How to know when to relate the tangents (Divide components)

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To determine when to divide components of vectors to relate tangents, it's essential to recognize scenarios involving vector equations, such as inelastic collisions. The discussion highlights the lack of clarity in textbooks and instruction on this method, leading to confusion among students. An example involving two cars colliding at a 55-degree angle illustrates how to set up the equations for momentum in both x and y directions. By dividing the equations, students can derive the tangent of the angle, simplifying the problem-solving process. Understanding this technique is crucial for effectively eliminating variables and solving vector-related problems.
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Homework Statement


What ques does one need to recognize they need to divide components of two vectors together to "relate the tangents?"

The Attempt at a Solution


Alright, so I've seen this come up a LOT in the semester-dividing components or sums of components together to work out an equation in terms of tangents and to eliminate variables. The thing is my professor and the book we're using don't explain this, and when it's used nobody elaborates. Getting my professor to respond to a question results in a "google it" response. So tell me what I'm missing guys. When are you even supposed to use this trick.

I know there needs to be some indication of work, but all I can say is I've googled, asked around, and done plenty of logistics trying to figure this out, and it's too frustrating to continue on my own.

If you're wondering how I've gotten this far into the semester without knowing this, it hasn't come up on tests, only on one or two homework assignments. But it bugs me.
 
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Cake said:

Homework Statement


What ques does one need to recognize they need to divide components of two vectors together to "relate the tangents?"

The Attempt at a Solution


Alright, so I've seen this come up a LOT in the semester-dividing components or sums of components together to work out an equation in terms of tangents and to eliminate variables. The thing is my professor and the book we're using don't explain this, and when it's used nobody elaborates. Getting my professor to respond to a question results in a "google it" response. So tell me what I'm missing guys. When are you even supposed to use this trick.

I know there needs to be some indication of work, but all I can say is I've googled, asked around, and done plenty of logistics trying to figure this out, and it's too frustrating to continue on my own.

If you're wondering how I've gotten this far into the semester without knowing this, it hasn't come up on tests, only on one or two homework assignments. But it bugs me.
How about giving a specific example ?
 
Alright so, let's say I've got an inelastic collision in two dimensions. Equal mass cars. One is traveling east at 13 m/s, the other north at an unknown speed. When they collide they bounce off together at a 55 degree angle. The solution looks something like this:

##p_y=mv_?=2mv_f sin(55)##
##p_x=13m=2mv_f cos(55)##
[book notes to divide these without justification]
##\frac {v} {13} = tan (55)##
##v=41.5 mi/hr##

Edit: Fixed units
 
Last edited:
Cake said:
Alright so, let's say I've got an inelastic collision in two dimensions. Equal mass cars. One is traveling east at 13 m/s, the other north at an unknown speed. When they collide they bounce off together at a 55 degree angle. The solution looks something like this:

##p_y=mv_?=2mv_f sin(55)##
##p_x=13m=2mv_f cos(55)##
[book notes to divide these without justification]
##\frac {v} {13} = tan (55)##
##v=41.5 m/s##
It's basic algebra.

Essentially you are saying ##13m=2mv_f \cos(55^\circ)##, so the left side is the same thing as the right side, at least numerically..

Then all you're doing in this method is dividing both sides of the first equation by the same thing.You can get the same result by taking a few more steps.

Solve the first equation for ##\ \sin(55^\circ\,,\ ## and solve the second for ##\ \cos(55^\circ\ . ##

Then get ##\ \tan(55^\circ,\ ## by dividing the two results.
 
SammyS said:
It's basic algebra.

Essentially you are saying ##13m=2mv_f \cos(55^\circ)##, so the left side is the same thing as the right side, at least numerically..

Then all you're doing in this method is dividing both sides of the first equation by the same thing.You can get the same result by taking a few more steps.

Solve the first equation for ##\ \sin(55^\circ\,,\ ## and solve the second for ##\ \cos(55^\circ\ . ##

Then get ##\ \tan(55^\circ,\ ## by dividing the two results.
That's so obnoxiously simple. Thank you though.
 
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