Why does energy conservation give me the wrong answer in this inelastic collision problem?Collision Between Two Particles: Writing the Mass As A Function of The AngleLoss of kinetic energy in inelastic collisionElastic collision of point particle and rodIn a CMCS 2-body system, why does the speed of the particles after collision stay the same?Conservation of relativistic momentumSpecial Relativity problem using conservation of energy and momentumWhat's the physical reason behind the increment of total kinetic energy of a system after an inelastic collision?How does the mass and velocity affect the elasticity of a collision?Why isn't energy conserved in this collision problem?Predicting elastic collisions
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Why does energy conservation give me the wrong answer in this inelastic collision problem?
Collision Between Two Particles: Writing the Mass As A Function of The AngleLoss of kinetic energy in inelastic collisionElastic collision of point particle and rodIn a CMCS 2-body system, why does the speed of the particles after collision stay the same?Conservation of relativistic momentumSpecial Relativity problem using conservation of energy and momentumWhat's the physical reason behind the increment of total kinetic energy of a system after an inelastic collision?How does the mass and velocity affect the elasticity of a collision?Why isn't energy conserved in this collision problem?Predicting elastic collisions
$begingroup$
Suppose there are two objects. The first object is stationary and has the next parameters: v1=0 m/s, m1= 4kg. The second object is heading towards the first one with speed v2= 8m/s and mass m2= 60 kg. Due to collision, objects merge into one and continue to move with the velocity v3. Suppose all the energy gets converted into speed. I want to know the magnitude of velocity v3. My question is : Why do we get slightly different results when using kinetic energy compared to conservation of momentum when solving for v3. If we assume no energy gets converted into heat, couldn't we set up equations like : Ek1=0 J and Ek2=$fracm_2*v_2^22$ . E3=E2+E1=$fracv_3^22*(m_1+m_2)$ . v3=$sqrtfrac(E_1+E_2)*2(m_1+m_2)$ ? Why do we get different result this way comparing to the result when we calculate v3 using conservation of momentum?
newtonian-mechanics energy momentum energy-conservation collision
edited yesterday
knzhou
45.1k11122219
asked yesterday
ToTheSpace 2ToTheSpace 2
221
$endgroup$
add a comment |
$begingroup$
Suppose there are two objects. The first object is stationary and has the next parameters: v1=0 m/s, m1= 4kg. The second object is heading towards the first one with speed v2= 8m/s and mass m2= 60 kg. Due to collision, objects merge into one and continue to move with the velocity v3. Suppose all the energy gets converted into speed. I want to know the magnitude of velocity v3. My question is : Why do we get slightly different results when using kinetic energy compared to conservation of momentum when solving for v3. If we assume no energy gets converted into heat, couldn't we set up equations like : Ek1=0 J and Ek2=$fracm_2*v_2^22$ . E3=E2+E1=$fracv_3^22*(m_1+m_2)$ . v3=$sqrtfrac(E_1+E_2)*2(m_1+m_2)$ ? Why do we get different result this way comparing to the result when we calculate v3 using conservation of momentum?
newtonian-mechanics energy momentum energy-conservation collision
edited yesterday
knzhou
45.1k11122219
asked yesterday
ToTheSpace 2ToTheSpace 2
221
$endgroup$
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
yesterday
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
yesterday
1
$begingroup$
@Jasper Not at all! If somebody doesn't understand why inelastic collisions (i.e. sticking together) don't conserve energy, the title edit doesn't change anything. All it does is make the answer easier to see for people who do know mechanics.
$endgroup$
– knzhou
yesterday
$begingroup$
Imagine the two objct being cars. When they collide, they change their outer form (alas!), and that takes up some energy you don't get back. That's where the "missing energy" from the correct momentum-based calculation goes into.
$endgroup$
– Ralf Kleberhoff
yesterday
2
$begingroup$
@RalfKleberhoff This is where the energy could go, but in general the objects to not need to deform for the collision to be perfectly inelastic.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
$begingroup$
Suppose there are two objects. The first object is stationary and has the next parameters: v1=0 m/s, m1= 4kg. The second object is heading towards the first one with speed v2= 8m/s and mass m2= 60 kg. Due to collision, objects merge into one and continue to move with the velocity v3. Suppose all the energy gets converted into speed. I want to know the magnitude of velocity v3. My question is : Why do we get slightly different results when using kinetic energy compared to conservation of momentum when solving for v3. If we assume no energy gets converted into heat, couldn't we set up equations like : Ek1=0 J and Ek2=$fracm_2*v_2^22$ . E3=E2+E1=$fracv_3^22*(m_1+m_2)$ . v3=$sqrtfrac(E_1+E_2)*2(m_1+m_2)$ ? Why do we get different result this way comparing to the result when we calculate v3 using conservation of momentum?
