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:
And so the analysis continues.
So I am trying to set up a similar pair of diagrams for the following question:
Here's my impulse diagram:
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.
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?