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Why do we prefer momentum conservation to energy conservation?
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I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question.
The problem is as follows:
A bullet of mass $25,g$ is fired horizontally into a ballistic pendulum of mass $5.0,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10,cm$, find the speed of the bullet.
(H.C. Verma, Centre of Mass, Q47)
Let the mass of the bullet be $m=25,g$.
Let the mass of the pendulum be $M=5,kg$.
Let the peak height be $h=0.1,m$.
Let the initial and final velocities be $u$ and $v$ respectively.
Method 1: (Momentum Conservation)
$$mu=(M+m)v implies v=fracmuM+m\ textAlso, frac12(M+m)v^2=(M+m)gh\ implies u^2=2biggl(fracM+mmbiggr)^2gh\ implies u=201sqrt2 text m/s$$
Method 2: (Energy Conservation)
Since both the masses move together after the collision, and because the velocity at highest point is null, therefore:
$$frac12mu^2=frac12(M+m)v^2=(M+m)gh\ implies u^2=2ghbiggl(fracM+mmbiggr) implies u=sqrt402$$
To summarize myself, I am curious about the following:
Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations?
homework-and-exercises newtonian-mechanics momentum energy-conservation collision
edited 1 hour ago
Ben Crowell
46.2k3147279
asked 2 hours ago
Utkarsh Verma
326
add a comment |Â
up vote
1
down vote
favorite
I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question.
The problem is as follows:
A bullet of mass $25,g$ is fired horizontally into a ballistic pendulum of mass $5.0,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10,cm$, find the speed of the bullet.
(H.C. Verma, Centre of Mass, Q47)
Let the mass of the bullet be $m=25,g$.
Let the mass of the pendulum be $M=5,kg$.
Let the peak height be $h=0.1,m$.
Let the initial and final velocities be $u$ and $v$ respectively.
Method 1: (Momentum Conservation)
$$mu=(M+m)v implies v=fracmuM+m\ textAlso, frac12(M+m)v^2=(M+m)gh\ implies u^2=2biggl(fracM+mmbiggr)^2gh\ implies u=201sqrt2 text m/s$$
Method 2: (Energy Conservation)
Since both the masses move together after the collision, and because the velocity at highest point is null, therefore:
$$frac12mu^2=frac12(M+m)v^2=(M+m)gh\ implies u^2=2ghbiggl(fracM+mmbiggr) implies u=sqrt402$$
To summarize myself, I am curious about the following:
Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations?
homework-and-exercises newtonian-mechanics momentum energy-conservation collision
edited 1 hour ago
Ben Crowell
46.2k3147279
asked 2 hours ago
Utkarsh Verma
326
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
– Ben Crowell
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question.
The problem is as follows:
A bullet of mass $25,g$ is fired horizontally into a ballistic pendulum of mass $5.0,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10,cm$, find the speed of the bullet.
(H.C. Verma, Centre of Mass, Q47)
Let the mass of the bullet be $m=25,g$.
Let the mass of the pendulum be $M=5,kg$.
Let the peak height be $h=0.1,m$.
Let the initial and final velocities be $u$ and $v$ respectively.
Method 1: (Momentum Conservation)
$$mu=(M+m)v implies v=fracmuM+m\ textAlso, frac12(M+m)v^2=(M+m)gh\ implies u^2=2biggl(fracM+mmbiggr)^2gh\ implies u=201sqrt2 text m/s$$
Method 2: (Energy Conservation)
Since both the masses move together after the collision, and because the velocity at highest point is null, therefore:
$$frac12mu^2=frac12(M+m)v^2=(M+m)gh\ implies u^2=2ghbiggl(fracM+mmbiggr) implies u=sqrt402$$
To summarize myself, I am curious about the following:
Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations?
homework-and-exercises newtonian-mechanics momentum energy-conservation collision
edited 1 hour ago
Ben Crowell
46.2k3147279
asked 2 hours ago
Utkarsh Verma
326
I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question.
The problem is as follows:
A bullet of mass $25,g$ is fired horizontally into a ballistic pendulum of mass $5.0,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10,cm$, find the speed of the bullet.
(H.C. Verma, Centre of Mass, Q47)
Let the mass of the bullet be $m=25,g$.
Let the mass of the pendulum be $M=5,kg$.
Let the peak height be $h=0.1,m$.
Let the initial and final velocities be $u$ and $v$ respectively.
