Conservation of momentum - with understanding

[mathchimp]

I just want to state that i DID solve the problem. I just seek understanding of it.

I'd be really grateful if someone could answer two of my questions at the end of this post. The problem I've solved here is just to show what I'm dealing with.

1. Homework Statement

Object 1 is moving towards object 2 with a velocity of v1=10m/s.
Object 2 is standing still.

Mass of object 1 is m1=4kg, and for object 2 it's m2=2kg.

If all kinetic energy is conserved after collision, determine v1f and v2f (final velocities).

Homework Equations

So far, what I know:

CONSERVATION OF MOMENTUM:
m1*v1+m2*v2 = m1*v1f+m2*v2f

CONSERVATION OF KINETIC ENERGY (in this case all of it is conserved)
1/2*(m1*v1^2)+1/2*(m2*v2^2) = 1/2*(m1*v1f^2)+1/2*(m2*v2f^2)

The Attempt at a Solution

What first I did was use the equation for conservation of momentum:

m1*v1+m2*v2 = m1*v1f+m2*v2f

Using this equation, and assuming that all kinetic energy was conserved, using

1/2*(m1*v1^2)+1/2*(m2*v2^2) = 1/2*(m1*v1f^2)+1/2*(m2*v2f^2)

I get two equations, two unknowns, and end up with:

2*1v1f+v2f = 20
2*v1f^2 + v2f^2 = 200

I plugged v2f into the second equation and ended up with 2 solutions:

v1f = 10m/s or 3.33 m/s

In which case I get

v2f = 0 or 13.34 m/s

There are 2 things that confuse me:

1) How do I actually pick the right velocity? In this case, v1f=3.33 and v2f=13.34 makes more sense to me (and it is the right answer), but I'm unsure how to explain it "physically". Is a good way to do this to just plug the solutions into the formula for coefficient of restitution and see whether it makes sense? I get 1 for the right solution here and -1 for the other one, can I conclude the other one isn't possible because they wouldn't actually hit each other (and e=1 means it's an elastic collision)?

2)In our physics class, the teacher encourages us to use the conservation of momentum formula which would, if these objects were moving one towards another, look like this:

m1*v1-m2*v2 = -(m1*v1f+m2*v2)f

Does it matter if I use the equation above or the other one I've solved my problem with?

I've actually done quite a number of these problems and it always somehow turns out to be unimportant. Obviously, if one object is moving to the left, its velocity is negative, so I'd just plug in a negative value, but in that case I'm getting something like:

m1*v1-m2*v2 = m1*v1f+m2*v2f

Which is, well... a different equation. They usually don't trouble us with getting the direction of these objects in the end (negative or positive velocities, doesn't matter) but I'm wondering if both would provide me with correct (identical) answers.

 

[Orodruin]

1) If the objects do not hit each other, then both energy and momentum is conserved and they are left with the same velocities as they had from the beginning. Kinematically, this will always be allowed. Now, you are looking at a one-dimensional problem, but if you considered more than one dimension you would kinematically obtain an entire family of allowed solutions (that can be parametrised by the scattering angle). There is no way of telling which of these will be the actual result unless you obtain more information about the collision (such as the scattering angle).

2) I would strongly discourage that use. In general, using velocity and velocity components is much less confusing than using speeds.

 

[Stephen Tashi]

It's clearer to use subscripts to denote different variables - e.g. "$mv_{1f}$" instead of $mv1f$.

but I'm unsure how to explain it "physically". Is a good way to do this to just plug the solutions into the formula for coefficient of restitution and see whether it makes sense? I get 1 for the right solution here and -1 for the other one, can I conclude the other one isn't possible because they wouldn't actually hit each other (and e=1 means it's an elastic collision)?

It seems to me you are asking how to explain the answer "mathematically". As I see it, the physical explanation comes from literary interpretation of the problem. The mention of "collision" and our common experience of collisions indicates that the objects would undergo a change in velocity.

Consider a more complicated problem where you are given the velocities and masses of several objects and then given some of their velocities "20 seconds later" and asked to find the velocities that are not given. Such a problem does not specify which objects may have collided. If you are not given the initial positions of the objects, you cannot compute which objects have collided. So there is no physical information that let's you discard solutions where some of the objects keep the same velocity.

Does it matter if I use the equation above or the other one I've solved my problem with?

There are two different styles to setting up equations involving directed quantities.

The "human friendly" style is to draw arrows (vectors) in different directions (to hint how things are actually moving or accelerating or pushing etc.) and to use minus signs in equations. The difficulty with that style is that sometimes the directions of the arrows representing the answers are unknown. You have to keep in mind that if you get a negative answer, it indicates the arrow you drew for the answer points the other way. In the problem you solved, this style could give you expressions like $m_1v_1 - m_2v_2$ for total momentum and you would represent all given velocities as positive numbers.

The theoretical style is to treat all arrows as pointing in the "positive" direction. In the case of a 1-dimensonal problem all arrows point to the right. In the case of a 2-dimensional problem, an arrow is drawn so its x and y components point in the positive directions. When using this style, diagrams are often omitted because they would be confusing. We just write equations. In your problem, this style represents total momentum with expressions like $m_1v_1 + m_2v_2$ and you represent given velocities using both positive and negative numbers.

but I'm wondering if both would provide me with correct (identical) answers.

Yes, both styles provide identical answers if done correctly - and both styles can be confusing! For example, in the theoretical style, we can still have equations with minus signs in them. If we know two things act "in opposite directions" we may need an equation with a minus sign in it. This highlights the distinction between the operation of subtraction "minus" versus the concept of a negative quantity.