newtonian-mechanics energy momentum energy-conservation collision
edited yesterday
knzhou
45.1k11122219
asked yesterday
ToTheSpace 2ToTheSpace 2
221
$endgroup$
Suppose there are two objects. The first object is stationary and has the next parameters: v1=0 m/s, m1= 4kg. The second object is heading towards the first one with speed v2= 8m/s and mass m2= 60 kg. Due to collision, objects merge into one and continue to move with the velocity v3. Suppose all the energy gets converted into speed. I want to know the magnitude of velocity v3. My question is : Why do we get slightly different results when using kinetic energy compared to conservation of momentum when solving for v3. If we assume no energy gets converted into heat, couldn't we set up equations like : Ek1=0 J and Ek2=$fracm_2*v_2^22$ . E3=E2+E1=$fracv_3^22*(m_1+m_2)$ . v3=$sqrtfrac(E_1+E_2)*2(m_1+m_2)$ ? Why do we get different result this way comparing to the result when we calculate v3 using conservation of momentum?
newtonian-mechanics energy momentum energy-conservation collision
newtonian-mechanics energy momentum energy-conservation collision
edited yesterday
knzhou
45.1k11122219
asked yesterday
ToTheSpace 2ToTheSpace 2
221
edited yesterday
knzhou
45.1k11122219
asked yesterday
ToTheSpace 2ToTheSpace 2
221
edited yesterday
knzhou
45.1k11122219
edited yesterday
knzhou
45.1k11122219
edited yesterday
knzhou
45.1k11122219
45.1k11122219
asked yesterday
ToTheSpace 2ToTheSpace 2
221
asked yesterday
ToTheSpace 2ToTheSpace 2
221
asked yesterday
ToTheSpace 2ToTheSpace 2
221
221
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
yesterday
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
yesterday
1
$begingroup$
@Jasper Not at all! If somebody doesn't understand why inelastic collisions (i.e. sticking together) don't conserve energy, the title edit doesn't change anything. All it does is make the answer easier to see for people who do know mechanics.
$endgroup$
– knzhou
yesterday
$begingroup$
Imagine the two objct being cars. When they collide, they change their outer form (alas!), and that takes up some energy you don't get back. That's where the "missing energy" from the correct momentum-based calculation goes into.
$endgroup$
– Ralf Kleberhoff
yesterday
2
$begingroup$
@RalfKleberhoff This is where the energy could go, but in general the objects to not need to deform for the collision to be perfectly inelastic.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
yesterday
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
yesterday
1
$begingroup$
@Jasper Not at all! If somebody doesn't understand why inelastic collisions (i.e. sticking together) don't conserve energy, the title edit doesn't change anything. All it does is make the answer easier to see for people who do know mechanics.
$endgroup$
– knzhou
yesterday
$begingroup$
Imagine the two objct being cars. When they collide, they change their outer form (alas!), and that takes up some energy you don't get back. That's where the "missing energy" from the correct momentum-based calculation goes into.
$endgroup$
– Ralf Kleberhoff
yesterday
2
$begingroup$
@RalfKleberhoff This is where the energy could go, but in general the objects to not need to deform for the collision to be perfectly inelastic.
$endgroup$
– Aaron Stevens
yesterday
3
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
yesterday
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
yesterday
2
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
yesterday
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
yesterday
1
1
$begingroup$
@Jasper Not at all! If somebody doesn't understand why inelastic collisions (i.e. sticking together) don't conserve energy, the title edit doesn't change anything. All it does is make the answer easier to see for people who do know mechanics.
$endgroup$
– knzhou
yesterday
$begingroup$
@Jasper Not at all! If somebody doesn't understand why inelastic collisions (i.e. sticking together) don't conserve energy, the title edit doesn't change anything. All it does is make the answer easier to see for people who do know mechanics.
$endgroup$
– knzhou
yesterday
$begingroup$
Imagine the two objct being cars. When they collide, they change their outer form (alas!), and that takes up some energy you don't get back. That's where the "missing energy" from the correct momentum-based calculation goes into.
$endgroup$
– Ralf Kleberhoff
yesterday
$begingroup$
Imagine the two objct being cars. When they collide, they change their outer form (alas!), and that takes up some energy you don't get back. That's where the "missing energy" from the correct momentum-based calculation goes into.
$endgroup$
– Ralf Kleberhoff
yesterday
2
2
$begingroup$
@RalfKleberhoff This is where the energy could go, but in general the objects to not need to deform for the collision to be perfectly inelastic.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@RalfKleberhoff This is where the energy could go, but in general the objects to not need to deform for the collision to be perfectly inelastic.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Momentum is conserved if there are no external forces involved. This is the case in your setup, so the final velocity you get from conservation of momentum is right.
