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March 14th, 2017, 12:26 PM  #1 
Member Joined: Feb 2014 Posts: 91 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: 1,301 Thanks: 664 
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 $v \sin(208.63^\circ) = 10$ $v \approx 20.87$ 
March 14th, 2017, 01:22 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,611 Thanks: 1295 
the statement, "One piece flies out along the positive xaxis 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. 
March 14th, 2017, 01:37 PM  #4  
Member Joined: Feb 2014 Posts: 91 Thanks: 1  Quote:
Quote:
 
March 14th, 2017, 02:02 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,611 Thanks: 1295  Quote:
 
March 14th, 2017, 02:16 PM  #6  
Senior Member Joined: Sep 2015 From: CA Posts: 1,301 Thanks: 664  Quote:
I would have called the other direction "another flies in at $30^\circ$ to the x direction".  
March 14th, 2017, 02:53 PM  #7  
Math Team Joined: Jul 2011 From: Texas Posts: 2,611 Thanks: 1295  Quote:
If that's the case, wouldn't the component equations for velocity be $v_x  20\cos(30) + 1 = 0 \implies v_x=10\sqrt{3}1$ $v_y20\sin(30) = 0 \implies v_y=10$ Meh ... poorly worded problem, so who friggin' knows.  
March 14th, 2017, 03:28 PM  #8 
Senior Member Joined: Jun 2015 From: England Posts: 644 Thanks: 184 
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. 
March 14th, 2017, 04:27 PM  #9  
Math Team Joined: Jul 2011 From: Texas Posts: 2,611 Thanks: 1295  Quote:
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 306090 triangle. Recall of values is accomplished by visualizing one of the circles shown ...  
March 14th, 2017, 04:49 PM  #10 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,289 Thanks: 441 Math Focus: Yet to find out.  

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