# Impulsive Tensions involving 3 particles

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1. Oct 2, 2015

### Physgeek64

1. The problem statement, all variables and given/known data
Please help; I've been attempting this question for days now and cannot get very far.... Im desperate now

Three identical particles X, Y, and Z of mass m are placed on a smooth horizontal
table. X is joined to Y and Z by light (can be thought of as massless) inextensible
strings XY and XZ. The angle XYZ is 60. An impulse I is applied to X in the
direction YX. The strings act as constraints so that the initial motions of Y
and Z must be the same as the components of the initial motion of X along YX
and ZX, respectively. Determine the initial velocities of the particles.

2. Relevant equations
So I managed to get

Vxcos(alpha)= Vy where alpha is the angle made by the horizontal and velocity of x
Vxsin(alpha)= Vzsin(theta)
I=2mVy+ mVzcos(theta)
Vxcos(theta + alpha) = Vz
3. The attempt at a solution

2. Oct 2, 2015

### TSny

I suspect that your equations are correct. But, you didn't specify the meaning of the angle theta in your equations. It would have been helpful if you had stated that you are taking the string between X and Y to be horizontal.

3. Oct 2, 2015

### Physgeek64

Oops- I forgot to mention that. Changing the orientation of it just makes the maths a bit easier I find. Do you have any advice as to where to go from here?

4. Oct 2, 2015

### Staff: Mentor

I would write out the 2nd law equations as if there were a time-dependent force F(t) acting on X , rather than an impulsive force. Then I would integrate as if the force F were very large and imposed for a very short time.

Chet

5. Oct 2, 2015

### TSny

OK, your equations look correct. You have four equations for four unknowns. So now it's algebra.
You might consider writing the equations in terms of the horizontal and vertical components of the velocity of X rather than the magnitude and direction of the velocity of X.

6. Oct 2, 2015

### haruspex

Did you mean YXZ?

7. Oct 3, 2015

### Physgeek64

No it's the angle xyz. If it were yxz I could do it :/

8. Oct 3, 2015

### haruspex

If the only known angle is XYZ then you do not have enough information. Consider the case where XZ is extremely long. The angle YXZ could be obtuse, meaning that Z will not immediately move (the string XZ will go slack). Conversely, if XZ=XY then Y and Z could be at the same point.

9. Oct 3, 2015

### Staff: Mentor

Yeah. The XYZ must be a typo.

Physgeek64: Let's see what you come up with if you solve it with YXZ. Now that's an interesting (and challenging) problem.

Chet

10. Oct 3, 2015

### Physgeek64

I've solved it if I let yxz be 60 degrees. My big problem is if xyz is 60 because, as you've said, resolving momentum in the Xz direction is near impossible

11. Oct 3, 2015

### haruspex

If the given angle is XYZ then, as I have shown, solving the question is not near impossible - it is completely impossible. It has to be a typo.
With YXZ as 60, you get fractions like 4/9 in the answer, yes?

12. Oct 3, 2015

### Physgeek64

I got answers such as 7/15

13. Oct 3, 2015

### Staff: Mentor

Can you please tell us what your results were for the x and y components of the three particle velocities. We would like to compare them with our results. Thanks.

14. Oct 3, 2015

### Staff: Mentor

I get 7/15 also.