If you start with conservation of energy, you'll see that you get a different velocity.
Since physicists strongly believe in conservation of energy, this means that some of the mechanical energy went somewhere. This is a strong hint that you can not assume conservation of mechanical energy at will. If you assume both conservation of momentum and energy, the bodies can't stick together and move as one.
The "sticking" eats up some energy that is lost on the mechanical side.
answered yesterday
JasperJasper
1,0941517
$endgroup$
2
$begingroup$
Since physicists strongly believe in conservation of energy... I don't think the belief in it makes the conclusions true. Conservation of energy isn't like Tinker Bell :)
$endgroup$
– Aaron Stevens
yesterday
add a comment |
$begingroup$
The other answers are correct in saying that energy is not conserved, but I feel like they are lacking in truly showing why this is the case. So let's assume energy and momentum are both conserved in this process and see that we arrive at a contradiction.
So we have object 1 at rest and object 2 hits it and they stick together. Using momentum conservation:
$$m_2v_2=m_1v+m_2v$$
Using energy conservation (cancelling $1/2$ from each term):
$$m_2v_2^2=m_1v^2+m_2v^2$$
Now for fun tricks. In each of the above equations let's get all terms involving object 2 on the left and all terms involving object 1 on the right and then divide the energy equation by the momentum equation:
$$fracm_2(v_2^2-v^2)m_2(v_2-v)=fracm_1v^2m_1v$$
Now, $v_2^2-v^2=(v_2-v)(v_2+v)$, therefore (cancelling many things)
$$v_2+v=v$$
or
$$v_2=0$$
Uh oh. We see that before the collision object 2 is at rest. But object 1 was at rest also$^*$. Therefore no collision with these properties could have occurred. Therefore something is wrong with our assumptions of both energy and momentum conservation.
Momentum has to be conserved since there are no external forces on the system. Therefore it must be that energy is not conserved. There is nothing saying energy must be conserved here, but we have now shown that it actually cannot be conserved.
$^*$ In general if we had specified the velocity of $m_1$ by $v_1$ we would have arrived at $v_1=v_2$ using similar methods used above. This just means the objects started out with the same velocity and never actually collide. Therefore, the only way our system can conserve both energy and momentum and have equal "final" velocities is if no collision happened.
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
$endgroup$
add a comment |
$begingroup$
The objects are merged, and an inelastic collision is one in which objects stick together after impact: kinetic energy is not conserved. Therefore, you can't assume that kinetic energy is conserved; merging takes a little bit of energy.
If this system is isolated, and no force acted upon neither of the two objects, then, momentum is conserved; hence, you can calculate the final velocity using the momentum's equation.
answered yesterday
Busy MinderBusy Minder
213
New contributor
Busy Minder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Suppose all the energy gets converted into speed.
Uh oh! That's your error. You can't just suppose that something that doesn't happen, happens.
The collision was inelastic. That means some of the energy gets converted to heat, as the objects merge. This is the difference between your total before and after energy.
answered yesterday
Neil_UKNeil_UK
1605
$endgroup$
add a comment |
$begingroup$
I think that the hypothesis of momentum conservation is flawed since the problem really doesn't state that. Also there's nothing saying that the system is isolated at all.
Hence you calculated what you can (assuming that "energy" in the problem statement refers only to the kinetic energy itself).
If the hypothesis of kinetic energy conversion is correct (I would assume so since it's stated) then your energy-based calculation is correct :) and the missing momentum is due to an external source.
answered yesterday
user3155984user3155984
1
New contributor
user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Momentum is conserved in collisions. Why would you assume external forces? It seems like you are the one making additional assumptions, not the OP.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@Aaron: I know what is momentum/energy conservation. I'm not making any additional assumptions and just using the conditions stated in the problem. If the problem is badly formulated this is another situation.
$endgroup$
– user3155984
yesterday
1
$begingroup$
The problem isn't poorly formulated. Just before and just after the collision momentum has to be conserved. If there are external forces then you just have to look just before and just after. Momentum will not be lost due to the collision, and it seems like the OP is interested in the collision, not unspecified external forces that aren't involved with the collision.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@AaronStevens The problem is badly formulated in the sense that it imposes two assumptions that contradict each other (given that there are no external forces), namely "objects merge into one" and "all the energy gets converted into speed."
