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July 7, 2015

July 7, 2015

Posted by **Maria** on Monday, February 25, 2008 at 6:43pm.

- Physics -
**drwls**, Monday, February 25, 2008 at 7:22pmThe center of mass travels to the right with velocity (13/23)*9 = 5.087 m/s

In a coordinate system moving with the center of mass, in an elastic collision, both masses merely change directions while keeping the same speed. The 10 kg object ends up with a speed, in earth-based coordinates, of

5.087 + 5.087 = 10.174 m/s

"Center of mass coordinate system" is the shortcut way of doing this kind of elastic oollision problem. You would get the answer by solving simultaneous equations of conservation of momentum and kinetic energy, with 2 unknowns, but it would take several more steps.

- Physics -
**Count Iblis**, Monday, February 25, 2008 at 7:45pmYou can solve this problem using conservation of energy and momentum. However, just writing down these equations and solving for the speed is a tedious way to solve collision problems.

Usually it is easier to transform to the "zero momentum frame" a.k.a. "center of mass frame". In this frame the total momentum is zero (the center of mass then doesn't move in this frame hence the name "center of mass frame").

Let's first look at the general problem from the perspective of an observer in the center of mass frame. The total momentum before the collsion is zero:

p1 + p2 = 0

The total momentum after the collision must thus also be zero:

p1' + p2' = 0

kinetic energy can be written as:

1/2 m v^2 = p^2/(2m).

The total kinetic energy before the collision is:

p1^2/(2 m1) + p2^2/(2 m2) =

(using p1 + p2 = 0)

p1^2 (1/(2 m1) + 1/(2m2))

The total kinetic energy after the collision is:

p1'^2/(2 m1) + p2'^2/(2 m2) =

(using p1' + p2' = 0)

p1'^2 (1/(2 m1) + 1/(2m2))

Equating the kinetic energy before and after the collision yields:

p1^2 = p1'^2 --->

p1 = p1' or p1 = - p1'

If p1 = p1' then the objects don't collide, they just move through each other. So, p1 = - p1' represents the collision.

So, in the center of mass frame, conservation of momentum and energy simply means that the momenta of the two objects change signs which means that the velocities change sign.

Since the momentum sign change in the center of mass frame is such a trivial thing, it is worthwhile to transform to this frame, change the sign of the momenta and then transform back to the original frame. These manipulations do not require much effort.

In this case, you consider an observer moving at velocity w to the right. We are going to choose w such that the total momentum in the frame of this observer is zero:

[13 (9 m/s - w) - 10 w ] kg = 0 --->

23 kg w = 117 kg m/s --->

w = 5.087 m/s

Now, since the 10 kg object was at rest in the original frame it is moving with velocity - w in the center of mass frame. After the collision the velocity changes sign, so the 10 kg object will be moving with velocity w in the center of mass frame. But that means that it will be moving with velocity 2 w = 10.17 m/s in the original frame.