Momentum Problem: Find Velocities After Collision (2M, M, 5v, 3v)

[Gregg]

Homework Statement

Two particles, A, of mass 2M, and B, of mass M , whilst moving on a smooth horizontal
table in opposite directions with speeds of 5v and 3v respectively, collide directly.

Find their velocities after the collision in terms of v and the coefficient of
restitution e.

The Attempt at a Solution

Since I am not given the co-efficient of restitution. It seems that all I have is the conservation of momentum equation. An equation with 2 unknowns, how is it possible to get the velocities in terms of v?

 

[Filip Larsen]

You have two unknowns, speed of A and B after collision, and two knowns, the restitution coefficient and the (reference) speed v. To relate these, you need one equation for conservation of momentum and one equation for definition of the restitution coefficient.

 

[Gregg]

...Does the question ask me to give the velocities in terms of e? Not find e, after the velocities?

 

[Filip Larsen]

...Does the question ask me to give the velocities in terms of e? Not find e, after the velocities?

Yes, you should find the velocities of A and B after collision as a function of the known (but unspecified) values v and e.

 

[Gregg]

ok

e = \frac{v_2 - v_1}{8v}

26v=4v_1+2v_2

v_1 = \frac{13v-8ve}{3}

v_2 = \frac{13v+16ve}{3}

I think that's it.

Next part:

b) Find the value of e for which the speed of B after the collision is 3v.

By conservation of momentum:

v_1 = 5v

e = \frac{v_2 - v_1}{u_1 - u_2}

e = 1 ?

I'm not entirely sure about the signs for e, in some ways which I did it I get negative e. Which is wrong. What can I assume about the direction of the particles, is v_1 = -5v? With e as 1, is the loss in K.E. 0?

 

[Filip Larsen]

26v=4v_1+2v_2

This does not look right. Check your signs for the left hand side of that equation.

What can I assume about the direction of the particles, is v_1 = -5v?

From the two equations in question alone, you cannot in general determine the direction of travel of each mass after the collision, except in the special case where e is exactly zero and both masses moves as one after the collision. But this should not be a problem as long as you keep it simple for yourself and use a consistent direction as the "positive" direction in all equations. For instance, if mass A initially moves with 5v, the velocity of mass B should be modeled as -3v, that is, reverse sign relative to A.

With e as 1, is the loss in K.E. 0?

Yes, that is a fully elastic collision.

 

[Benson]

Hi,

I have a similar question but with actual values of e and initial velocities given.

The part I don't understand is, strictly speaking, e = |v2-v1|/|u2-u1|. If I know the relative velocity before collision, and I know e, then |v2-v1| = e*|u2-u1|. So, for example, in my quesion, e=0.4, u1 = 2m/s to the right, and u2 = 5m/s to the left. Defining right as positive, that gives me a relative velocity before collision of 7m/s, and thus |v2-v1|=0.4*7 = 2.8. Then how do I know whether v2 = v1+2.8, or v1 = v2 +2.8?

With m1 = 2kg and m2 = 4kg, both cases give yield reasonable answers (if v2 = v1+2.8, v1 = 0.8m/s left, and v2 = 3.6m/s left; if v1 = v2 +2.8, v1 = 4.56m/s left, and v2 = 1.73m/s left).

How do I know which one to choose?

Thanks,
Benson

 

[Filip Larsen]

That is a good question. The equations alone allow both solutions, so to pick one you will have to consider the geometry of the collision, which is easiest to picture if you consider the situation in the so-called centre of mass coordinate system where the total momentum is zero.

Assume that in the centre of mass coordinate system m1 moves right and m2 moves left before the collision. After the collision we have two solutions corresponding to the two solution you mention: either m1 moves left and m2 moves right, or m1 moves right and m2 moves left. The two situations therefore corresponds to whether the masses hit each other and reverse direction or if they just pass each other without reversing direction.

In collisions between simple rigid bodies I believe the solution that reverses the direction of each mass in the centre of mass system is the solution most naturally occurring. For the masses to pass each other would mean they would have to have some kind of sticky surface that allow them to sling each other around a bit or that they are allowed to pass through each other in some way.

 

[Benson]

Filip,

What you're saying about the two cases makes sense, but both of my solutions have both masses going left. So neither case has the masses passing through each other. Also, both have the mass that was originally moving left slowing down somewhat; one case more so than the other, so there is no illogical solution where mass continues traveling in the same direction and speeds up. So I still don't know how to separate the solutions.

Benson

 

[Filip Larsen]

The condition of going left or right is relative to each other, which is why I was referring to the coordinate system where the centre of mass is at rest.

In your case you have two solutions. First one has v1=-0.8 and v2=-2.8, so here m2 is moving faster to the left than m1 is, thus m1 is moving right relative to m2. Second solution has v1=-4.53 and v2=-1.73 so here m1 is moving faster left than m2, thus m1 is moving left relative to m2. Originally, m1 was moving right relative to m2, so the second solution is the solution where the masses reverse direction and the first is where they pass each other.

If you transform the speeds to the coordinate system of the centre of mass (where the momentum is zero; in your case add 2.67 m/s to each speed), you will find that the two solutions end up as plus/minus some value for each mass, i.e. as seen from the centre of mass, the two masses will move in opposite directions of each other, which, since they have to move either left or right in a one-dimensional model, means two solutions. If the masses could move in two dimensions, we would (with the same information as in your problem) have infinitely many solutions where the masses moved in opposite directions.