Class 11 Exam > Class 11 Videos > Rocket Problem (Part - 11) - Laws of Motion, Physics
Rocket Problem (Part - 11) - Laws of Motion, Physics Video Lecture - Class 11
FAQs on Rocket Problem (Part - 11) - Laws of Motion, Physics Video Lecture - Class 11
| 1. What are Newton's laws of motion and how do they apply to rocket propulsion? |  |
Ans. Newton's laws of motion are three fundamental principles that describe the motion of objects. In the context of rocket propulsion, these laws can be applied as follows:
- Newton's First Law (Law of Inertia) states that an object at rest will stay at rest, and an object in motion will stay in motion, unless acted upon by an external force. In the case of a rocket, the force of the expelled gases propels it forward.
- Newton's Second Law (Law of Acceleration) states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In other words, a rocket can accelerate by expelling more gas or by reducing its mass.
- Newton's Third Law (Law of Action-Reaction) states that for every action, there is an equal and opposite reaction. This law explains how a rocket moves forward by expelling gases in one direction, causing a reaction force that propels the rocket in the opposite direction.
| 2. How does a rocket overcome gravity to reach space? | |
Ans. A rocket overcomes gravity to reach space through a combination of Newton's laws of motion and the principle of thrust. The thrust generated by the rocket's engines creates a force that is greater than the force of gravity pulling it down. This allows the rocket to accelerate upward, gradually overcoming the force of gravity. As the rocket ascends, it continues to expel gases at high speeds, generating thrust and counteracting gravity until it reaches the desired altitude.
| 3. What is the role of the rocket's fuel in its propulsion? | |
Ans. The rocket's fuel plays a crucial role in its propulsion. The fuel, usually a combination of a fuel and an oxidizer, undergoes a chemical reaction known as combustion. During combustion, the fuel and oxidizer react, releasing large amounts of energy in the form of hot gases. These gases are expelled through the rocket's engine nozzle at high speeds, creating a force called thrust. According to Newton's third law, this thrust propels the rocket in the opposite direction, enabling it to move forward and overcome gravity.
| 4. How does the mass of a rocket affect its acceleration? | |
Ans. According to Newton's second law of motion, the acceleration of an object is inversely proportional to its mass. Therefore, a lighter rocket will experience a greater acceleration compared to a heavier one, assuming the same amount of force is applied. This is because a lighter rocket has less mass to overcome, allowing it to achieve higher accelerations. In practical terms, rockets strive to reduce their mass by using lightweight materials and efficiently expelling exhaust gases, as a higher acceleration can lead to faster ascent and more efficient space travel.
| 5. Can rockets operate in the absence of air or in space? | |
Ans. Yes, rockets can operate in the absence of air or in space. Unlike aircraft that rely on air for lift, rockets operate on the principle of thrust, which doesn't require air. In fact, rockets are designed to operate in a vacuum, such as the conditions found in space. The lack of air resistance in space actually allows rockets to reach higher speeds and travel more efficiently. However, rockets do require an initial push from the atmosphere to overcome gravity, which is typically achieved through a combination of atmospheric flight and staging.
Text Transcript from Video
hello friends this video on lots of
motion parts 11 is brought to you by
example.com no more fear from exam
please make sure that you have watched
all the videos from part 1 - Jen before
going ahead with part 11 let us look at
this problem a rocket is fired
vertically up during burning of engine
which lasts for 50 seconds a constant
acceleration of 20 g was observed little
suppose that the rocket was fired from
this point on the ground vertically
upwards okay now it says that the engine
burning of engine continues till 50
seconds that is 18 50 seconds it moves
till this point a and let us suppose the
ground point is o so that means
initially its velocity was zero because
it was at rest till this point the
acceleration was 20 G the question asks
what is the speed of the rocket at the
end of 50 seconds that means the
question asks what put with the speed at
this point right so let us try to
calculate the same for the first part we
are given time is equal to 50 seconds we
are given acceleration is equal to 20 G
and we know that initial velocity U is
equal to zero now from the first
equation of kinematics we know V is
equal to u plus 18 so we can write it as
u with zero so a we can write as 20 G
into T so this will come out to be 20
into 9.8 into 50 so this comes out to be
nine eight zero zero meters per second
so that means this would be the velocity
at the end of five seconds that means at
Point a now the second part of the
question asks calculate the maximum
height to which the rocket will rise
before it begins to fall back now as
soon as the
of the engine stops what will happen the
rocket will still move some distance
upwards and only after that it will
start falling down right that is what
normally happens so what would be the
difference in the difference would be
that from o to a it had a constant
acceleration of 20 G but from a - let us
say the maximum point where it reaches
that is being so from A to B what would
be the acceleration the acceleration
would be nothing but the normal
acceleration due to gravity correct so
in this case what would be the velocity
at point B the velocity of at point V
let us suppose B - that would be equal
to 0 and what would be what is the
velocity at Point a this V we have
already calculated it as 9 8 0 0 meter
per second the question asks what is the
maximum height that the rocket reaches
that means the question asks us to find
the distance OB so what we do let us
calculate it part wise let us say this
distance is s 1 and this distance is s 2
so let us first calculate for journey
from o to a so in that case what would
be s 1 let us apply the kinematic second
equation which say is s is equal to UT
plus half ay T Square so what is Q in
this case from o to a you will zero so
this is left with half into a what is a
for this journey o to a is equal to 20g
and what is T T is 50 seconds so this is
50 squared now if you calculate this
this comes out to be 2 4 5 0 0 0 meters
or we can write it as 2 45 kilometres
find so this is the distance from o to a
