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

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#1
1
5 years ago
#2
(Original post by Bwp12)
I believe the arrows give you the hint required.
0
5 years ago
#3
(Original post by Bwp12)
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!
0
5 years ago
#4
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!
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.
0
5 years ago
#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
0
5 years ago
#6
(Original post by 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.
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.
0
5 years ago
#7
Momentum is not conserved because:
Taking moving right as positive
change in momentum
= change in velocity * mass
= (v -(-v))* mass
= -2v *mass
3
5 years ago
#8
(Original post by lyamlim97)
Momentum is not conserved because:
Taking moving right as positive
change in momentum
= change in velocity * mass
= (v -(-v))* mass
= -2v *mass
You have calculated the change in momentum (impulse) of the object and not momentum 2
5 years ago
#9
(Original post by Bwp12)
Speed remains contant, as v = v HOWEVER the velocity changes as there is a change in direction. Mass doesn't change either.

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.
1
5 years ago
#10
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.
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, shouldn't 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.
1
5 years ago
#11
(Original post by 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.
Correct, but the direction of the velocities are represented by the arrows .
0
5 years ago
#12
You have calculated the change in momentum (impulse) of the object and not momentum Exactly. There is a change in momentum. Hence momentum is not conserved
0
5 years ago
#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.
1
5 years ago
#14
(Original post by lyamlim97)
Momentum is not conserved because:
Taking moving right as positive
change in momentum
= change in velocity * mass
= (v -(-v))* mass
= -2v *mass
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. 0
5 years ago
#15
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. 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...
0
5 years ago
#16
(Original post by 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...
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.
0
5 years ago
#17
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.
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.
0
5 years ago
#18
(Original post by 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.
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?
0
5 years ago
#19
(Original post by 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.
Yeah but this looks like a GCSE question. They will only be taught about elastic collisions (although they will never be specifically told this). So any kid that has any kind of practical awareness of say, a bouncing tennis ball, will know things don;t bounce forever and make the link that connective energy is not conserved. But stop thinking about things! No marks.

Where as fast forward to A-level and beyond the questions normally say something like "assume all collisions are elastic"
0
5 years ago
#20
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?
The momentum of the system of the two objects would be conserved - the sum of the momenta of the objects before the collision would equal the sum of the momenta of the objects after the collision so there is no net change in momentum.

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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