$endgroup$
– Andreas Blass
yesterday
$begingroup$
No... The OP is incorrectly assuming that energy is conserved when the objects stick together. The "problem" (or you could say system) is just a perfectly inelastic collision where the objects stick together. Any incorrect assumptions come from the OP, not from the "problem" or system.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Momentum is conserved if there are no external forces involved. This is the case in your setup, so the final velocity you get from conservation of momentum is right.
If you start with conservation of energy, you'll see that you get a different velocity.
Since physicists strongly believe in conservation of energy, this means that some of the mechanical energy went somewhere. This is a strong hint that you can not assume conservation of mechanical energy at will. If you assume both conservation of momentum and energy, the bodies can't stick together and move as one.
The "sticking" eats up some energy that is lost on the mechanical side.
answered yesterday
JasperJasper
1,0941517
$endgroup$
2
$begingroup$
Since physicists strongly believe in conservation of energy... I don't think the belief in it makes the conclusions true. Conservation of energy isn't like Tinker Bell :)
$endgroup$
– Aaron Stevens
yesterday
add a comment |
$begingroup$
Momentum is conserved if there are no external forces involved. This is the case in your setup, so the final velocity you get from conservation of momentum is right.
If you start with conservation of energy, you'll see that you get a different velocity.
Since physicists strongly believe in conservation of energy, this means that some of the mechanical energy went somewhere. This is a strong hint that you can not assume conservation of mechanical energy at will. If you assume both conservation of momentum and energy, the bodies can't stick together and move as one.
The "sticking" eats up some energy that is lost on the mechanical side.
answered yesterday
JasperJasper
1,0941517
$endgroup$
2
$begingroup$
Since physicists strongly believe in conservation of energy... I don't think the belief in it makes the conclusions true. Conservation of energy isn't like Tinker Bell :)
$endgroup$
– Aaron Stevens
yesterday
add a comment |
$begingroup$
Momentum is conserved if there are no external forces involved. This is the case in your setup, so the final velocity you get from conservation of momentum is right.
If you start with conservation of energy, you'll see that you get a different velocity.
Since physicists strongly believe in conservation of energy, this means that some of the mechanical energy went somewhere. This is a strong hint that you can not assume conservation of mechanical energy at will. If you assume both conservation of momentum and energy, the bodies can't stick together and move as one.
The "sticking" eats up some energy that is lost on the mechanical side.
answered yesterday
JasperJasper
1,0941517
$endgroup$
Momentum is conserved if there are no external forces involved. This is the case in your setup, so the final velocity you get from conservation of momentum is right.
If you start with conservation of energy, you'll see that you get a different velocity.
Since physicists strongly believe in conservation of energy, this means that some of the mechanical energy went somewhere. This is a strong hint that you can not assume conservation of mechanical energy at will. If you assume both conservation of momentum and energy, the bodies can't stick together and move as one.
The "sticking" eats up some energy that is lost on the mechanical side.
answered yesterday
JasperJasper
1,0941517
answered yesterday
JasperJasper
1,0941517
answered yesterday
JasperJasper
1,0941517
answered yesterday
JasperJasper
1,0941517
1,0941517
2
$begingroup$
Since physicists strongly believe in conservation of energy... I don't think the belief in it makes the conclusions true. Conservation of energy isn't like Tinker Bell :)
$endgroup$
– Aaron Stevens
yesterday
add a comment |
2
$begingroup$
Since physicists strongly believe in conservation of energy... I don't think the belief in it makes the conclusions true. Conservation of energy isn't like Tinker Bell :)
$endgroup$
– Aaron Stevens
yesterday
2
2
$begingroup$
Since physicists strongly believe in conservation of energy... I don't think the belief in it makes the conclusions true. Conservation of energy isn't like Tinker Bell :)
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
Since physicists strongly believe in conservation of energy... I don't think the belief in it makes the conclusions true. Conservation of energy isn't like Tinker Bell :)
$endgroup$
– Aaron Stevens
yesterday
add a comment |
$begingroup$
The other answers are correct in saying that energy is not conserved, but I feel like they are lacking in truly showing why this is the case. So let's assume energy and momentum are both conserved in this process and see that we arrive at a contradiction.
So we have object 1 at rest and object 2 hits it and they stick together. Using momentum conservation:
$$m_2v_2=m_1v+m_2v$$
Using energy conservation (cancelling $1/2$ from each term):
$$m_2v_2^2=m_1v^2+m_2v^2$$
Now for fun tricks. In each of the above equations let's get all terms involving object 2 on the left and all terms involving object 1 on the right and then divide the energy equation by the momentum equation:
$$fracm_2(v_2^2-v^2)m_2(v_2-v)=fracm_1v^2m_1v$$
Now, $v_2^2-v^2=(v_2-v)(v_2+v)$, therefore (cancelling many things)
$$v_2+v=v$$
or
$$v_2=0$$
Uh oh. We see that before the collision object 2 is at rest. But object 1 was at rest also$^*$. Therefore no collision with these properties could have occurred. Therefore something is wrong with our assumptions of both energy and momentum conservation.