now we have to calculate the distance
from A to B so let us consider the
journey from A to B here let us apply
the third equation of
we see it says be square that is the
final velocity here what is the final
velocity here initial point is a and
final point is B so the final velocity
is B dash that means V dash square is
equal to u square U is the initial
velocity here initial velocity is the
velocity at Point a that is V so B dash
square is equal to v square plus 2a s2
right so in this we find that V dash is
equal to zero so we can say this is zero
what is V squared that is nine eight
zero zero whole square plus 2 into a is
G into s2 then so now this G again in
this case will act in the negative
direction because the rocket is moving
up by the acceleration due to gravity
will act in this direction so this will
have a negative sign now if you
calculate this you find s2 comes out to
be 4 9 0 0 km/h
the maximum height that the rocket can
reach will be equal to s that is equal
to s1 plus s2 which is equal to 245 plus
4 9 0 0 km/h 6 5 1 4 5 kilometers that
means this is the maximum height till
which the rocket will reach and after
that it will start falling back to the
ground so I hope it is clear and please
know don't get confused with the sign of
acceleration that why it is plus G
somewhere why it is minus G somewhere
whatever convention you take you take
the same convention for everywhere so
you will get the same result like if you
consider that the acceleration is
positive you consider it positive in
both the places if you consider it
negative you consider negative in both
the places in
in both the scenarios you will get the
same result thank you
visit exam viacom to watch free
educational videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks once again
motion parts 11 is brought to you by
example.com no more fear from exam
please make sure that you have watched
all the videos from part 1 - Jen before
going ahead with part 11 let us look at
this problem a rocket is fired
vertically up during burning of engine
which lasts for 50 seconds a constant
acceleration of 20 g was observed little
suppose that the rocket was fired from
this point on the ground vertically
upwards okay now it says that the engine
burning of engine continues till 50
seconds that is 18 50 seconds it moves
till this point a and let us suppose the
ground point is o so that means
initially its velocity was zero because
it was at rest till this point the
acceleration was 20 G the question asks
what is the speed of the rocket at the
end of 50 seconds that means the
question asks what put with the speed at
this point right so let us try to
calculate the same for the first part we
are given time is equal to 50 seconds we
are given acceleration is equal to 20 G
and we know that initial velocity U is
equal to zero now from the first
equation of kinematics we know V is
equal to u plus 18 so we can write it as
u with zero so a we can write as 20 G
into T so this will come out to be 20
into 9.8 into 50 so this comes out to be
nine eight zero zero meters per second
so that means this would be the velocity
at the end of five seconds that means at
Point a now the second part of the
question asks calculate the maximum
height to which the rocket will rise
before it begins to fall back now as
soon as the
of the engine stops what will happen the
rocket will still move some distance
upwards and only after that it will
start falling down right that is what
normally happens so what would be the
difference in the difference would be
that from o to a it had a constant
acceleration of 20 G but from a - let us
say the maximum point where it reaches
that is being so from A to B what would
be the acceleration the acceleration
would be nothing but the normal
acceleration due to gravity correct so
in this case what would be the velocity
at point B the velocity of at point V
let us suppose B - that would be equal
to 0 and what would be what is the
velocity at Point a this V we have
already calculated it as 9 8 0 0 meter
per second the question asks what is the
maximum height that the rocket reaches
that means the question asks us to find
the distance OB so what we do let us
calculate it part wise let us say this
distance is s 1 and this distance is s 2
so let us first calculate for journey
from o to a so in that case what would
be s 1 let us apply the kinematic second
equation which say is s is equal to UT
plus half ay T Square so what is Q in
this case from o to a you will zero so
this is left with half into a what is a
for this journey o to a is equal to 20g
and what is T T is 50 seconds so this is
50 squared now if you calculate this
this comes out to be 2 4 5 0 0 0 meters
or we can write it as 2 45 kilometres
find so this is the distance from o to a
now we have to calculate the distance
from A to B so let us consider the
journey from A to B here let us apply
the third equation of
we see it says be square that is the
final velocity here what is the final
velocity here initial point is a and
final point is B so the final velocity
is B dash that means V dash square is
equal to u square U is the initial
velocity here initial velocity is the
velocity at Point a that is V so B dash
square is equal to v square plus 2a s2
right so in this we find that V dash is
equal to zero so we can say this is zero
what is V squared that is nine eight
zero zero whole square plus 2 into a is
G into s2 then so now this G again in
this case will act in the negative
direction because the rocket is moving
up by the acceleration due to gravity
will act in this direction so this will
have a negative sign now if you
calculate this you find s2 comes out to
be 4 9 0 0 km/h
the maximum height that the rocket can
reach will be equal to s that is equal
to s1 plus s2 which is equal to 245 plus
4 9 0 0 km/h 6 5 1 4 5 kilometers that
means this is the maximum height till
which the rocket will reach and after
that it will start falling back to the
ground so I hope it is clear and please
know don't get confused with the sign of
acceleration that why it is plus G
somewhere why it is minus G somewhere
whatever convention you take you take
the same convention for everywhere so
you will get the same result like if you
consider that the acceleration is
positive you consider it positive in
both the places if you consider it
negative you consider negative in both
the places in
in both the scenarios you will get the
same result thank you
visit exam viacom to watch free
educational videos try free 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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Nov 29, 2024 Last updated |
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