Momentum

[AmirTimur]

---

Members don't see this ad.

Hi all I hope you can help me with this problem.

I pulled this from the wiki premed website. Please see attached picture.

I don't understand why the answer is not A. Vectors clearly support total momentum before = total momentum after. Since F ball 1 on 2 equals to F ball 2 on 1, then change in momentum of ball 1 = change in momentum of ball 2. Since ball 2 is less massive, it has a greater change in velocity, which is clearly indicated by the velocity vector.

The answer is : C <---- please highlight to see....

Does anyone have any idea about how to do this?

Thanks so much in advance!

 

[rjosh33]

Hi all I hope you can help me with this problem.

I pulled this from the wiki premed website. Please see attached picture.

I don't understand why the answer is not A. Vectors clearly support total momentum before = total momentum after. Since F ball 1 on 2 equals to F ball 2 on 1, then change in momentum of ball 1 = change in momentum of ball 2. Since ball 2 is less massive, it has a greater change in velocity, which is clearly indicated by the velocity vector.

The answer is : C <---- please highlight to see....

Does anyone have any idea about how to do this?

Thanks so much in advance!

Without getting into the details of vectors and forces, you can eliminate options A and B immediately. It says "a large mass collides elastically in one dimension with a smaller mass at rest." In an elastic collision of this sort, the only way for the large moving mass to bounce backwards (choice A) or stop in its tracks (choice B) is if it hits a stationary mass that is larger or identical to itself, respectively. But the moving mass hits a smaller mass, so we know it must continue moving to the right. This eliminates options A and B.

However, given that the smaller mass isn't negligible, the original moving mass will lose some of its forward velocity (and, hence, momentum). Option D shows the original moving mass retaining all of its original velocity, which is wrong. We can thus eliminate option D too. This leaves C, the choice in which the larger moving mass retains its rightward motion, albeit at a reduced velocity (and momentum), which is correct in this scenario.

 

[AmirTimur]

Without getting into the details of vectors and forces, you can eliminate options A and B immediately. It says "a large mass collides elastically in one dimension with a smaller mass at rest." In an elastic collision of this sort, the only way for the large moving mass to bounce backwards (choice A) or stop in its tracks (choice B) is if it hits a stationary mass that is larger or identical to itself, respectively. But the moving mass hits a smaller mass, so we know it must continue moving to the right. This eliminates options A and B.

However, given that the smaller mass isn't negligible, the original moving mass will lose some of its forward velocity (and, hence, momentum). Option D shows the original moving mass retaining all of its original velocity, which is wrong. We can thus eliminate option D too. This leaves C, the choice in which the larger moving mass retains its rightward motion, albeit at a reduced velocity (and momentum), which is correct in this scenario.

Thanks for your reasoning, it makes sense. I guess POE works really well here. If someone has an explanation with vectors and forces, I would be really grateful. I just want to make sure I really understand this concept

 

[milski]

Thanks for your reasoning, it makes sense. I guess POE works really well here. If someone has an explanation with vectors and forces, I would be really grateful. I just want to make sure I really understand this concept

You cannot get by only with vectors/forces/momentum - there is an unlimited number of states which have the same momentum as the initial state in the problem. The only way to resolve which one is the resulting state is by using the energies and taking into account if energy is lost, preserved or added during the collision.

Let's have an example with some numbers. If the heavy object has a mass 2m and velocity 2v before the collision and the small object has mass m and is stationary, v=0. The total momentum of the system is 2m*2v+m*0=4mv, total kinetic energy is 2m*(2v)^2/2=4mv^2

One possibility after a collision is to have the heavy ball move back to the left at velocity -v, that would give it a momentum of -2mv and the small ball will need to have a momentum of 6mv to preserve the total momentum or in other words will have to move at 6v to the right. The total energy of this new state is 2m*(-v)^2/2+m*(6v)^2/2=19mv^2. As you can see, for that to happen a lot of energy would have to be added to the system during the collision.

