Class 11 Exam > Class 11 Videos > Tips to solve problems (Part - 17) - Laws of Motion, Physics
Tips to solve problems (Part - 17) - Laws of Motion, Physics Video Lecture - Class 11
FAQs on Tips to solve problems (Part - 17) - Laws of Motion, Physics Video Lecture - Class 11
| 1. What are the three laws of motion? |  |
Ans. The three laws of motion are:
- Newton's First Law of Motion: An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity unless acted upon by an external force.
- Newton's Second Law of Motion: The acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. It can be mathematically represented as F = ma, where F is the net force, m is the mass, and a is the acceleration.
- Newton's Third Law of Motion: For every action, there is an equal and opposite reaction. This means that whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object.
| 2. How can the laws of motion be applied in everyday life? | |
Ans. The laws of motion are applicable in various aspects of everyday life. Some examples include:
- When a car accelerates, it experiences a force in the forward direction (Newton's Second Law).
- When a person pushes a door, the door pushes back with an equal and opposite force (Newton's Third Law).
- When a ball is thrown in the air, it follows a parabolic trajectory due to the combination of its initial velocity and the force of gravity (Newton's First Law).
- When riding a bicycle, the force applied on the pedals causes the bike to accelerate (Newton's Second Law).
| 3. What is inertia? How does it relate to Newton's First Law of Motion? | |
Ans. Inertia is the property of an object to resist any change in its state of motion. It relates to Newton's First Law of Motion because according to the law, an object at rest will remain at rest, and an object in motion will continue moving with a constant velocity unless acted upon by an external force. This behavior of objects is a result of their inertia. Objects with more mass have greater inertia, making it harder to change their state of motion.
| 4. How does force affect the motion of an object? | |
Ans. Force affects the motion of an object by causing it to accelerate. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. Therefore, a greater force will result in a larger acceleration, while a smaller force will result in a smaller acceleration. If the net force acting on an object is zero, the object will either remain at rest or continue moving with a constant velocity.
| 5. Can an object be in motion without any force acting on it? | |
Ans. No, an object cannot be in motion without any force acting on it. According to Newton's First Law of Motion, an object at rest will remain at rest, and an object in motion will continue moving with a constant velocity unless acted upon by an external force. Therefore, if an object is already in motion, it must be experiencing a force to maintain that motion. In the absence of any external force, an object will eventually come to rest due to the presence of other forces like friction or air resistance.
Text Transcript from Video
Allu friends this video on laws of
motion part 17 is brought to you by exam
feel calm no more fear from exam please
make sure that you have watched all the
videos from part 1 to 16 before going
ahead with 17 now that we have almost
covered
oil our studies and the topics related
to friction we will have to look at
certain problems now before going
in-depth to solve numerical problems I
will give you certain tips to solve
problems related to laws of motion as
well as friction so quickly recall the
topics which we studied that is the
three laws of motion first law second or
third law friction types of friction
because the numericals will be based
upon all these topics so let us quickly
have a look at the tips the first thing
which you need to do while solving a
problem related to motion or kinematics
is draw a Freebody diagram what is a
Freebody diagram Freebody diagram is
nothing but whether you have a question
related to a truck moving a car moving
or a man moving whatever it is you just
draw a Freebody diagram you just
indicate their body that is the box or
the car or the plane or whatever you
just did you noticed with some box and
you indicate all the forces that is
acting on that body and also write the
magnitude as well as direction of the
forces so this rough diagram which you
make is referred to as the free body
diagram things will become clearer as we
keep solving numericals the next thing
is indicate the magnitude of the forces
acting on the bodies let us suppose a
book is lying on a table now if a book
is lying on the table there will be a
force which will be acting downwards
that is due to the weight of the book
that is mg there will be a normal
reaction force acting upward which will
arise due to the surfaces being in
contact so while drawing a Freebody
diagram you should
indicate the magnitude of the force
acting on the bodies indicate the
direction of the forces acting on the
bodies as I did just now magnitude as
well as direction of the forces right
equation of motion for each of the
bodies now let us suppose I am taking
this simple example in this case let us
say the body is at rest there is no
external force acting that means there
is no external force applied in the
horizontal direction and we already know
if a body is at rest and there is no
external force applied in that case
there will be no frictional force so in
this case the equation of motion in the
vertical direction would be n that is
the normal will be equal to mg so this
is how we wrote the equation of motion
