My Math Forum A bomb suddenly explodes. find velocity of a piece

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 March 14th, 2017, 12:26 PM #1 Member   Joined: Feb 2014 Posts: 84 Thanks: 1 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?
 March 14th, 2017, 01:20 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,201 Thanks: 613 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 $v \sin(208.63^\circ) = -10$ $v \approx 20.87$ Thanks from The_Ys_Guy
March 14th, 2017, 01:22 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.
Attached Images
 p_conserve.jpg (16.0 KB, 3 views)

March 14th, 2017, 01:37 PM   #4
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Quote:
 Originally Posted by skeeter 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 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?

March 14th, 2017, 02:02 PM   #5
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Quote:
 Originally Posted by The_Ys_Guy 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 $20\cos(30^\circ)=20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$

March 14th, 2017, 02:16 PM   #6
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Quote:
 Originally Posted by skeeter 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".

March 14th, 2017, 02:53 PM   #7
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Quote:
 Originally Posted by romsek 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.

March 14th, 2017, 03:28 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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 trigvals1.jpg (14.1 KB, 1 views)

March 14th, 2017, 04:27 PM   #9
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Quote:
 Originally Posted by studiot 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 ...
Attached Images
 pi_2&pi_4.jpg (14.9 KB, 2 views) pi_3&pi_6.jpg (16.1 KB, 1 views)

March 14th, 2017, 04:49 PM   #10
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Quote:
 Originally Posted by The_Ys_Guy 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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