Attached: Solving Homework Equations

[c0der]

Homework Statement

Attached

Homework Equations

Attached

The Attempt at a Solution

Attached. Couple of questions/confirmations:

it should be vbcos(45)i - vbsin(45)j, but this is the same thing as cos(45)=sin(45)
vb should be directed at B
How do you solve equations 1 and 2? Trial and error?
vb/r will always be 5m/s in this case?

 

[lewando]

Try squaring both sides of 1 and 2 then add the two equations together.

 

[c0der]

Thanks! Then I get vb^2 = 29 + 20sin(theta).

Plugging sin(theta) = [ vb^2 - 29 ] / 20 into (2) I get two solutions for vb, one being negative. I hope I am correct in saying that this is the one to discard as the magnitude of a vector cannot be negative.

However in a previous problem, the relative velocity magnitude comes out negative in the solution, which doesn't make sense

 

[lewando]

Thanks! Then I get vb^2 = 29 + 20sin(theta).

Not following how you arrived at that. Squaring eq. 1 should give you:

(0.707v_b)² = 25cos²θ

No substitution into (eq. 2)² needed. Just add the 2 equations.

However you did it, did you at least get v_b = 6.21 m/s?

 

[c0der]

Yes I got vb = 6.21 or vb = -3.382 from the quadratic. I substituted because there are 2 unknowns there.

When I add the squares of the equations,

vb^2 (cos^2(45) + sin^2(45)) = 25 ( cos^2(theta) + sin^2(theta) ) + 4 + 20*sin(theta)

vb^2 = 29 + 20sin(theta), still 2 unknowns

 

[lewando]

I guess I should have been clearer.


Sorry, big EDIT:
Before squaring eq. 1, isolate the sinθ on the right-hand side.

Before squaring eq. 2, isolate the cosθ on the right-hand side.

Now you can square them.

Adding the 2 equations gives you sin²θ + cos²θ on the right hand side.

And everyone knows what that is equal to. Now you have a quadratic in v_b only.

 

[c0der]

Excellent, that still works out the same, so I take the positive magnitude.

In other problems such as 12-227, the magnitude of the relative velocity comes out negative. If this is not wrong, this means that the overall sense of direction of the vector needs to be reversed.

In this problem here, we get a quadratic with a negative velocity magnitude. I assume we discard this not because it's negative, but because it's smaller than vb/r as vb = vr + vb/r. Hope this is correct.