
The SI unit of acceleration due to gravity is m/s2.
12.) Solution:
Ans:
The average value of acceleration due to gravity g on the earth’s surface is found to be 9.8m/s2.
13.) Solution:
Ans:
The acceleration due to gravity g on the earth’s surface is related to it’s mass M and radius R by the following equation as,
g = G*M/R2
14.) Solution:
Ans:
The acceleration due to gravity g and universal gravitational constant G are related with each other by the following equation,
g= G*M/R2
15.) Solution:
Ans:
If a body falls from rest freely from a height h under gravity then u= 0, and acceleration a is replaced by acceleration due to gravity g, then the height fallen by the body is given by,
h= 1/2gt2
16.) Solution:
Ans:
If a body is vertically thrown upward with initial velocity u to a height h then there will be retardation i.e a= -g, then the equation of motion is given by,
v2 = u2 – 2gh
But, when body reach at highest point of height h then v =0.
Thus, 0 = u2 – 2gh
Thus, maximum height attained by the body is given by, h = u2/2g
17.) Solution:
Ans:
Mass of any body is defined as it is the quantity of matter it contains.
While weight of any body is defined as it is the force with which earth attracts it.
18.) Solution:
Ans:
Mass:
Mass is the quantity of matter contained in a body when it is at rest.
Mass is the scalar quantity.
SI unit of mass is kg.
Mass is measured by beam balance.
The mass of each body remains constant everywhere and does not changes with the place.
Weight:
Weight is the force with which earth attracts any body.
Weight is the vector quantity.
The SI unit of weight is N.
Weight is measured by spring balance which gives readings in N.
The weight of the body does not remains constant but changes with place to place due to change in acceleration due to gravity.
19.) Solution:
Ans:
The SI unit of mass is kg.
While the SI unit of weight is N.
20.) Solution:
Ans:
We know that, the acceleration due to gravity at the centre of the earth is zero.
The weight of the body of mass m kg is given by,
Weight = m*g
But, at the centre of the earth g= 0
Thus, weight = 0
Hence, the weight of the body of mass m kg at the centre of the earth is zero.
21.) Solution:
Ans:
The mass of body does not changes if the place is changed because it doesn’t depends on the acceleration due to gravity.
But, weight of the body changes as place is changed. Because, weight of the body depends on the acceleration due to gravity.
22.) Solution:
1kgf = 9.8N
Ans:
The meaning of 1kgf= 9.8N is that, the one kilogram force is the force due to gravity on a mass of 1kg.
That is, 1kgf= mass 1 kg* acceleration due to gravity
1kgf= g N
1kgf = 9.8 N
Multiple choice type:
1.) The gravitational force between the two bodies is always
Ans: a) always attractive
2.) The value of G is
Ans: b) 6.7*10-11 Nm2kg-2
3.) The force of attraction between two masses each of 1kg kept at a separation of 1m is
Ans: d) 6.7*10-11 N
4.) A body is projected vertically upward with an initial velocity u. If the acceleration due to gravity is g then time for which it remains in air is
Ans: c) 2u/g
5.) An object falling freely from rest reaches ground in 2s. If acceleration due to gravity is 9.8m/s2. The velocity of the object on reaching the ground will be
Ans: c) 19.6 m/s
Numerical:
1.) Solution:
Ans:
We know that, the gravitational force of attraction between two bodies is inversely proportional to the square of the distance between them.
Hence, F a 1/r2
If separation is reduced to half then r = r/2
Then, force of attraction will be increased by 4 times.
Thus, 10N becomes 40N.
2.) Solution:
Ans:
We know that, the weight of the body is given by,
Weight= m*g = 5*10= 50N
Here, we assumed that g= 10m/s2
3.) Solution:
Ans:
Weight in kgf = 10*1kgf = 10kgf
Weight in newton = 10*9.8= 98N
4.) Solution:
Ans:
Force of gravity on the body is nothing but the weight of the body.
Weight = m*g = 5*9.8= 49N
Thus, the magnitude of force of gravity is 49N vertically downward.
5.) Solution:
Ans:
We know that,
Weight= m*g
2N = m*9.8
Thus, mass m = 2/9.8 = 0.204 kg
6.) Solution:
Ans:
We know that,
Weight= m*g
On the earth weight of body is 98N
Thus, weight=m*9.8
98=m*9.8
Hence, mass m = 98/9.8= 10kg
We know that, mass of the body does not changes with place because it doesn’t depends on the acceleration due to gravity.
On moon the mass of body will be 10kg and acceleration due to gravity is 1.6m/s2
Thus, weight of the body on moon is given by,
Weight =m*g = 10*1.6
Weight= 16N
Thus the weight of the body on moon is 16N.
7.) Solution:
Ans:
On the earth acceleration due to gravity is 9.8m/s2
Thus, weight of the man on earth is given by
Weight =m*g = m*9.8
600= m*9.8
Thus, mass m =600/9.8 = 61.22
And we know that, acceleration due to gravity on moon is 1.6m/s2
Thus, weight of man on the moon is given by
Weight =m*g
Weight= 61.22*1.6= 97.952~100N
Thus weight of the man in moon is 100N.
Because the acceleration due to gravity on moon is 1/6th time the acceleration due to gravity on earth’s surface.
8.) Solution:
Ans:
Given that, mass of block is 10.5 kg hence the force of gravity is given by
Force of gravity=m*g = 10.5*19= 105N
And weight of the block is given by
Weight= m*g = 10.5*10= 105N
Since, force of gravity is nothing but the weight of the body
9.) Solution:
Ans:
Given that,
Time t =3s
g=9.8m/s2
And initial velocity u = 0
Let s be the height of ball and v be the velocity with which it strike the ground.
