Borax Posted August 23, 2014 Report Share Posted August 23, 2014 The total momentum of the system needs to be zero, thus all horizontal and vertical components will cancel out... The answer is supposed to be D, but that doesn't make sense to me. Wouldn't the vertical components of V2 and V3 be opposite to one another? If so, they aren't equal.... Further, shouldn't both the horizontal components of V2 and V3 be EXACTLY equal (-ve 1/2 V1) since they need to balance out the horizontal component of V1? Thanks for the help guys Link to comment Share on other sites More sharing options...

thehead321 Posted August 23, 2014 Report Share Posted August 23, 2014 It says in the question "unknown speeds" so i'm pretty sure v_2 and v_3 are referring to magnitudes. Most velocity vectors have that little arrow thing on top of them as well. So, yes, they're opposite in direction, but same in magnitude. As for your second question, it's: v_2*cos(w) + v_3*cos(z) = v_1 [horizontal components]. You're falsely assuming that the first 2 terms are necessarily equal. They don't have to be. For example, 4 + 6 = 10, not necessarily just 5 + 5 = 10. (i.e. x + y = z doesn't mean x = y, it can be, but not necessarily) Hope that helps Link to comment Share on other sites More sharing options...

MathToMed Posted August 23, 2014 Report Share Posted August 23, 2014 As a rule of thumb, whenever you're using SOH-CAH-TOA or the pythagorean theorem (or by extension, the sine/cosine laws), such as when splitting a vector into its components you are not looking at the sides as vectors. You are treating them as scalars, and then making a realization that these scalars are then the lengths of the associated vectors. For this post, bolded v indicates a vector quantity, while unbolded indicates it is a scalar. Also note where applicable, the angle measures are in degrees and not radians (didn't feel like putting in degree symbols). Imagine breaking v2 and v3 into component vectors. Since v1 has absolutely no vertical component (note that this is always the case, since you can always rotate your diagram so one vector points to the right, and choosing an appropriate reference frame), then the vertical components of v2 and v3 must be equal and opposite, in order for their momentum vectors to sum to the zero vector 0 (prior to the collision). this means that their scalar values, or "side lengths" v2sin(w) and v3sin(z) must be equal.However the horizontal components need not be equal, as thehead321 mentioned. Don't be mislead by the diagram! It's entirely possible that, say v3 may point perfectly downward (and so angle z = 90). This would mean that v2 = (-v3) + (-v1), since they are perpendicular to each other, and v2 must "cancel out" the horizontal component of v1, and vertical component of v3 and so it would point up and to the left. In such a case, v2cos(w) = |v1|, but v3cos(z) = v3cos(90) = 0, so v2cos(w) and v3cos(z) are necessarily different.It's a bit difficult to follow what I'm saying without an explicit diagram, but hopefully that helps a bit. Let me know if you need any clarification, if I have access to a scanner I might draw some diagrams and post them if you need it. Link to comment Share on other sites More sharing options...

Borax Posted August 23, 2014 Author Report Share Posted August 23, 2014 As a rule of thumb, whenever you're using SOH-CAH-TOA or the pythagorean theorem (or by extension, the sine/cosine laws), such as when splitting a vector into its components you are not looking at the sides as vectors. You are treating them as scalars, and then making a realization that these scalars are then the lengths of the associated vectors. For this post, bolded v indicates a vector quantity, while unbolded indicates it is a scalar. Also note where applicable, the angle measures are in degrees and not radians (didn't feel like putting in degree symbols). Imagine breaking v2 and v3 into component vectors. Since v1 has absolutely no vertical component (note that this is always the case, since you can always rotate your diagram so one vector points to the right, and choosing an appropriate reference frame), then the vertical components of v2 and v3 must be equal and opposite, in order for their momentum vectors to sum to the zero vector 0 (prior to the collision). this means that their scalar values, or "side lengths" v2sin(w) and v3sin(z) must be equal. However the horizontal components need not be equal, as thehead321 mentioned. Don't be mislead by the diagram! It's entirely possible that, say v3 may point perfectly downward (and so angle z = 90). This would mean that v2 = (-v3) + (-v1), since they are perpendicular to each other, and v2 must "cancel out" the horizontal component of v1, and vertical component of v3 and so it would point up and to the left. In such a case, v2cos(w) = |v1|, but v3cos(z) = v3cos(90) = 0, so v2cos(w) and v3cos(z) are necessarily different. It's a bit difficult to follow what I'm saying without an explicit diagram, but hopefully that helps a bit. Let me know if you need any clarification, if I have access to a scanner I might draw some diagrams and post them if you need it. Makes much more sense this way. I didn't know to treat the values as scalars when using SOC CAH TOA, which lead me to believe choice C was the most correct. Thanks again Link to comment Share on other sites More sharing options...

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