so first we draw a Freebody diagram we
indicate the magnitude of the forces
acting on the body we also indicate the
direction of the forces then we write
the equation of motion for each of the
bodies now equate equal and opposite
forces along a specific axis x axis or y
axis for example this normal and this mg
they both are acting along y axis so we
equated them but let us suppose if there
is a force capital F is applied on the
body in the horizontal direction
we cannot this force with normal because
this is acting along x axis so we can
only equate the forces along one
particular axis if we are talking of x
axis we can equate all the forces which
is acting along x axis similarly if we
are talking of Y axis we can only lead
with all the forces which are acting
along the y axis now we have already
discussed one example here let us
discuss one more example which is very
similar to that a book of mass 1 kg is
at rest on the table let us make a
clearer diagram here let us suppose this
is a book of mass 1 kg so there may be a
force mg which will which will act in
this
direction the value of M is 1 that means
mg will be 1 into G that is G there will
be a normal reaction force which will
act in this direction since the body is
at rest nothing else with actor so
therefore n is equal to G this would be
the equation of motion for this
particular example now let us look at
few more tips to solve problems now we
will see the scenario of two bodies
which are connected by string through
pulley somewhat like this let us suppose
we have a pulley let us suppose we have
a pulley and we have two objects which
are attached by a string through this
pulley let us suppose this is m1 and
this is m2 so you will normally go come
across several types of numericals
related to this pulley things so we will
give you certain tips which will be
generalized and which will be true for
all such fully problems firstly if the
pulley is frictionless if the problem
says that this pulley is frictionless in
that case the tension of the two strings
will be equal before I say that let me
explain let us suppose if there are two
bodies which are attached to a pulley
now the forces which will be acting on
the two bodies are on this body there
will be a force that is a weight acting
downwards which is m1 G similarly for
this body there will be a force acting
downwards m2 G for this body there won't
be a tension force we meant you must
view heard of this film tension when I
was talking about contact force if this
is an example of a contact force this
force comes into play when two bodies
are in contact with each other this
string is in contact with this body and
this string passes over this pulley so
whenever you have a bully problem there
will be a tension force which will act
towards the center of the pulley or
towards the pulley so for this that this
is the tension force for this
do there will be another tension force
people now we come to the different
cases in the first case we say if the
pulley is frictionless in that case t 1
will be equal to t 2 that means the
value of the tensions will be equal
respective of even though M 1 and M 2
are different masses the value of the
tensions on the two masses will be
different also in case of a pulley
problem let us suppose since the weight
is acting in this direction so this body
will accelerate in this direction that
is this body will move in this direction
so it will have some acceleration say a
1 similarly if this body is coming down
then this will go up so this will have
an acceleration in the upward direction
let us see a 2 now if we see that the
pulley is frictionless and that is t1 is
equal to t2 and also a 1 will be equal
to a 2 that is the acceleration for the
two bodies however they will be
different opposite directions but the
values will be the same let us give us
consider the second case that is the
pulley is not frictionless now if the
bullet is not frictionless in that case
T 1 is not equal to t 2 and also a 1 is
not equal to a 2 let us take the third
case where m 1 is greater than M 2 now
if M 1 is greater than M 2 in that case
how would the figure be if m1 is equal
to m2 let us draw the figure separately
then you will get a better understanding
let us suppose M 1 is greater than M 2
in that case this force that means this
M 1 G will be greater than M 1 this mode
that is M 1 is heavier so it will have a
tendency to come downward because the
mass is more therefore in that case this
M 1 will come down it will move
downwards that means the acceleration ay
1 will be downwards
whereas M 2 will move up so the
acceleration will be upward so M 1 is
lift M 1 is greater
M 2 and that is M 1 moves vertically
downwards and M 2 moves vertically
upwards now let us take the fourth and
the last scenario where M 1 is less than
M 2 this is exactly the scenario M 1 is
less than M 2 so in this is M 2 moves
vertically downwards and M 1 moves
vertically upwards so these are the
certain points which are universally
true so you should always keep them in
mind before you go ahead to solve
numerical problems so this is one
scenario where two bodies are connected
by a string through a pulley now let us
look at another scenario which will be
helpful to solve the numerical problems
here we say two bodies are connected
through string such that one lies on the
smooth horizontal surface so here we are
considering the scenario something like
this let us suppose this is one object
m1 this object is lying on a surface and
this object is connected via a bully to
another object say M 2 so this is a
surface over which M 1 is line and it is
