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can anyone please help with the setup of a question.

Firstly, here is a similar question just to show the style of the solution expected:

enter image description here

And so the analysis continues.

So I am trying to set up a similar pair of diagrams for the following question:

enter image description here

Here's my impulse diagram:

enter image description here

But I can't think how to draw the velocity diagram. I was initially going to show B moving with a velocity u straight down but when I look at the answer to the problem the velocity of B is given as:

$\frac{2J\sqrt{21}}{15m}$ at $\arctan{3\sqrt{3}}$ to AB

And so my diagram with B heading straight down would have been wrong.

Thanks for any help, Mitch.

gnitsuk
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  • Did you pose this question on Physics SE? – John Aug 10 '18 at 15:58
  • No I didn't. Previously I have received very good help here with physics questions. I'll wait a while and see if any folks can help. Thanks. – gnitsuk Aug 10 '18 at 16:04
  • Yeah, we're a pretty helpful bunch over here. It just looks like setting up the problem requires physics more than math. The math is straightforward after setting up the problem, I think. – John Aug 10 '18 at 17:19

1 Answers1

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For Particle A:

$$J_2 = mv_a$$

For Particle B in the horizontal direction:

$$J_1 \cos 60^{\circ} - J_2 = J_1/2 - J_2 = mv_{b,x}$$

And in the vertical direction:

$$J_1 \sin 60^{\circ} - J = (\sqrt{3}J_1/2) - J = mv_{b,y}$$

For Particle C along the direction $CB$:

$$J_1 = mv_c$$

Following that, you probably have some statements about conservation of momentum of the whole system in the $x$ and $y$ directions, which will give you two more equations, allowing you to solve for the six unknowns $v_a, v_{b,x}, v_{b,y}, v_c, J_1$, and $J_2$, in terms of $m$ and $J$.

Can you try from here?

John
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    Thanks for your reply. It did indeed help, and I was able to solve the problem - and subsequent ones also. Thanks, Mitch. – gnitsuk Aug 17 '18 at 15:51