Momentum has to be conserved since there are no external forces on the system. Therefore it must be that energy is not conserved. There is nothing saying energy must be conserved here, but we have now shown that it actually cannot be conserved.
$^*$ In general if we had specified the velocity of $m_1$ by $v_1$ we would have arrived at $v_1=v_2$ using similar methods used above. This just means the objects started out with the same velocity and never actually collide. Therefore, the only way our system can conserve both energy and momentum and have equal "final" velocities is if no collision happened.
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
$endgroup$
add a comment |
$begingroup$
The other answers are correct in saying that energy is not conserved, but I feel like they are lacking in truly showing why this is the case. So let's assume energy and momentum are both conserved in this process and see that we arrive at a contradiction.
So we have object 1 at rest and object 2 hits it and they stick together. Using momentum conservation:
$$m_2v_2=m_1v+m_2v$$
Using energy conservation (cancelling $1/2$ from each term):
$$m_2v_2^2=m_1v^2+m_2v^2$$
Now for fun tricks. In each of the above equations let's get all terms involving object 2 on the left and all terms involving object 1 on the right and then divide the energy equation by the momentum equation:
$$fracm_2(v_2^2-v^2)m_2(v_2-v)=fracm_1v^2m_1v$$
Now, $v_2^2-v^2=(v_2-v)(v_2+v)$, therefore (cancelling many things)
$$v_2+v=v$$
or
$$v_2=0$$
Uh oh. We see that before the collision object 2 is at rest. But object 1 was at rest also$^*$. Therefore no collision with these properties could have occurred. Therefore something is wrong with our assumptions of both energy and momentum conservation.
Momentum has to be conserved since there are no external forces on the system. Therefore it must be that energy is not conserved. There is nothing saying energy must be conserved here, but we have now shown that it actually cannot be conserved.
$^*$ In general if we had specified the velocity of $m_1$ by $v_1$ we would have arrived at $v_1=v_2$ using similar methods used above. This just means the objects started out with the same velocity and never actually collide. Therefore, the only way our system can conserve both energy and momentum and have equal "final" velocities is if no collision happened.
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
$endgroup$
add a comment |
$begingroup$
The other answers are correct in saying that energy is not conserved, but I feel like they are lacking in truly showing why this is the case. So let's assume energy and momentum are both conserved in this process and see that we arrive at a contradiction.
So we have object 1 at rest and object 2 hits it and they stick together. Using momentum conservation:
$$m_2v_2=m_1v+m_2v$$
Using energy conservation (cancelling $1/2$ from each term):
$$m_2v_2^2=m_1v^2+m_2v^2$$
Now for fun tricks. In each of the above equations let's get all terms involving object 2 on the left and all terms involving object 1 on the right and then divide the energy equation by the momentum equation:
$$fracm_2(v_2^2-v^2)m_2(v_2-v)=fracm_1v^2m_1v$$
Now, $v_2^2-v^2=(v_2-v)(v_2+v)$, therefore (cancelling many things)
$$v_2+v=v$$
or
$$v_2=0$$
Uh oh. We see that before the collision object 2 is at rest. But object 1 was at rest also$^*$. Therefore no collision with these properties could have occurred. Therefore something is wrong with our assumptions of both energy and momentum conservation.
Momentum has to be conserved since there are no external forces on the system. Therefore it must be that energy is not conserved. There is nothing saying energy must be conserved here, but we have now shown that it actually cannot be conserved.
$^*$ In general if we had specified the velocity of $m_1$ by $v_1$ we would have arrived at $v_1=v_2$ using similar methods used above. This just means the objects started out with the same velocity and never actually collide. Therefore, the only way our system can conserve both energy and momentum and have equal "final" velocities is if no collision happened.
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
$endgroup$
The other answers are correct in saying that energy is not conserved, but I feel like they are lacking in truly showing why this is the case. So let's assume energy and momentum are both conserved in this process and see that we arrive at a contradiction.