connected by a string through a pulley
to another object M 2 so in this case
the first case we consider is if the
pulley is frictionless now if the pulley
is frictionless here again in this case
also we will have this tension T 2 or M
2 this will have the weight acting
downwards that is M 2 into G here also
we will have the tension T 1 acting in
this direction so if the pulley is
frictionless in that case T 1 will be
equal to t 2 and also a 1 will be equal
to
- later suppose a1 is the acceleration
associated with M 1 and a2 is the
acceleration associated with mg so if
the pulley is frictionless in that is t1
equal to t2 and a1 is equal to a2 if the
pulley is not frictionless in that case
obviously T 1 will not be equal to t2
and also K 1 will not be equal to a2 now
let's talk about the motion of M 1 and M
2 in this case the motion of M 2 will be
in the vertical direction whereas the
motion of M 1 will be in the horizontal
direction that is if M 2 moves downwards
so M 2 moves down then M 1 moves towards
right that means M 2 is moving like this
that is a 2 is like this then a 1 will
be this way that is it will move towards
right so this is how the motion will be
governed so I have already discussed
certain tips which you need to follow to
deal with the numerical problems which
we are going to discuss hereafter so now
let us look at the first problem thank
you visit exam viacom to watch key
educational videos try 3 online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks once again
motion part 17 is brought to you by exam
feel calm no more fear from exam please
make sure that you have watched all the
videos from part 1 to 16 before going
ahead with 17 now that we have almost
covered
oil our studies and the topics related
to friction we will have to look at
certain problems now before going
in-depth to solve numerical problems I
will give you certain tips to solve
problems related to laws of motion as
well as friction so quickly recall the
topics which we studied that is the
three laws of motion first law second or
third law friction types of friction
because the numericals will be based
upon all these topics so let us quickly
have a look at the tips the first thing
which you need to do while solving a
problem related to motion or kinematics
is draw a Freebody diagram what is a
Freebody diagram Freebody diagram is
nothing but whether you have a question
related to a truck moving a car moving
or a man moving whatever it is you just
draw a Freebody diagram you just
indicate their body that is the box or
the car or the plane or whatever you
just did you noticed with some box and
you indicate all the forces that is
acting on that body and also write the
magnitude as well as direction of the
forces so this rough diagram which you
make is referred to as the free body
diagram things will become clearer as we
keep solving numericals the next thing
is indicate the magnitude of the forces
acting on the bodies let us suppose a
book is lying on a table now if a book
is lying on the table there will be a
force which will be acting downwards
that is due to the weight of the book
that is mg there will be a normal
reaction force acting upward which will
arise due to the surfaces being in
contact so while drawing a Freebody
diagram you should
indicate the magnitude of the force
acting on the bodies indicate the
direction of the forces acting on the
bodies as I did just now magnitude as
well as direction of the forces right
equation of motion for each of the
bodies now let us suppose I am taking
this simple example in this case let us
say the body is at rest there is no
external force acting that means there
is no external force applied in the
horizontal direction and we already know
if a body is at rest and there is no
external force applied in that case
there will be no frictional force so in
this case the equation of motion in the
vertical direction would be n that is
the normal will be equal to mg so this
is how we wrote the equation of motion
so first we draw a Freebody diagram we
indicate the magnitude of the forces
acting on the body we also indicate the
direction of the forces then we write
the equation of motion for each of the
bodies now equate equal and opposite
forces along a specific axis x axis or y
axis for example this normal and this mg
they both are acting along y axis so we
equated them but let us suppose if there
is a force capital F is applied on the
body in the horizontal direction
we cannot this force with normal because
this is acting along x axis so we can
only equate the forces along one
particular axis if we are talking of x
axis we can equate all the forces which
is acting along x axis similarly if we
are talking of Y axis we can only lead
with all the forces which are acting
along the y axis now we have already
discussed one example here let us
discuss one more example which is very
similar to that a book of mass 1 kg is
at rest on the table let us make a
clearer diagram here let us suppose this
is a book of mass 1 kg so there may be a
force mg which will which will act in
this
direction the value of M is 1 that means
mg will be 1 into G that is G there will
be a normal reaction force which will
act in this direction since the body is
at rest nothing else with actor so
therefore n is equal to G this would be
the equation of motion for this
particular example now let us look at
few more tips to solve problems now we
will see the scenario of two bodies
which are connected by string through
pulley somewhat like this let us suppose
we have a pulley let us suppose we have
a pulley and we have two objects which
