Help with a momentum exchange please

In summary, the solution to the inelastic, collinear momentum exchange is finding the sum of the modulus of each speed.
  • #1
Quant ummm?
5
0
Hi there,

I am dealing with an inelastic, collinear momentum exchange of the form:

m1v1+m2v2=m1u1+m2u2

where m1 & m2 are known.

v1 & v2 are both 0

u1 & u2 are both unknown, however

MOD(u1)+MOD(u2) is known (ie the sum of modulus of each speed, I don't know how to do straight brackets here...)

which initial speed is regarded as + or - is irrelevant (to me).

I know that the exact speeds for u1 & u2 can be calculated, but I can't quite get me head around how (I'm more used to knowing one or the other, not their sum).

Although I can find the answer by gradually increasing one of the values on a spreadsheet, I'd like to see the actual solution. I'm guessing it can be solved either simultaneously or with a bit of calculus, but I'm not very good and working these things out.

Any help would be much appreciated.



p.s. I know this looks like homework, but its not. It really does though doesn't it. Real world though, honest.
 
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  • #2
Since v1 and v2 are both 0, you immediately have:
u1=-[m2/m1]u2

You the substitute into your expression for|u1| + |u2| to get |u1|

u1 = the result with a sign ambiguity.
 
  • #3
[tex]m_{1}u_{1}+m_{2}u_{2}=0[/tex] can be rearragned to [tex]\frac{u_{1}}{u_{2}}=-\frac{m_2}{m_1} [/tex]and taken the modulus we get [tex]\frac{|u_{1}|}{|u_{2}|}=\frac{m_{2}}{m_{1}}[/tex]
Now, Let's call [tex]|u_{1}|+|u_{2}|=x[/tex] where x is known, and if we divide by [tex]|u_{2}|[/tex] and rearrange we get [tex]\frac{|u_{1}|}{|u_{2}|}=\frac{x}{|u_{2}|}-1[/tex]
Therefore [tex]\frac{x}{|u_{2}|}-1=\frac{m_{2}}{m_{1}}[/tex] and rearragning for [tex]u_{2}[/tex] we get [tex]|u_{2}|=\frac{m_{1}}{m_{1}+m_{2}}x[/tex] and similarly for [tex]|u_{1}|=\frac{m_{2}}{m_{1}+m_{2}}x[/tex]
 
  • #4
That's brilliant, thank you very much for your help.
 

Related to Help with a momentum exchange please

1. What is momentum exchange?

Momentum exchange is a physical concept that describes the transfer of momentum between two objects or systems. It occurs when there is a force acting on one object that causes it to move and, as a result, causes the other object to move in the opposite direction with the same amount of momentum.

2. How is momentum exchange related to Newton's Third Law?

Newton's Third Law states that for every action, there is an equal and opposite reaction. Momentum exchange is directly related to this law as the force exerted on one object is the action, and the resulting movement of the other object is the reaction.

3. What are some examples of momentum exchange in everyday life?

Some examples of momentum exchange in everyday life include a billiard ball hitting another billiard ball, a car colliding with a wall, and a person pushing a shopping cart. In each of these scenarios, there is a transfer of momentum between two objects.

4. How does momentum exchange affect collisions?

Momentum exchange plays a crucial role in determining the outcome of collisions. When two objects collide, the momentum of the first object is transferred to the second object, causing it to move. The amount of momentum exchanged depends on the mass and velocity of the objects involved.

5. Can momentum exchange be manipulated or controlled?

Yes, momentum exchange can be manipulated or controlled by changing the mass or velocity of the objects involved. For example, in sports such as football or hockey, players can use their bodies to exert a greater force on the ball or puck, causing it to move in the desired direction with more momentum.

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