Elastic collision 2

1. May 17, 2014

Karol

1. The problem statement, all variables and given/known data
I made an example, two masses colliding elastically according to the drawing. i only know the direction of the first mass after collision (can i determine also the velocity?)
I try to solve but i go into difficulties.

2. Relevant equations
Conservation of momentum: $m_1v_1+m_2v_2=m_1u_1+m_2u_2$
Conservation of energy: $\frac{1}{2}mv^2=mgh$

3. The attempt at a solution
Conservation of momentum:
$4-3\cos 60^0=u_1\sin 30^0-u_2\sin\alpha$
$3\sin60^0=u_1\cos 30^0-u_2\cos\alpha$

Conservation of energy:
$\frac{1}{2}2\cdot 4+\frac{1}{2}1\cdot9=\frac{1}{2}2u_1^2+\frac{1}{2}1u_2^2$
$$\rightarrow u_2 = \sqrt{17-2u_1^2}$$
When i insert the last one into the previous ones i almost cannot solve.
It would be much easier if i were allowed to select in advance also the magnitude of u1

Attached Files:

• Snap1.jpg
File size:
11.2 KB
Views:
67
2. May 17, 2014

dauto

Double check the equation you provided for energy conservation. It makes no sense since there is no h in this problem

3. May 17, 2014

dauto

By almost cannot solve do you mean you solved it?

4. May 17, 2014

dauto

The equations for the momentum components are missing the masses, it seems.

5. May 17, 2014

paisiello2

Where did the 60° come from?

Last edited: May 17, 2014
6. May 17, 2014

vela

Staff Emeritus
The masses are missing from the righthand side of the momentum equations.

Try solving the two momentum equations for $u_2 \cos\alpha$ and $u_2 \sin\alpha$. Square and sum them to eliminate $\alpha$ and get $u_2^2$ in terms of $u_1$. Then plug that result into the energy equation.

7. May 19, 2014

Karol

I solved the equations:
$4-3\cos 60^0=2u_1\sin 30^0-u_2\sin\alpha$
$3\sin60^0=2u_1\cos 30^0-u_2\cos\alpha$
$\Rightarrow u_2^2=4u_1^2-11u_1+9.25$
I insert into this the result of the conservation of energy:
$$u_2 = \sqrt{17-2u_1^2}$$
And get a quadratic equation for u1 with the results: 2.38 and -0.54. i take only the positive result (right?) and get for u2 also close to 2.38 and for the angle $$\alpha=-2.89$$.
I checked the results of u1 and u2 with the conservation of momentum and energy and they agree, but when i checked the initial and final relative velocities:
$v_1-v_2 (x\ coordinate)=...=3.5$
$v_1-v_2 (y\ coordinate)=...=2.6$
$u_1-u_2 (x\ coordinate)=...=1.07$
$u_1-u_2 (y\ coordinate)=...=-0.32$
And i see that the equation for relative velocities before and after:
$\mathbf v_1-\mathbf v_2=-(\mathbf u_1-\mathbf u_2)$
Isn't fulfilled, and i posted a thread about it named Elastic Collision and i understood that also in 2 and 3-D it must be fulfilled, am i right?

8. May 19, 2014

vela

Staff Emeritus
It seems like you didn't read the rest of that other thread. That relationship doesn't hold in the 2D and 3D cases in general.

9. May 19, 2014

Karol

Thanks, i read but didn't understand for sure.
I was told that in the C.M. frame all collisions, including those from 3-D, are one dimensional. as i visualize it they are but the plane on which they occur changes. does in this case the formula hold? as i understand your answer it doesn't hold, it holds only for strictly one dimensional collisions, am i right?

Last edited: May 19, 2014
10. May 19, 2014

dauto

The correct formula is for the magnitude |v1 - v2| = |u1 - u2|. You miscalculated the components. I think you're missing some signs

11. May 22, 2014

vela

Staff Emeritus
In the CM frame, you can orient the coordinate system so that both objects approach each other along one axis. However, when they bounce off each other, they're no longer constrained to that axis. I think this is what you're describing, and you're correct. This would be a two-dimensional collision. The relative velocity doesn't simply change sign, as you verified in this specific case. That relationship between the relative velocities only holds for strictly one-dimensional collisions.