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?










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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














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?










share|cite|improve this question























  • 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












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?










share|cite|improve this question















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






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edited 1 hour ago









Ben Crowell

46.2k3147279




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asked 2 hours ago









Utkarsh Verma

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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
















  • 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










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.







share|cite|improve this answer
















  • 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

















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.






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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.
























    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.






    share|cite|improve this answer




















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      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.







      share|cite|improve this answer
















      • 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














      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.







      share|cite|improve this answer
















      • 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












      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.







      share|cite|improve this answer













      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.








      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      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












      • 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










      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.






      share|cite|improve this answer








      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.





















        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.






        share|cite|improve this answer








        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.



















          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.






          share|cite|improve this answer








          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.







          share|cite|improve this answer








          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.









          share|cite|improve this answer



          share|cite|improve this answer






          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




          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.




















              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.






              share|cite|improve this answer
























                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.






                share|cite|improve this answer






















                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Murillo Spadin

                  564




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