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March 14th, 2017, 01:26 PM   #1
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A bomb suddenly explodes. find velocity of a piece

I'm sure I did this right but I can't take any chances.... not after last test 78.59% grade. So could someone verify I did this correctly?




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March 14th, 2017, 02:20 PM   #2
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it's tough to say but I don't think you've done this correctly

You have 2 equations, 1 for each cardinal direction

Assume the initial mass is 3 so each piece has mass 1.

In the x direction

$1 + 20\cos(30^\circ) + v \cos(\theta) = 0$

In the y direction

$20 \sin(30^\circ) + v \sin(\theta) = 0$

So

$1 + 10\sqrt{3} + v \cos(\theta) = 0$

$v \cos(\theta) = -(1+10\sqrt{3})$

$v \sin(\theta) = -10$

Dividing

$\tan(\theta) = \dfrac{10}{1+10\sqrt{3}}$

$\theta = \arctan\left( \dfrac{10}{1+10\sqrt{3}}\right) \approx (28.63^\circ, 208.63^\circ)$

Common sense tells us that $208.63^\circ$ must be the correct answer if momentum is to be conserved.

Then $v \sin(208.63^\circ) = -10$

$v \approx 20.87$
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March 14th, 2017, 02:22 PM   #3
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the statement, "One piece flies out along the positive x-axis at a speed of 1 m/s." is crystal clear, however, what does "another flies in at $30^\circ$ to the x direction" mean?

Is there an a diagram given for this problem?

see attached for two ways one could interpret the statement about the 20 m/s piece ... magnitude of the 3rd piece's velocity will be the same in either case, but directions will be different.
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March 14th, 2017, 02:37 PM   #4
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Quote:
Originally Posted by skeeter View Post
the statement, "One piece flies out along the positive x-axis at a speed of 1 m/s." is crystal clear, however, what does "another flies in at $30^\circ$ to the x direction" mean?

Is there an a diagram given for this problem?

see attached for two ways one could interpret the statement about the 20 m/s piece ... magnitude of the 3rd piece's velocity will be the same in either case, but directions will be different.
No graph was given. I'll make an assumption,

Quote:
Originally Posted by romsek View Post
it's tough to say but I don't think you've done this correctly

You have 2 equations, 1 for each cardinal direction

Assume the initial mass is 3 so each piece has mass 1.

In the x direction

$1 + 20\cos(30^\circ) + v \cos(\theta) = 0$

In the y direction

$20 \sin(30^\circ) + v \sin(\theta) = 0$

So

$1 + 10\sqrt{3} + v \cos(\theta) = 0$

$v \cos(\theta) = -(1+10\sqrt{3})$

$v \sin(\theta) = -10$

Dividing

$\tan(\theta) = \dfrac{10}{1+10\sqrt{3}}$

$\theta = \arctan\left( \dfrac{10}{1+10\sqrt{3}}\right) \approx (28.63^\circ, 208.63^\circ)$

Common sense tells us that $208.63^\circ$ must be the correct answer if momentum is to be conserved.

Then $v \sin(208.63^\circ) = -10$

$v \approx 20.87$
I do see exactly how you did everything except how did you convert 20cos(30) = 10\sqrt{3}. I wanna learn how to do that too. What's the name of the technique?
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March 14th, 2017, 03:02 PM   #5
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Quote:
Originally Posted by The_Ys_Guy View Post
I do see exactly how you did everything except how did you convert 20cos(30) = 10\sqrt{3}. I wanna learn how to do that too. What's the name of the technique?
romsek knows the unit circle, i.e. $\cos(30^\circ) = \dfrac{\sqrt{3}}{2}$, so $20\cos(30^\circ)=20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$

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March 14th, 2017, 03:16 PM   #6
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Quote:
Originally Posted by skeeter View Post
the statement, "One piece flies out along the positive x-axis at a speed of 1 m/s." is crystal clear, however, what does "another flies in at $30^\circ$ to the x direction" mean?

Is there an a diagram given for this problem?

see attached for two ways one could interpret the statement about the 20 m/s piece ... magnitude of the 3rd piece's velocity will be the same in either case, but directions will be different.
I assumed it to be positive $30^\circ$.

I would have called the other direction "another flies in at $-30^\circ$ to the x direction".
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March 14th, 2017, 03:53 PM   #7
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Quote:
Originally Posted by romsek View Post
I assumed it to be positive $30^\circ$.

I would have called the other direction "another flies in at $-30^\circ$ to the x direction".
makes sense ... so does "in" mean the vector points in from quad 1 to the origin, making it actually point in the direction of quad 3? ... if the word "out" were used, I could see its direction pointing toward quad 1.

If that's the case, wouldn't the component equations for velocity be

$v_x - 20\cos(30) + 1 = 0 \implies v_x=10\sqrt{3}-1$

$v_y-20\sin(30) = 0 \implies v_y=10$

Meh ... poorly worded problem, so who friggin' knows.
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March 14th, 2017, 04:28 PM   #8
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Skeeter's circle is very pretty but hard to reproduce in an exam.

I must say I prefer this version that puts all the values in place for you.
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March 14th, 2017, 05:27 PM   #9
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Skeeter's circle is very pretty but hard to reproduce in an exam.
The idea is not to "reproduce" the unit circle in its entirety during an exam, rather to learn the values of $(\cos{\theta},\sin{\theta})$ with respect to their respective multiples of $\dfrac{\pi}{2}$, $\dfrac{\pi}{4}$, $\dfrac{\pi}{3}$, and $\dfrac{\pi}{6}$. The circle's symmetry allows one to visualize quadrants and the respective signs for those trig values.

Trig values for multiples of $\dfrac{\pi}{2}$ are based on the intersections of the unit circle and the coordinate axes.

Trig values for specific multiples of $\dfrac{\pi}{4}$ are based on a right isosceles triangle with a unit hypotenuse.

Trig values for specific multiples of $\dfrac{\pi}{3}$ & $\dfrac{\pi}{6}$ are based on the 30-60-90 triangle.

Recall of values is accomplished by visualizing one of the circles shown ...
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File Type: jpg pi_3&pi_6.jpg (16.1 KB, 1 views)
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March 14th, 2017, 05:49 PM   #10
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Originally Posted by The_Ys_Guy View Post
I'm sure I did this right but I can't take any chances.... not after last test 78.59% grade.
78.59%! Well done! Always plenty of time to make up for lost marks in the future .
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