# Impulse/momentum problem

1. Jun 8, 2004

### hobo

I have no clue on where to start on this. I have spent time playing with that cart thing and the momemtum balls and I have ideas floating around, but nothing concrete.

There is a 248.2kg Soviet satellite travelling due east at a velocity of 1025km/hr and a 149.9kg US satellite travelling due west at a velocity of 1181.8km/hr. The colission is perfectly elastic. What happens afterwards?

2. Jun 8, 2004

### arildno

Use conservation of momentum (no external forces) and conservation of energy (fully elastic collision)

3. Jun 9, 2004

### hobo

From what you told me, I used

m1v1 + m2v2 = m1v1' + m2v2'
(mv1^2)/2 + (mv2^2)/2 = (mv1'^2)/2 + (mv2'^2)/2

I solved both for V1' then I plugged in the numbers.After that, it came out to

5364792208 - 35930354.61v2' +59675.19v2'^2

I then plugged it into the quadratic equation, getting 328.08333m/s which is v2, and 274.0153711m/s, which is not the answer my teacher gave me

4. Jun 9, 2004

### arildno

Did you take into account that the velocities are in different directions?

5. Jun 9, 2004

### arildno

Now, let's solve this problem in full:
I will assume that you gave me the right information, in particular that the collision was fully ELASTIC.

The equations for two objects, 1 and 2, to be solved are:
$$m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}$$

This is conservation of momentum, as you had; b and a refer to initial and final velocities (before and after collision).

Energy conservation:
$$\frac{1}{2}m_{1}v_{1,a}^{2}+\frac{1}{2}m_{2}v_{2,a}^{2}=\frac{1}{2}m_{1}v_{1,b}^{2}+\frac{1}{2}m_{2}v_{2,b}^{2}$$

This is also, I believe, what you meant.
Note that the energy equation can be simplified in the following manner:
$$m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a})+m_{2}(v_{2,b}-v_{2,a})(v_{2,b}+v_{2,a})=0$$

Rewrite the momentum equation as:
$$m_{2}(v_{2,b}-v_{2,a})=-m_{1}(v_{1,b}-v_{1,a})$$

Hence, the energy equation may be written as:
$$m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a}-(v_{2,b}+v_{2,a}))=0$$

The root $$v_{1,b}=v_{1,a}$$ corresponds to no collision, so we have the following system to solve:
$$v_{1,b}+v_{1,a}=v_{2,b}+v_{2,a}$$
$$m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}$$

Solving for $$v_{1,a},v_{2,a}$$ yields:
$$v_{1,a}=\frac{(m_{1}-m_{2})v_{1,b}+2m_{2}v_{2,b}}{m_{1}+m_{2}}$$
$$v_{2,a}=\frac{(m_{2}-m_{1})v_{2,b}+2m_{1}v_{1,b}}{m_{1}+m_{2}}$$

Now, we have simplified the symbolic expressions maximally; it is time to enter in the values!
We set:
$$m_{1}=149.9,v_{1,b}=1181.8$$
$$m_{2}=248.2,v_{2,b}=-1025$$

The answers should be quite different from the ones you gave