So we have object 1 at rest and object 2 hits it and they stick together. Using momentum conservation:
$$m_2v_2=m_1v+m_2v$$
Using energy conservation (cancelling $1/2$ from each term):
$$m_2v_2^2=m_1v^2+m_2v^2$$
Now for fun tricks. In each of the above equations let's get all terms involving object 2 on the left and all terms involving object 1 on the right and then divide the energy equation by the momentum equation:
$$fracm_2(v_2^2-v^2)m_2(v_2-v)=fracm_1v^2m_1v$$
Now, $v_2^2-v^2=(v_2-v)(v_2+v)$, therefore (cancelling many things)
$$v_2+v=v$$
or
$$v_2=0$$
Uh oh. We see that before the collision object 2 is at rest. But object 1 was at rest also$^*$. Therefore no collision with these properties could have occurred. Therefore something is wrong with our assumptions of both energy and momentum conservation.
Momentum has to be conserved since there are no external forces on the system. Therefore it must be that energy is not conserved. There is nothing saying energy must be conserved here, but we have now shown that it actually cannot be conserved.
$^*$ In general if we had specified the velocity of $m_1$ by $v_1$ we would have arrived at $v_1=v_2$ using similar methods used above. This just means the objects started out with the same velocity and never actually collide. Therefore, the only way our system can conserve both energy and momentum and have equal "final" velocities is if no collision happened.
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
edited yesterday
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
answered yesterday
Aaron StevensAaron Stevens
13.1k42248
13.1k42248
add a comment |
add a comment |
$begingroup$
The objects are merged, and an inelastic collision is one in which objects stick together after impact: kinetic energy is not conserved. Therefore, you can't assume that kinetic energy is conserved; merging takes a little bit of energy.
If this system is isolated, and no force acted upon neither of the two objects, then, momentum is conserved; hence, you can calculate the final velocity using the momentum's equation.
answered yesterday
Busy MinderBusy Minder
213
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add a comment |
$begingroup$
The objects are merged, and an inelastic collision is one in which objects stick together after impact: kinetic energy is not conserved. Therefore, you can't assume that kinetic energy is conserved; merging takes a little bit of energy.
If this system is isolated, and no force acted upon neither of the two objects, then, momentum is conserved; hence, you can calculate the final velocity using the momentum's equation.
answered yesterday
Busy MinderBusy Minder
213
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$endgroup$
add a comment |
$begingroup$
The objects are merged, and an inelastic collision is one in which objects stick together after impact: kinetic energy is not conserved. Therefore, you can't assume that kinetic energy is conserved; merging takes a little bit of energy.
If this system is isolated, and no force acted upon neither of the two objects, then, momentum is conserved; hence, you can calculate the final velocity using the momentum's equation.
answered yesterday
Busy MinderBusy Minder
213
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Busy Minder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
The objects are merged, and an inelastic collision is one in which objects stick together after impact: kinetic energy is not conserved. Therefore, you can't assume that kinetic energy is conserved; merging takes a little bit of energy.
If this system is isolated, and no force acted upon neither of the two objects, then, momentum is conserved; hence, you can calculate the final velocity using the momentum's equation.
answered yesterday
Busy MinderBusy Minder
213
New contributor
Busy Minder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Busy MinderBusy Minder
213
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answered yesterday
Busy MinderBusy Minder
213
answered yesterday
Busy MinderBusy Minder
213
213
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$begingroup$
Suppose all the energy gets converted into speed.
Uh oh! That's your error. You can't just suppose that something that doesn't happen, happens.
The collision was inelastic. That means some of the energy gets converted to heat, as the objects merge. This is the difference between your total before and after energy.
answered yesterday
Neil_UKNeil_UK
1605
$endgroup$
add a comment |
$begingroup$
Suppose all the energy gets converted into speed.
Uh oh! That's your error. You can't just suppose that something that doesn't happen, happens.
The collision was inelastic. That means some of the energy gets converted to heat, as the objects merge. This is the difference between your total before and after energy.
answered yesterday
Neil_UKNeil_UK
1605
$endgroup$
add a comment |
$begingroup$
Suppose all the energy gets converted into speed.
Uh oh! That's your error. You can't just suppose that something that doesn't happen, happens.
The collision was inelastic. That means some of the energy gets converted to heat, as the objects merge. This is the difference between your total before and after energy.
answered yesterday
Neil_UKNeil_UK
1605
$endgroup$
Suppose all the energy gets converted into speed.
Uh oh! That's your error. You can't just suppose that something that doesn't happen, happens.
The collision was inelastic. That means some of the energy gets converted to heat, as the objects merge. This is the difference between your total before and after energy.
answered yesterday
Neil_UKNeil_UK
1605
answered yesterday
Neil_UKNeil_UK
1605
answered yesterday
Neil_UKNeil_UK
1605
answered yesterday
Neil_UKNeil_UK
1605
1605
add a comment |
add a comment |
$begingroup$
I think that the hypothesis of momentum conservation is flawed since the problem really doesn't state that. Also there's nothing saying that the system is isolated at all.