- a)
We have,
s=u*t + 1/2*at2
s= 0 + 0.5*g*9
Thus, s= 0.5*9.8*9 = 44.1m
Thus, the height from which the ball is released is 44.1m
- b) we have
v = u+a*t
V = 0+g*t
Thus, v = 9.8*3 = 29.4 m/s
Thus, the velocity with which the ball will strike the ground is 29.4m/s
10.) Solution:
Ans:
Given that,
Mass m = 5kg
Thus, the force is given by
Force=m*g=5*9.8= 49N
Thus 49N is the force needed to muscles apply to hold mass of 5kg.
Here the assumption is g= 9.8m/s2
11.) Solution:
Ans:
Given that,
Height h=20m
Acceleration due to gravity g= 10m/s2
a) s=20m
At highest point v=0 and a=-g = -10m/s2
We know that,
v2 = u2 + 2as
0 = u2 -2gs
Thus, u2 = 2*10*20= 400
Hence, initial velocity u = 20m/s
b)the final velocity of the ball on reaching the ground will be 20m/s
c) we know that,
s = ut + 1/2at2
Here, the total displacement covered will be zero.
s=0
u = 20m/s
And a= -g = -10
Thus, s = ut + 1/2at2
0= 20*t – 0.5*10*t2
Thus, 20t = 5t2
Thus, 5t = 20
Hence, t = 4s
Thus, the total time of journey of ball is 4s.
12.) Solution:
Ans:
Given that,
Initial velocity u= 0
Final velocity v = 20m/s
Acceleration due to gravity g= 10m/s2
Height h= ?
We know that,
v2 = u2 + 2as
Here, v2 = u2 + 2*g*h
400= 0+ 2*10*h
Thus, h= 400/20= 20m
Thus, the height of the tower will be 20m
13.) Solution:
Ans:
Given that,
Time of journey of ball t= 6s
And g= 9.8m/s2
We know that,
Time of ascent t = 6/2= 3s
And for ascent u=0
We have,
h= ut + 1/2at2
h= 0+1/2*10*9
h=45 m
Hence, the greatest height reached by the ball is 45m.
To calculate the initial velocity,
at that point h=0
Hence, h=ut+1/2at2
Thus, 0= u*6-1/2*10*36
6u= 5*36
6u= 180
Thus, u= 30m/s
Hence, the initial velocity of the ball is 30m/s.
14.) Solution:
Ans:
Given that, u= 20m/s
Time t= 2s
We know that,
h= ut+1/2*a*t2
h= 20*2 – 1/2*10*4
Thus, h= 40 – 20 = 20m
15.) a) Solution:
Ans:
Given that,
Initial velocity u = 0
h= 80m
g= 10m/s2
- a)
We know that,
h = ut + 1/2*a*t2
80= 0+ 1/2*10*t2
5t2 = 80
Thus, t2 = 80/5= 16
Hence, time t = 4s
- b) we know that,
v= u + at
v= 0+ 10*4= 40m/s
16.) Solution:
Ans:
Given that,
Initial velocity u= 0
Time t= 2.5s
And g=10m/s2
We know that,
h= ut + 1/2*a*t2
h= 0+ 1/2*10*6.25
Thus, h= 5*6.25= 31.25m
Thus, the height of the building is 31.25m
17.) Solution:
Ans:
Given that,
Initial velocity u= 49m/s
Final velocity v= 0
We know that,
v2 = u2 + 2as
0= 2401- 2*9.8*s
Thus, 19.6*s= 2401
Thus, height s= 2401/19.6 = 122.5m
Thus, the maximum height attained will be 122.5m
Now, we know that
v=u + at
0= 49 – 9.8*t
9.8*t= 49
Thus, t= 49/9.8= 5s
But, the total time taken by ball to reach the ground is the sum of ascent time and descent time.
And time of ascent= time of descent = 5s
Thus, total time = 5+5= 10s
18.) Solution:
Ans:
Given that,
Initial velocity u=0
Time t= 4s
And g= 10m/s2
We know that,
h= ut + 1/2at2
Thus, h= 0+1/2*10*16
Thus, h= 80m
Hence, the height of the tower will be 80m.
19.) Solution:
Ans:
Given that,
Speed of sound= 330m/s
Initial velocity u= 0
Time t= 20s
We know that,
h= ut + 1/2*a*t2
h= 0+ 1/2*10*400
Thus, h= 2000m
Hence, depth of the water surface is 2000m.
Now, given time is time of pebble to reach the water.
Let t1=20s
And let t2 be the time taken by sound.
As the velocity of sound is 330m/s and depth of water surface is 2000m.
The time taken by sound = 2000/330= 6.06s
Hence, the total time when echo is heard after the pebble is dropped is given by 20s+6.06s= 26.06s~26.1s
Hence, the time when echo is heard after the pebble is dropped is 26.1s.
20.) Solution:
Ans:
Given that,
Initial velocity u= 19.6m/s
Time t= 5s
And g= 9.8m/s2
We know that,
h= ut + 1/2*a*t2
h= 19.6*5 – 1/2*9.8*25
Thus, h= 98- 122.5
Hence, h= -24.5m
Thus, the height of the tower is 24.5m
Now, we know that
v= u + at
v = 19.6 – 9.8*5 = -29.4m/s
Thus, the velocity of the ball on reaching the ground is 29.4m/s