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 Mar 16 at 12:44
knzhou
45.4k11122219
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
$endgroup$
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 Mar 16 at 12:44
knzhou
45.4k11122219
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
$endgroup$
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
Mar 16 at 9:30
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
Mar 16 at 17:40
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
Mar 16 at 17:41
$begingroup$
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
Mar 16 at 18:03
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
Mar 16 at 19:34
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 Mar 16 at 12:44
knzhou
45.4k11122219
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
$endgroup$
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 Mar 16 at 12:44
knzhou
45.4k11122219
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
edited Mar 16 at 12:44
knzhou
45.4k11122219
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
edited Mar 16 at 12:44
knzhou
45.4k11122219
edited Mar 16 at 12:44
knzhou
45.4k11122219
edited Mar 16 at 12:44
knzhou
45.4k11122219
45.4k11122219
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
asked Mar 16 at 9:02
ToTheSpace 2ToTheSpace 2
251
251
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
Mar 16 at 9:30
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
Mar 16 at 17:40
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
Mar 16 at 17:41
$begingroup$
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
Mar 16 at 18:03
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
Mar 16 at 19:34
add a comment |
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
Mar 16 at 9:30
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
Mar 16 at 17:40
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
Mar 16 at 17:41
$begingroup$
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
Mar 16 at 18:03
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
Mar 16 at 19:34
3
3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
Mar 16 at 9:30
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
Mar 16 at 9:30
2
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
Mar 16 at 17:40
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
Mar 16 at 17:40
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
Mar 16 at 17:41
$begingroup$
@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
Mar 16 at 17:41
$begingroup$
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
Mar 16 at 18:03
$begingroup$
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
Mar 16 at 18:03
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
Mar 16 at 19:34
$begingroup$
@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
Mar 16 at 19:34
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
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 Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
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 Mar 16 at 10:09
JasperJasper
1,1031517
$endgroup$
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
Mar 16 at 19:36
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 Mar 16 at 9:56
Busy MinderBusy Minder
213
$endgroup$
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 Mar 16 at 15:08
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 Mar 16 at 19:24
user3155984user3155984
1
$endgroup$
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
Mar 16 at 19:31
$begingroup$
@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
Mar 16 at 19:39
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
Mar 16 at 19:42
$begingroup$
@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
Mar 17 at 1:46
$begingroup$
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
Mar 17 at 2:00
add a comment |
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5 Answers
5
active
oldest
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5 Answers
5
active
oldest
votes
active
oldest
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active
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$begingroup$
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 Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
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 Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
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 Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
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 Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
edited Mar 16 at 18:53
answered Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
answered Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
answered Mar 16 at 10:29
Aaron StevensAaron Stevens
13.5k42250
13.5k42250
add a comment |
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 Mar 16 at 10:09
JasperJasper
1,1031517
$endgroup$
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
Mar 16 at 19:36
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 Mar 16 at 10:09
JasperJasper
1,1031517
$endgroup$
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
Mar 16 at 19:36
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 Mar 16 at 10:09
JasperJasper
1,1031517
$endgroup$
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 Mar 16 at 10:09
JasperJasper
1,1031517
answered Mar 16 at 10:09
JasperJasper
1,1031517
answered Mar 16 at 10:09
JasperJasper
1,1031517
answered Mar 16 at 10:09
JasperJasper
1,1031517
1,1031517
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
Mar 16 at 19:36
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
Mar 16 at 19:36
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
Mar 16 at 19:36
$begingroup$
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
Mar 16 at 19:36
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 Mar 16 at 9:56
Busy MinderBusy Minder
213
$endgroup$
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 Mar 16 at 9:56
Busy MinderBusy Minder
213
$endgroup$
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 Mar 16 at 9:56
Busy MinderBusy Minder
213
$endgroup$
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 Mar 16 at 9:56
Busy MinderBusy Minder
213
answered Mar 16 at 9:56
Busy MinderBusy Minder
213
answered Mar 16 at 9:56
Busy MinderBusy Minder
213
answered Mar 16 at 9:56
Busy MinderBusy Minder
213
213
add a comment |
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 Mar 16 at 15:08
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 Mar 16 at 15:08
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 Mar 16 at 15:08
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 Mar 16 at 15:08
Neil_UKNeil_UK
1605
answered Mar 16 at 15:08
Neil_UKNeil_UK
1605
answered Mar 16 at 15:08
Neil_UKNeil_UK
1605
answered Mar 16 at 15:08
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 Mar 16 at 19:24
user3155984user3155984
1
$endgroup$
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
Mar 16 at 19:31
$begingroup$
@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
Mar 16 at 19:39
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
Mar 16 at 19:42
$begingroup$
@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
Mar 17 at 1:46
$begingroup$
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
Mar 17 at 2:00
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 Mar 16 at 19:24
user3155984user3155984
1
$endgroup$
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
Mar 16 at 19:31
$begingroup$
@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
Mar 16 at 19:39
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
Mar 16 at 19:42
$begingroup$
@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
Mar 17 at 1:46
$begingroup$
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
Mar 17 at 2:00
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 Mar 16 at 19:24
user3155984user3155984
1
$endgroup$
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 Mar 16 at 19:24
user3155984user3155984
1
answered Mar 16 at 19:24
user3155984user3155984
1
answered Mar 16 at 19:24
user3155984user3155984
1
answered Mar 16 at 19:24
user3155984user3155984
1
1
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
Mar 16 at 19:31
$begingroup$
@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
Mar 16 at 19:39
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
Mar 16 at 19:42
$begingroup$
@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
Mar 17 at 1:46
$begingroup$
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
Mar 17 at 2:00
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
Mar 16 at 19:31
$begingroup$
@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
Mar 16 at 19:39
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
Mar 16 at 19:42
$begingroup$
@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
Mar 17 at 1:46
$begingroup$
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
Mar 17 at 2:00
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
Mar 16 at 19:31
$begingroup$
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
Mar 16 at 19:31
$begingroup$
@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
Mar 16 at 19:39
$begingroup$
@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
Mar 16 at 19:39
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
Mar 16 at 19:42
$begingroup$
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
Mar 16 at 19:42
$begingroup$
@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
Mar 17 at 1:46
$begingroup$
@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
Mar 17 at 1:46
$begingroup$
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
Mar 17 at 2:00
$begingroup$
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
Mar 17 at 2:00
add a comment |
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3
$begingroup$
Energy is not conserved in your setup.
$endgroup$
– Jasper
Mar 16 at 9:30
2
$begingroup$
@knzhou With this edit, the answer is right in the title...
$endgroup$
– Jasper
Mar 16 at 17:40
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
Mar 16 at 17:41
$begingroup$
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
Mar 16 at 18:03
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
Mar 16 at 19:34