2d Elastic collision concept and formula

[sl235]

Would the 2 items displayed in the image below act as I expect in a zero G environment?

Assuming that item A had a mass of 100kg and was stationary, Items B have a mass of 1kg and travel at 10m/s.

what formula would I need to use to calculate the velocities and vectors of the items after the collision?

Thanks

 

[mathman]

The basic equations would be conservation of energy and conservation of momentum.

 

[mfb]

+ conservation of angular momentum
+ momentum transfer occurs perpendicular to the contact line.
The B-items will keep a small fraction of their horizontal movement. This is easier to see with a very small mass A: They would just push it away.

 

[sl235]

thanks for the replies. Mfb, would the lighter part A move upward at a 90 degree angle as depicted?

 

[voko]

Due to symmetry, balls B will move identically after the collision. So you could simplify things by assuming an imaginary ball of mass m = 2kg moving upward at 10 m/s instead of the two balls. Conservation of momentum and energy gives you two equations, and you have two unknowns: velocity of A and velocity of the imaginary ball after the collision.

 

[sl235]

Thanks Voko, again I appologise for my lack of understanding, but from what you are saying I understand that the fact that the 2 balls striking at a 45 degree angle is not significant?

 

[voko]

The 45 degree angle is very significant: it ensures that the balls will then move downward, which is why they can be replaced with the imaginary ball I mentioned.

 

[mfb]

Mfb, would the lighter part A move upward at a 90 degree angle as depicted?

It would (for any mass), due to symmetry.

After taking symmetry into account, you have 3 unkown values after the collision:
1) upwards velocity of A
2) downwards velocity of Bs (same value for both B)
3) sidewards velocity of Bs (same magnitude, opposite direction)

Momentum conservation in vertical direction gives you a relation between 1 and 2, energy conservation a relation between all 3. This leaves one degree of freedom, which corresponds to the angle at the impact points: the transferred momentum in vertical and horizontal direction has to be the same, this gives a third equation and allows to solve the system.@voko: I can't see how a ball moving upwards would give any similar system.

 

[voko]

@voko: I can't see how a ball moving upwards would give any similar system.

And I can't see how the balls could retain any horizontal velocity. The system is modeled to undergo an instantaneous absolutely elastic collision. Which means the normal momentum is reflected, and the tangential momentum is preserved. Which means the angle of reflection = angle of incidence, which in this case means the reflected motion is strictly downward.

 

[voko]

To supplement that. In a less than fully elastic collision, there will be some horizontal velocity. But there is not enough information to model that.

 

[mfb]

And I can't see how the balls could retain any horizontal velocity. The system is modeled to undergo an instantaneous absolutely elastic collision. Which means the normal momentum is reflected, and the tangential momentum is preserved.

The wall has a finite mass, and will move a bit after the collision. Therefore, the transferred momentum is smaller - the downwards movement will be slower than the initial velocity (should be obvious from momentum conservation) and some horizontal movement remains.