are attached by a string through this
pulley let us suppose this is m1 and
this is m2 so you will normally go come
across several types of numericals
related to this pulley things so we will
give you certain tips which will be
generalized and which will be true for
all such fully problems firstly if the
pulley is frictionless if the problem
says that this pulley is frictionless in
that case the tension of the two strings
will be equal before I say that let me
explain let us suppose if there are two
bodies which are attached to a pulley
now the forces which will be acting on
the two bodies are on this body there
will be a force that is a weight acting
downwards which is m1 G similarly for
this body there will be a force acting
downwards m2 G for this body there won't
be a tension force we meant you must
view heard of this film tension when I
was talking about contact force if this
is an example of a contact force this
force comes into play when two bodies
are in contact with each other this
string is in contact with this body and
this string passes over this pulley so
whenever you have a bully problem there
will be a tension force which will act
towards the center of the pulley or
towards the pulley so for this that this
is the tension force for this
do there will be another tension force
people now we come to the different
cases in the first case we say if the
pulley is frictionless in that case t 1
will be equal to t 2 that means the
value of the tensions will be equal
respective of even though M 1 and M 2
are different masses the value of the
tensions on the two masses will be
different also in case of a pulley
problem let us suppose since the weight
is acting in this direction so this body
will accelerate in this direction that
is this body will move in this direction
so it will have some acceleration say a
1 similarly if this body is coming down
then this will go up so this will have
an acceleration in the upward direction
let us see a 2 now if we see that the
pulley is frictionless and that is t1 is
equal to t2 and also a 1 will be equal
to a 2 that is the acceleration for the
two bodies however they will be
different opposite directions but the
values will be the same let us give us
consider the second case that is the
pulley is not frictionless now if the
bullet is not frictionless in that case
T 1 is not equal to t 2 and also a 1 is
not equal to a 2 let us take the third
case where m 1 is greater than M 2 now
if M 1 is greater than M 2 in that case
how would the figure be if m1 is equal
to m2 let us draw the figure separately
then you will get a better understanding
let us suppose M 1 is greater than M 2
in that case this force that means this
M 1 G will be greater than M 1 this mode
that is M 1 is heavier so it will have a
tendency to come downward because the
mass is more therefore in that case this
M 1 will come down it will move
downwards that means the acceleration ay
1 will be downwards
whereas M 2 will move up so the
acceleration will be upward so M 1 is
lift M 1 is greater
M 2 and that is M 1 moves vertically
downwards and M 2 moves vertically
upwards now let us take the fourth and
the last scenario where M 1 is less than
M 2 this is exactly the scenario M 1 is
less than M 2 so in this is M 2 moves
vertically downwards and M 1 moves
vertically upwards so these are the
certain points which are universally
true so you should always keep them in
mind before you go ahead to solve
numerical problems so this is one
scenario where two bodies are connected
by a string through a pulley now let us
look at another scenario which will be
helpful to solve the numerical problems
here we say two bodies are connected
through string such that one lies on the
smooth horizontal surface so here we are
considering the scenario something like
this let us suppose this is one object
m1 this object is lying on a surface and
this object is connected via a bully to
another object say M 2 so this is a
surface over which M 1 is line and it is
connected by a string through a pulley
to another object M 2 so in this case
the first case we consider is if the
pulley is frictionless now if the pulley
is frictionless here again in this case
also we will have this tension T 2 or M
2 this will have the weight acting
downwards that is M 2 into G here also
we will have the tension T 1 acting in
this direction so if the pulley is
frictionless in that case T 1 will be
equal to t 2 and also a 1 will be equal
to
- later suppose a1 is the acceleration
associated with M 1 and a2 is the
acceleration associated with mg so if
the pulley is frictionless in that is t1
equal to t2 and a1 is equal to a2 if the
pulley is not frictionless in that case
obviously T 1 will not be equal to t2
and also K 1 will not be equal to a2 now
let's talk about the motion of M 1 and M
2 in this case the motion of M 2 will be
in the vertical direction whereas the
motion of M 1 will be in the horizontal
direction that is if M 2 moves downwards
so M 2 moves down then M 1 moves towards
right that means M 2 is moving like this
that is a 2 is like this then a 1 will
be this way that is it will move towards
right so this is how the motion will be
governed so I have already discussed
certain tips which you need to follow to
deal with the numerical problems which
we are going to discuss hereafter so now
let us look at the first problem thank
you visit exam viacom to watch key
educational videos try 3 online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks once again
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Dec 25, 2024 Last updated |
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