As you vary the change in energy, you are going to vary the outcome of the collision. The more energy you add, the faster will the balls run away from each other. If you want to write a general equation, it will be something like this:

m1*(V1f-V1i)+m2*(V2f-V2i)=0
m1(V1f^2-V1i^2)/2+m2(V1f^2-V1i^2)=&#916;E

Where m1 and m2 are the two masses, V1i and V2i the velocities before the collision and V1f and V2f after the collision.

The biggest takeway from this is already in what rjosh33 said:
In an elastic collision, for the moving object to bounce back, it has to be lighter than the stationary object.

 

[AmirTimur]

You cannot get by only with vectors/forces/momentum - there is an unlimited number of states which have the same momentum as the initial state in the problem. The only way to resolve which one is the resulting state is by using the energies and taking into account if energy is lost, preserved or added during the collision.

Let's have an example with some numbers. If the heavy object has a mass 2m and velocity 2v before the collision and the small object has mass m and is stationary, v=0. The total momentum of the system is 2m*2v+m*0=4mv, total kinetic energy is 2m*(2v)^2/2=4mv^2

One possibility after a collision is to have the heavy ball move back to the left at velocity -v, that would give it a momentum of -2mv and the small ball will need to have a momentum of 6mv to preserve the total momentum or in other words will have to move at 6v to the right. The total energy of this new state is 2m*(-v)^2/2+m*(6v)^2/2=19mv^2. As you can see, for that to happen a lot of energy would have to be added to the system during the collision.

As you vary the change in energy, you are going to vary the outcome of the collision. The more energy you add, the faster will the balls run away from each other. If you want to write a general equation, it will be something like this:

m1*(V1f-V1i)+m2*(V2f-V2i)=0
m1(V1f^2-V1i^2)/2+m2(V1f^2-V1i^2)=&#916;E

Where m1 and m2 are the two masses, V1i and V2i the velocities before the collision and V1f and V2f after the collision.

The biggest takeway from this is already in what rjosh33 said:
In an elastic collision, for the moving object to bounce back, it has to be lighter than the stationary object.

Thanks, milski. Now that you plugged in the numbers I can clearly see that KE is not conserved in choice A. I totally wouldn't see that have I not tried to plug in the numbers to see what's going on. Thanks!

 

[Gibbs]

You cannot get by only with vectors/forces/momentum - there is an unlimited number of states which have the same momentum as the initial state in the problem. The only way to resolve which one is the resulting state is by using the energies and taking into account if energy is lost, preserved or added during the collision.

Let's have an example with some numbers. If the heavy object has a mass 2m and velocity 2v before the collision and the small object has mass m and is stationary, v=0. The total momentum of the system is 2m*2v+m*0=4mv, total kinetic energy is 2m*(2v)^2/2=4mv^2

One possibility after a collision is to have the heavy ball move back to the left at velocity -v, that would give it a momentum of -2mv and the small ball will need to have a momentum of 6mv to preserve the total momentum or in other words will have to move at 6v to the right. The total energy of this new state is 2m*(-v)^2/2+m*(6v)^2/2=19mv^2. As you can see, for that to happen a lot of energy would have to be added to the system during the collision.

As you vary the change in energy, you are going to vary the outcome of the collision. The more energy you add, the faster will the balls run away from each other. If you want to write a general equation, it will be something like this:

m1*(V1f-V1i)+m2*(V2f-V2i)=0
m1(V1f^2-V1i^2)/2+m2(V1f^2-V1i^2)=&#916;E

Where m1 and m2 are the two masses, V1i and V2i the velocities before the collision and V1f and V2f after the collision.

The biggest takeway from this is already in what rjosh33 said:
In an elastic collision, for the moving object to bounce back, it has to be lighter than the stationary object.

Hmmm...
http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html#c5

Sent using SDN Mobile

 

[milski]

Hmmm...
http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html#c5

Sent using SDN Mobile

Yes, same thing there. Or have you spotted any disagreements?