Stuff You Need To Know
In a problem, you are often given directions such as north (N), east (E), south (S), and west (W). To use the Two Dimensional Inelastic Collison Simulator (TDICS), you must use the convention shown to the right. Therefore, N would be 000°, E would be 090°, etc. A bearing of 135° would be drawn such that the vector would be 45° below the horizontal (090°). Do not confuse the diagram to the right with the one shown below.
The diagram to the left is probably familiar to you if you have used the sine (sin), cosine (cos) and tangent (tan) functions in math. Do not forget these! These will also be used in combination with the one above but not interchangeably. You will need to use the one on the left for sign (+ or -) conventions. The sign refers to positive or negative and the sin refers to the sine function or what we do best.
In case you were wondering why the first letter is underlined in each quadrant, it helps you to remember which trig function is positive in a particular quadrant. The first quadrant is from 0° – 90° and all the trig functions are positive, the the second quadrant is from 90° – 180° and the sin is positive while the others are negative, etc. In case you are wondering why I did not mention the three remaining trig functions it is because we don’t use them. In case you were not wondering, do not worry about it, because there are times you wonder and times you don’t. The simplest case to solve is when the two momentum vectors are at right angles to each other. In this case you can use the pythagorean theorem to find the magnitude of the resultant as shown below: c2 = a2 + b2 where a and b are the legs of the triangle and c is the hypotenuse. To find the direction and ultimately the bearing, you can use sinθ, cosθ, or the tanθ as shown below: sinθ = opp/hyp or cosθ = adj/hyp or tanθ = opp/adj Remember a couple of things: (1) The adjacent side can never be the hypotenuse. (2) SOH, CAH, TOA helps to remember the trig functions. Things get slightly more complicated when the angle is acute (less than 90°) or obtuse (greater than 180°). One method of solution is to resolve each momentum vector into two right angle components. This allows for using the sin, cos, or tan shown above because you will end up with a right triangle. You can vectorially add the components in the same or opposite direction and proceed as if these componets were given to you at the outset of the problem. An alternative solution is to use the two momentum vectors to construct a parallelogram. You can choose either triangle from the parallelogram and use the law of cosines to determine the resultant as shown below: c2 = a2 + b2 -2abcosC This formula is not as intimidating as it first appears because it simply extends the pythagorean theorem because the included angle between the two momentum vectors is not 90°. If the included angle was 90°, the cos90° = 0, and you are left with the pythagorean theorem. A word of caution is very important here: if the angle between the two momentum vectors is acute (less than 90°), the included angle used in the law of cosines will be greater than 90°. As shown in the second diagram above, the cos will be negative in the second quadrant and you must include the negative sign in the law of cosines. To find the direction and ultimately the bearing, you use the law of sines as shown below: a/sin A = b/sin B = c/sin C where a, b, and c represent the magnitudes of the vector and A, B, and C are the angles opposite those vectors. All of this “stuff” will become a lot clearer as you work through the animation and the problems following the animation.