Hence you calculated what you can (assuming that "energy" in the problem statement refers only to the kinetic energy itself).
If the hypothesis of kinetic energy conversion is correct (I would assume so since it's stated) then your energy-based calculation is correct :) and the missing momentum is due to an external source.
answered yesterday
user3155984user3155984
1
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$endgroup$
2
$begingroup$
Momentum is conserved in collisions. Why would you assume external forces? It seems like you are the one making additional assumptions, not the OP.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@Aaron: I know what is momentum/energy conservation. I'm not making any additional assumptions and just using the conditions stated in the problem. If the problem is badly formulated this is another situation.
$endgroup$
– user3155984
yesterday
1
$begingroup$
The problem isn't poorly formulated. Just before and just after the collision momentum has to be conserved. If there are external forces then you just have to look just before and just after. Momentum will not be lost due to the collision, and it seems like the OP is interested in the collision, not unspecified external forces that aren't involved with the collision.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@AaronStevens The problem is badly formulated in the sense that it imposes two assumptions that contradict each other (given that there are no external forces), namely "objects merge into one" and "all the energy gets converted into speed."
$endgroup$
– Andreas Blass
yesterday
$begingroup$
No... The OP is incorrectly assuming that energy is conserved when the objects stick together. The "problem" (or you could say system) is just a perfectly inelastic collision where the objects stick together. Any incorrect assumptions come from the OP, not from the "problem" or system.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
$begingroup$
I think that the hypothesis of momentum conservation is flawed since the problem really doesn't state that. Also there's nothing saying that the system is isolated at all.
Hence you calculated what you can (assuming that "energy" in the problem statement refers only to the kinetic energy itself).
If the hypothesis of kinetic energy conversion is correct (I would assume so since it's stated) then your energy-based calculation is correct :) and the missing momentum is due to an external source.
answered yesterday
user3155984user3155984
1
New contributor
user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Momentum is conserved in collisions. Why would you assume external forces? It seems like you are the one making additional assumptions, not the OP.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@Aaron: I know what is momentum/energy conservation. I'm not making any additional assumptions and just using the conditions stated in the problem. If the problem is badly formulated this is another situation.
$endgroup$
– user3155984
yesterday
1
$begingroup$
The problem isn't poorly formulated. Just before and just after the collision momentum has to be conserved. If there are external forces then you just have to look just before and just after. Momentum will not be lost due to the collision, and it seems like the OP is interested in the collision, not unspecified external forces that aren't involved with the collision.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@AaronStevens The problem is badly formulated in the sense that it imposes two assumptions that contradict each other (given that there are no external forces), namely "objects merge into one" and "all the energy gets converted into speed."
$endgroup$
– Andreas Blass
yesterday
$begingroup$
No... The OP is incorrectly assuming that energy is conserved when the objects stick together. The "problem" (or you could say system) is just a perfectly inelastic collision where the objects stick together. Any incorrect assumptions come from the OP, not from the "problem" or system.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
$begingroup$
I think that the hypothesis of momentum conservation is flawed since the problem really doesn't state that. Also there's nothing saying that the system is isolated at all.
Hence you calculated what you can (assuming that "energy" in the problem statement refers only to the kinetic energy itself).
If the hypothesis of kinetic energy conversion is correct (I would assume so since it's stated) then your energy-based calculation is correct :) and the missing momentum is due to an external source.
answered yesterday
user3155984user3155984
1
New contributor
user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I think that the hypothesis of momentum conservation is flawed since the problem really doesn't state that. Also there's nothing saying that the system is isolated at all.
Hence you calculated what you can (assuming that "energy" in the problem statement refers only to the kinetic energy itself).
If the hypothesis of kinetic energy conversion is correct (I would assume so since it's stated) then your energy-based calculation is correct :) and the missing momentum is due to an external source.
answered yesterday
user3155984user3155984
1
New contributor
user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
user3155984user3155984
1
New contributor
user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
user3155984user3155984
1
answered yesterday
user3155984user3155984
1
1
New contributor
user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user3155984 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Momentum is conserved in collisions. Why would you assume external forces? It seems like you are the one making additional assumptions, not the OP.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@Aaron: I know what is momentum/energy conservation. I'm not making any additional assumptions and just using the conditions stated in the problem. If the problem is badly formulated this is another situation.
$endgroup$
– user3155984
yesterday
1
$begingroup$
The problem isn't poorly formulated. Just before and just after the collision momentum has to be conserved. If there are external forces then you just have to look just before and just after. Momentum will not be lost due to the collision, and it seems like the OP is interested in the collision, not unspecified external forces that aren't involved with the collision.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@AaronStevens The problem is badly formulated in the sense that it imposes two assumptions that contradict each other (given that there are no external forces), namely "objects merge into one" and "all the energy gets converted into speed."