Method 1: (Momentum Conservation)
$$mu=(M+m)v implies v=fracmuM+m\ textAlso, frac12(M+m)v^2=(M+m)gh\ implies u^2=2biggl(fracM+mmbiggr)^2gh\ implies u=201sqrt2 text m/s$$
Method 2: (Energy Conservation)
Since both the masses move together after the collision, and because the velocity at highest point is null, therefore:
$$frac12mu^2=frac12(M+m)v^2=(M+m)gh\ implies u^2=2ghbiggl(fracM+mmbiggr) implies u=sqrt402$$
To summarize myself, I am curious about the following:
Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations?
homework-and-exercises newtonian-mechanics momentum energy-conservation collision
homework-and-exercises newtonian-mechanics momentum energy-conservation collision
edited 1 hour ago
Ben Crowell
46.2k3147279
asked 2 hours ago
Utkarsh Verma
326
edited 1 hour ago
Ben Crowell
46.2k3147279
asked 2 hours ago
Utkarsh Verma
326
edited 1 hour ago
Ben Crowell
46.2k3147279
edited 1 hour ago
Ben Crowell
46.2k3147279
edited 1 hour ago
Ben Crowell
46.2k3147279
46.2k3147279
asked 2 hours ago
Utkarsh Verma
326
asked 2 hours ago
Utkarsh Verma
326
asked 2 hours ago
Utkarsh Verma
326
326
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
– Ben Crowell
1 hour ago
add a comment |Â
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
– Ben Crowell
1 hour ago
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
– Ben Crowell
1 hour ago
I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
– Ben Crowell
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Shouldn't the energy be conserved as well as the momentum?
But this is a perfectly inelastic collision, i.e., kinetic energy is not conserved (maximally not conserved in fact).
From the Wikipedia article Inelastic collision:
An inelastic collision, in contrast to an elastic collision, is a
collision in which kinetic energy is not conserved due to the action
of internal friction.
...
A perfectly inelastic collision occurs when the maximum amount of
kinetic energy of a system is lost. In a perfectly inelastic
collision, i.e., a zero coefficient of restitution, the colliding
particles stick together.
answered 2 hours ago
Alfred Centauri
46.9k345141
1
Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks.
– Utkarsh Verma
1 hour ago
@UtkarshVerma think of the heat produced and absorbed in "embedding"
– anna v
26 mins ago
add a comment |Â
up vote
1
down vote
As in another answer, it's just that some of the kinetic energy is spent on heating the bob of the pendulum.
Energy conservation figures in the calculations when either there is effectively zero energy converted into heat, as in, say, the gravitational interaction of celestial bodies; or the heat figures in the dynamics, such as in the propagation of a shock, or flow of gas along a duct.
answered 1 hour ago
AmbretteOrrisey
1043
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
Energy is conserved only in elastic collisions, which is not this case. You can think it as energy is dissipated as heat when the bullet hits the pendulum, thus it's conserved too, though that's not seems so clear.
The point of using momentum is precisely that we don't need worry about the heat transference of the system and get along using just the given data. Also, momentum gives us very usefull information over the direction of the resulting velocity, which isn't present at energy calculations.
Be aware that some problems must be solved using both momentum and energy conservation.
answered 1 hour ago
Murillo Spadin
564
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Shouldn't the energy be conserved as well as the momentum?
But this is a perfectly inelastic collision, i.e., kinetic energy is not conserved (maximally not conserved in fact).
From the Wikipedia article Inelastic collision:
An inelastic collision, in contrast to an elastic collision, is a
collision in which kinetic energy is not conserved due to the action
of internal friction.
...
A perfectly inelastic collision occurs when the maximum amount of
kinetic energy of a system is lost. In a perfectly inelastic
collision, i.e., a zero coefficient of restitution, the colliding
particles stick together.
answered 2 hours ago
Alfred Centauri
46.9k345141
1
Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks.
– Utkarsh Verma
1 hour ago
@UtkarshVerma think of the heat produced and absorbed in "embedding"
– anna v
26 mins ago
add a comment |Â
up vote
3
down vote
accepted
Shouldn't the energy be conserved as well as the momentum?
But this is a perfectly inelastic collision, i.e., kinetic energy is not conserved (maximally not conserved in fact).
From the Wikipedia article Inelastic collision:
An inelastic collision, in contrast to an elastic collision, is a
collision in which kinetic energy is not conserved due to the action
of internal friction.
...
A perfectly inelastic collision occurs when the maximum amount of
kinetic energy of a system is lost. In a perfectly inelastic
collision, i.e., a zero coefficient of restitution, the colliding
particles stick together.
answered 2 hours ago
Alfred Centauri
46.9k345141
1
Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks.
– Utkarsh Verma
1 hour ago
@UtkarshVerma think of the heat produced and absorbed in "embedding"
– anna v
26 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Shouldn't the energy be conserved as well as the momentum?
But this is a perfectly inelastic collision, i.e., kinetic energy is not conserved (maximally not conserved in fact).
From the Wikipedia article Inelastic collision:
An inelastic collision, in contrast to an elastic collision, is a
collision in which kinetic energy is not conserved due to the action
of internal friction.
...
A perfectly inelastic collision occurs when the maximum amount of
kinetic energy of a system is lost. In a perfectly inelastic
collision, i.e., a zero coefficient of restitution, the colliding
particles stick together.
answered 2 hours ago
Alfred Centauri
46.9k345141
Shouldn't the energy be conserved as well as the momentum?
But this is a perfectly inelastic collision, i.e., kinetic energy is not conserved (maximally not conserved in fact).
From the Wikipedia article Inelastic collision:
An inelastic collision, in contrast to an elastic collision, is a
collision in which kinetic energy is not conserved due to the action
of internal friction.
...
