Collision with orientation, finding velocity

AI Thread Summary
The discussion centers on a physics problem involving a collision between two skaters, Debi Thomas and Katerina Witt, where momentum conservation principles are applied to find Katerina's initial velocity. The calculations show that after the collision, the two skaters move together at an angle of 37 degrees relative to Debi's original direction. The initial momentum equations are correctly set up, but the explanation of the process could be clearer, particularly in how momentum components are treated. The feedback emphasizes the importance of clearly stating the conservation of momentum principles rather than separating the momentum terms too simplistically. Overall, the calculations yield Katerina's initial velocity as approximately 3.48 m/s.
SuperHero
Messages
46
Reaction score
0

Homework Statement


During the warm-up for the 1990 Olympics free skating competition, Debi Thomas, mass 60kg, moving at 5 m/s and looking intently at her coach, collides with Katerina Witt, mass 65 kg,who is smiling at a photographer. The two skaters were moving at right angles to each other.After the collision, they hang on to each other and jointly move at an angle of 37 relative to Debi's original direction. How fast was Katerina traveling before the collision?


The Attempt at a Solution


http://s1302.beta.photobucket.com/user/Rameel17/media/Untitled_zps4d1ae6dc.png.html

This is the diagram that i came up with and here is my calculation:

m1v1 + m2v2 = mtv0
(60kg)(5m/s[E]) + (65kg)(v2) = (125kg)(vo)[E 37 N]

So what i did was i divided the m1v1 and m2v2 seperately with the total something like this:

m2v1 = mtv0
(60kg)(5m/s[E]) = (125kg)(vo)[E 37 N]
300 kgm/s [E]/125kg = (vo)(cos 37[E])
2.4 m/s[E]/ (cos 37[E]) = vo
vo = 3.00512558m/s

so i found the final total velocity, now to find the intial katerine's velocity

m2v2[N] = mtv0[E 37 N]
(65kg)(v2) = (125kg)(3.00512558m/s)(sin 37)
v2 = 3.4779177 m/s

Thats the answer.
is my process correct?
 
Physics news on Phys.org
The equations and the thinking behind them are all correct. The verbal explanation could be better. It's more usual to phrase it in terms of components of momentum in the two directions.
 
haruspex said:
The equations and the thinking behind them are all correct. The verbal explanation could be better. It's more usual to phrase it in terms of components of momentum in the two directions.

Hello, Thank you so much for the reply. What do you mean by conservation of momentum phrases?
 
In the OP you wrote
SuperHero said:
So what i did was i divided the m1v1 and m2v2 seperately with the total something like this:
You would not in general be able to separate them so easily, so the reader might suspect you don't understand what you are doing. It would be more convincing to write "by conservation of momentum in the direction of v1..." etc.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top