# can anyone please explain? How is C the answer? isn't momentum always conserved?

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#3

(Original post by

**Bwp12**)
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#4

(Original post by

Well if the answer is C then that's an error from the mark scheme but, as the above user stated, the arrows representing the direction of the velocity should be of help. However, I would be interested to know why you initially chose the kinetic energy as not being conserved. Thanks!

**Mehrdad jafari**)Well if the answer is C then that's an error from the mark scheme but, as the above user stated, the arrows representing the direction of the velocity should be of help. However, I would be interested to know why you initially chose the kinetic energy as not being conserved. Thanks!

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#5

The conversion rule of momentum is correct, however the force that the momentum gives bounds back as energy into the driver therefore losing momentum, if im correct...I'm doing GCSE physics atm so XD

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#6

(Original post by

I'm not him, but I suspect that it would be because some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect. This is called inelastic collision.

**Rek'Sa**)I'm not him, but I suspect that it would be because some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect. This is called inelastic collision.

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#7

Momentum is not conserved because:

Taking moving right as positive

change in momentum

= change in velocity * mass

= (v -(-v))* mass

= -2v *mass

Taking moving right as positive

change in momentum

= change in velocity * mass

= (v -(-v))* mass

= -2v *mass

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#8

(Original post by

Momentum is not conserved because:

Taking moving right as positive

change in momentum

= change in velocity * mass

= (v -(-v))* mass

= -2v *mass

**lyamlim97**)Momentum is not conserved because:

Taking moving right as positive

change in momentum

= change in velocity * mass

= (v -(-v))* mass

= -2v *mass

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#9

(Original post by

**Bwp12**)KE is a scalar quantity, so it does not have direction. KE is maintained because KE = 1/2mv^2. Therefore, KE is constant.

Momentum has direction, as it depends on the sign of "v." Although the size of "v" is constant in the diagram, the direction isn't. Therefore, the momentum changes.

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#10

(Original post by

Yeah, that could have been the case only if the velocities were not both represented as V, because now it implies that the object was bounced with the same velocity, meaning that no energy was dissipated during the collision.

**Mehrdad jafari**)Yeah, that could have been the case only if the velocities were not both represented as V, because now it implies that the object was bounced with the same velocity, meaning that no energy was dissipated during the collision.

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#11

(Original post by

So wouldn't that lead to a bigger question? As the question states, v represents velocity, which is a vector quantity therefore have both magnitude and direction. If we were to take v mathematically, should v be -v in the second diagram or vice versa? So that means mathematically, velocity of the object is the same in both diagrams as v stays v and assuming the mass doesn't change, the momentum is conserved.

**Rek'Sa**)So wouldn't that lead to a bigger question? As the question states, v represents velocity, which is a vector quantity therefore have both magnitude and direction. If we were to take v mathematically, should v be -v in the second diagram or vice versa? So that means mathematically, velocity of the object is the same in both diagrams as v stays v and assuming the mass doesn't change, the momentum is conserved.

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#12

(Original post by

You have calculated the change in momentum (impulse) of the object and not momentum

**Mehrdad jafari**)You have calculated the change in momentum (impulse) of the object and not momentum

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#13

Momentum is conserved when there are no external forces. Here there is an external force (whatever force is holding the wall still), so it need not be conserved. You can also see that there is a change in momentum (goes from mv to -mv) so by definition it is not conserved.

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#14

**lyamlim97**)

Momentum is not conserved because:

Taking moving right as positive

change in momentum

= change in velocity * mass

= (v -(-v))* mass

= -2v *mass

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#15

(Original post by

I should admit that you are correct because the system is not in isolation as the wall is attached to the ground and so there is an external force acting on the object. Theoretically, even the momentum should be conserved as the theoretical rebound speed of the object will not be the same as its original (and that the earth will move in the direction of the original velocity of the object with a negligible speed), but here it's assumed that the speed of the object remains constant.

**Mehrdad jafari**)I should admit that you are correct because the system is not in isolation as the wall is attached to the ground and so there is an external force acting on the object. Theoretically, even the momentum should be conserved as the theoretical rebound speed of the object will not be the same as its original (and that the earth will move in the direction of the original velocity of the object with a negligible speed), but here it's assumed that the speed of the object remains constant.

**ball**is not conserved, not what property of this entire system is not conserved...

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#16

(Original post by

The momentum of the system of the ball, wall and earth is conserved, correct. However, the question asks what property of the

**studentro**)The momentum of the system of the ball, wall and earth is conserved, correct. However, the question asks what property of the

**ball**is not conserved, not what property of this entire system is not conserved...
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#17

(Original post by

You are correct, but, theoretically, even the momentum of the ball would be conserved as some of it's momentum would be transferred to the earth, in which case the change in momentum would be zero and not -2v.

**Mehrdad jafari**)You are correct, but, theoretically, even the momentum of the ball would be conserved as some of it's momentum would be transferred to the earth, in which case the change in momentum would be zero and not -2v.

**system of the ball and earth**is conserved. The momentum of the

**ball alone**is clearly not conserved - it changes...

Conservation of momentum really isn't a hard concept, but so many people seem to misunderstand it.

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#18

(Original post by

No... That means the momentum of the

Conservation of momentum really isn't a hard concept, but so many people seem to misunderstand it.

**studentro**)No... That means the momentum of the

**system of the ball and earth**is conserved. The momentum of the**ball alone**is clearly not conserved - it changes...Conservation of momentum really isn't a hard concept, but so many people seem to misunderstand it.

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#19

**Rek'Sa**)

I'm not him, but I suspect that it would be because some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect. This is called inelastic collision.

Where as fast forward to A-level and beyond the questions normally say something like "assume all collisions are elastic"

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#20

(Original post by

That's cool. I might not know the concept quite well, but would you say the same thing if the object was collided with a heaver object than itself in the absence of external forces? Why would the momentum of the ball not be conserved?

**Mehrdad jafari**)That's cool. I might not know the concept quite well, but would you say the same thing if the object was collided with a heaver object than itself in the absence of external forces? Why would the momentum of the ball not be conserved?

However, when talking about just the momentum of one object (the ball), we are implicitly saying that the system is of the ball alone, so any forces exerted by the other object are external forces. The momentum of the ball is not the same before and after the collision, hence momentum is not conserved in the system of the ball.

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