A perfectly inelastic collision occurs when the maximum amount of
kinetic energy of a system is lost. In a perfectly inelastic
collision, i.e., a zero coefficient of restitution, the colliding
particles stick together.
answered 2 hours ago
Alfred Centauri
46.9k345141
answered 2 hours ago
Alfred Centauri
46.9k345141
answered 2 hours ago
Alfred Centauri
46.9k345141
answered 2 hours ago
Alfred Centauri
46.9k345141
46.9k345141
1
Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks.
– Utkarsh Verma
1 hour ago
@UtkarshVerma think of the heat produced and absorbed in "embedding"
– anna v
26 mins ago
add a comment |Â
1
Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks.
– Utkarsh Verma
1 hour ago
@UtkarshVerma think of the heat produced and absorbed in "embedding"
– anna v
26 mins ago
1
1
Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks.
– Utkarsh Verma
1 hour ago
Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks.
– Utkarsh Verma
1 hour ago
@UtkarshVerma think of the heat produced and absorbed in "embedding"
– anna v
26 mins ago
@UtkarshVerma think of the heat produced and absorbed in "embedding"
– anna v
26 mins ago
add a comment |Â
up vote
1
down vote
As in another answer, it's just that some of the kinetic energy is spent on heating the bob of the pendulum.
Energy conservation figures in the calculations when either there is effectively zero energy converted into heat, as in, say, the gravitational interaction of celestial bodies; or the heat figures in the dynamics, such as in the propagation of a shock, or flow of gas along a duct.
answered 1 hour ago
AmbretteOrrisey
1043
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
As in another answer, it's just that some of the kinetic energy is spent on heating the bob of the pendulum.
Energy conservation figures in the calculations when either there is effectively zero energy converted into heat, as in, say, the gravitational interaction of celestial bodies; or the heat figures in the dynamics, such as in the propagation of a shock, or flow of gas along a duct.
answered 1 hour ago
AmbretteOrrisey
1043
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As in another answer, it's just that some of the kinetic energy is spent on heating the bob of the pendulum.
Energy conservation figures in the calculations when either there is effectively zero energy converted into heat, as in, say, the gravitational interaction of celestial bodies; or the heat figures in the dynamics, such as in the propagation of a shock, or flow of gas along a duct.
answered 1 hour ago
AmbretteOrrisey
1043
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
As in another answer, it's just that some of the kinetic energy is spent on heating the bob of the pendulum.
Energy conservation figures in the calculations when either there is effectively zero energy converted into heat, as in, say, the gravitational interaction of celestial bodies; or the heat figures in the dynamics, such as in the propagation of a shock, or flow of gas along a duct.
answered 1 hour ago
AmbretteOrrisey
1043
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
AmbretteOrrisey
1043
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
AmbretteOrrisey
1043
answered 1 hour ago
AmbretteOrrisey
1043
1043
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
AmbretteOrrisey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
0
down vote
Energy is conserved only in elastic collisions, which is not this case. You can think it as energy is dissipated as heat when the bullet hits the pendulum, thus it's conserved too, though that's not seems so clear.
The point of using momentum is precisely that we don't need worry about the heat transference of the system and get along using just the given data. Also, momentum gives us very usefull information over the direction of the resulting velocity, which isn't present at energy calculations.
Be aware that some problems must be solved using both momentum and energy conservation.
answered 1 hour ago
Murillo Spadin
564
add a comment |Â
up vote
0
down vote
Energy is conserved only in elastic collisions, which is not this case. You can think it as energy is dissipated as heat when the bullet hits the pendulum, thus it's conserved too, though that's not seems so clear.
The point of using momentum is precisely that we don't need worry about the heat transference of the system and get along using just the given data. Also, momentum gives us very usefull information over the direction of the resulting velocity, which isn't present at energy calculations.
Be aware that some problems must be solved using both momentum and energy conservation.
answered 1 hour ago
Murillo Spadin
564
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Energy is conserved only in elastic collisions, which is not this case. You can think it as energy is dissipated as heat when the bullet hits the pendulum, thus it's conserved too, though that's not seems so clear.
The point of using momentum is precisely that we don't need worry about the heat transference of the system and get along using just the given data. Also, momentum gives us very usefull information over the direction of the resulting velocity, which isn't present at energy calculations.
Be aware that some problems must be solved using both momentum and energy conservation.
answered 1 hour ago
Murillo Spadin
564
Energy is conserved only in elastic collisions, which is not this case. You can think it as energy is dissipated as heat when the bullet hits the pendulum, thus it's conserved too, though that's not seems so clear.
The point of using momentum is precisely that we don't need worry about the heat transference of the system and get along using just the given data. Also, momentum gives us very usefull information over the direction of the resulting velocity, which isn't present at energy calculations.
Be aware that some problems must be solved using both momentum and energy conservation.
answered 1 hour ago
Murillo Spadin
564
answered 1 hour ago
Murillo Spadin
564
answered 1 hour ago
Murillo Spadin
564
answered 1 hour ago
Murillo Spadin
564
564
add a comment |Â
add a comment |Â
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I've added the homework-and-exercises tag. In the future, please use this tag on this type of question.
– Ben Crowell
1 hour ago