$endgroup$
– Andreas Blass
yesterday
$begingroup$
No... The OP is incorrectly assuming that energy is conserved when the objects stick together. The "problem" (or you could say system) is just a perfectly inelastic collision where the objects stick together. Any incorrect assumptions come from the OP, not from the "problem" or system.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
2
$begingroup$
Momentum is conserved in collisions. Why would you assume external forces? It seems like you are the one making additional assumptions, not the OP.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@Aaron: I know what is momentum/energy conservation. I'm not making any additional assumptions and just using the conditions stated in the problem. If the problem is badly formulated this is another situation.
$endgroup$
– user3155984
yesterday
1
$begingroup$
The problem isn't poorly formulated. Just before and just after the collision momentum has to be conserved. If there are external forces then you just have to look just before and just after. Momentum will not be lost due to the collision, and it seems like the OP is interested in the collision, not unspecified external forces that aren't involved with the collision.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@AaronStevens The problem is badly formulated in the sense that it imposes two assumptions that contradict each other (given that there are no external forces), namely "objects merge into one" and "all the energy gets converted into speed."
$endgroup$
– Andreas Blass
yesterday
$begingroup$
No... The OP is incorrectly assuming that energy is conserved when the objects stick together. The "problem" (or you could say system) is just a perfectly inelastic collision where the objects stick together. Any incorrect assumptions come from the OP, not from the "problem" or system.
$endgroup$
– Aaron Stevens
yesterday
2
2
$begingroup$
Momentum is conserved in collisions. Why would you assume external forces? It seems like you are the one making additional assumptions, not the OP.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
Momentum is conserved in collisions. Why would you assume external forces? It seems like you are the one making additional assumptions, not the OP.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@Aaron: I know what is momentum/energy conservation. I'm not making any additional assumptions and just using the conditions stated in the problem. If the problem is badly formulated this is another situation.
$endgroup$
– user3155984
yesterday
$begingroup$
@Aaron: I know what is momentum/energy conservation. I'm not making any additional assumptions and just using the conditions stated in the problem. If the problem is badly formulated this is another situation.
$endgroup$
– user3155984
yesterday
1
1
$begingroup$
The problem isn't poorly formulated. Just before and just after the collision momentum has to be conserved. If there are external forces then you just have to look just before and just after. Momentum will not be lost due to the collision, and it seems like the OP is interested in the collision, not unspecified external forces that aren't involved with the collision.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
The problem isn't poorly formulated. Just before and just after the collision momentum has to be conserved. If there are external forces then you just have to look just before and just after. Momentum will not be lost due to the collision, and it seems like the OP is interested in the collision, not unspecified external forces that aren't involved with the collision.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
@AaronStevens The problem is badly formulated in the sense that it imposes two assumptions that contradict each other (given that there are no external forces), namely "objects merge into one" and "all the energy gets converted into speed."
$endgroup$
– Andreas Blass
yesterday
$begingroup$
@AaronStevens The problem is badly formulated in the sense that it imposes two assumptions that contradict each other (given that there are no external forces), namely "objects merge into one" and "all the energy gets converted into speed."
$endgroup$
– Andreas Blass
yesterday
$begingroup$
No... The OP is incorrectly assuming that energy is conserved when the objects stick together. The "problem" (or you could say system) is just a perfectly inelastic collision where the objects stick together. Any incorrect assumptions come from the OP, not from the "problem" or system.
$endgroup$
– Aaron Stevens
yesterday
$begingroup$
No... The OP is incorrectly assuming that energy is conserved when the objects stick together. The "problem" (or you could say system) is just a perfectly inelastic collision where the objects stick together. Any incorrect assumptions come from the OP, not from the "problem" or system.
$endgroup$
– Aaron Stevens
yesterday
add a comment |
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3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
yesterday
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
yesterday
1
$begingroup$
@Jasper Not at all! If somebody doesn't understand why inelastic collisions (i.e. sticking together) don't conserve energy, the title edit doesn't change anything. All it does is make the answer easier to see for people who do know mechanics.
$endgroup$
– knzhou
yesterday
$begingroup$
Imagine the two objct being cars. When they collide, they change their outer form (alas!), and that takes up some energy you don't get back. That's where the "missing energy" from the correct momentum-based calculation goes into.
$endgroup$
– Ralf Kleberhoff
yesterday
2
$begingroup$
@RalfKleberhoff This is where the energy could go, but in general the objects to not need to deform for the collision to be perfectly inelastic.
$endgroup$
– Aaron Stevens
yesterday