Momentum Conservation in Elastic Collisions

[ShazK]

Hey, I was looking for a little bit of clarification. Nothing big, but here's the problem...

I have a diagram of two objects colliding at an angle, I created a timescale and defined velocities for each object. The collision is fully elastic. I was given the mass of one object, and need to solve for the other...

My question was... do I need to determine the components of each object's momentum before and after and use two separate equations for momentum, one of x and one of y momentum, or can I simply use the velocities and say that total momentum is conserved? And why?

 

[xanthym]

Momentum is a vector quantity. Thus, in general, it's easiest to resolve the initial & final total (system) Momentum vectors into orthogonal components ("x", "y", & "z" if necessary). Then you can equate the final Momentum vector components to the initial Momentum vector components on a component-by-component basis. (This assumes only internal system forces affect the entities.)

If the collision is also completely elastic, then final total Kinetic Energy will equal initial total Kinetic Energy. Kinetic Energy is a scalar quantity involving only the square of the magnitude of the various velocity vectors. Thus, for elastic collisions, the Kinetic Energy conservation portion does not require vector components (only velocity magnitudes).

Usually elastic collision problems are structured so you need every bit of info you can get about initial and final conditions. Hence, be prepared to used both sets of conditions, one vector & the other scalar, to develop the equations necessary for problem solution.


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[ShazK]

Here's something odd, I used just the x components and solved for m and got .2, but for some reason, when I just use KE I get m to be .01.. strange.

 

[xanthym]

Here's something odd, I used just the x components and solved for m and got .2, but for some reason, when I just use KE I get m to be .01.. strange.

Remember, if initial Momentum has both "x" and "y" components, both these components must be used in determining final momentum. The situation would be clearer if you state the actual problem and your work here.


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[ShazK]

Lets see, we had a diagram, from which I determined the velocities of the two pucks, both before and after their collisions. Dark puck before .813 and after .825 (unit doesn't matter in this case since its a scale). The light puck before is 1.175 and after .723.

Dark puck's mass is .4kg

What I did with KE was use 1/2mv^2 for the white and dark pucks before and add them, and set that equal to the sum of the KE's of the pucks after and solved for mass of the light puck and got .1.

When I used the x-component, I measure the angle and determined the x components of each velocity, and used those to have the sum of the x momentum before the collision = the sum of the x momentum after the collision, and solved for mass of the light puck and got .2. The angles: Dark puck before 0, after 45, light before 77, after 3, all towards the right and off of the x-axis.

 

[xanthym]

Lets see, we had a diagram, from which I determined the velocities of the two pucks, both before and after their collisions. Dark puck before .813 and after .825 (unit doesn't matter in this case since its a scale). The light puck before is 1.175 and after .723.

Dark puck's mass is .4kg

What I did with KE was use 1/2mv^2 for the white and dark pucks before and add them, and set that equal to the sum of the KE's of the pucks after and solved for mass of the light puck and got .1.

Your Kinetic Energy calculations don't seem correct. See below. (Of course, the following is valid because problem specifies collision to be ELASTIC.)
{D Speed Before} = (0.813)
{D Speed After} = (0.825)
{L Speed Before} = (1.175)
{L Speed After} = (0.723)
{D Mass) = (0.4 kg)
{L Mass} = M

{Initial Kinetic Energy} = (1/2)(0.4)(0.813)^2 + (1/2)(M)(1.175)^2
{Final Kinetic Energy} = (1/2)(0.4)(0.825)^2 + (1/2)(M)(0.723)^2

{Initial Kinetic Energy} = {Final Kinetic Energy}
::: ⇒ (1/2)(0.4)(0.813)^2 + (1/2)(M)(1.175)^2 = (1/2)(0.4)(0.825)^2 + (1/2)(M)(0.723)^2
::: ⇒ (M)(1.175)^2 - (M)(0.723)^2 = (0.4)(0.825)^2 - (0.4)(0.813)^2
::: ⇒ (M)(0.858) = {7.86e(-3)}
::: ⇒ (M) = {9.16e(-3) kg}
Above "M" value doesn't match your calculation of "M". Did you make a mistake copying some data??


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[ShazK]

I made a mistake in copying my answer in that case... I did get the same M value as you... so does that mean my second method was incorrect, and the KE one is correct? My velocities seem to be correct as well.

 

[xanthym]

I made a mistake in copying my answer in that case... I did get the same M value as you... so does that mean my second method was incorrect, and the KE one is correct? My velocities seem to be correct as well.

If the collision is indeed ELASTIC (which you say the problem indicates), results obtained from conservation of Kinetic Energy should be consistent with those obtained from conservation of Momentum. If they are not, then something within the problem itself is bad. This would not be the first time "the book" is wrong. Unfortunately, unless there's further information, not much more can be done here. Suggest you bring this to your teacher's attention.


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