Hi guys, Matt from Australia here. Brushing up on course material before beginning Bachelor of Science in Physics next year through an Australian university. With this subject, I found that they have chopped up the old lectures from 1999 a bit. I have instead downloaded all 20-something lectures and am watching them in order. It seems to be working fine. I'd like to get this subject knocked over in the next 6 weeks or so, I'd be interested to hear from anyone on the same timeline as me. Two questions at the moment...
OCW Scholar - Physics I: Classical Mechanics
Stacey Warren - Expert brainly.com
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1) in the challenge problems for vectors (topic 3), question 1.4 shows the answer to be 63 deg... I got 56 deg. Anyone have this issue also?
2) same topic (vectors, topic 3) challenge problems, problem 2.1. The notes show that the denominator of the function is magnitude of vector A * magnitude of vector B, yet when he calculates it out, he uses root3 * 2 (i.e. root3 + root 3, instead of root 3 * root 3 = 3). Hope this makes sense.
3) and another one... maybe I'm just not getting it... see attachment and red circled areas. Cross product formula for Bx should be -2?
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For your question 1) the challenge problems for vectors (topic 3), question 1.4 .
I also get answer 56 degree.
I think the solution in the pdf has mistake.
For your question 2 about vectors( topic 3) challenge problems, problem 2.1,
I think the solution in the pdf has mistake again.
The angle should equal to : arc cos ( -1 / ( 3^0.5 X 3^ 0.5) )
= arc cos ( -1/3 )
= 109 degree
I am programmer and computer science usually use ^ to represent power.
e.g. square root of 3 = 3 power 0.5 = 3^0.5
For your question 3 about vectors( topic 3) challenge problems, problem 3.1,
you are ture that the suggested answer has mistake.
Bx should be -2.
A x B
= ((1)(3)-(-1)(-1)) i + ((-1)(-2)-(1)(3)) j+ ((1)(-1)-(1)(-2))k
= 2i -j +k ............ (ans1)
Check the intermediate answer ans1:
Let C = A x B,
i) dot product of A and C is 0 ( becasue they are prependicular )
ii) dot product of B and C is also 0 ( becasue they are prependicular )
product of C and A
= (Ax )(Cx) + (Ay)(Cy) + (Az)(Cz)
= (1)(2) + (1)(-1) + (-1)(1)
product of C and B
= (Bx)(Cx) + (By)(Cy) + (Bz)(Cz)
= (-2)(2) + (-1)(-1) + (3)(1)
If you use the pdf's answer C = A x B = 2 i- 5 j- 3 k
to calcualte product of B and C, you will find that it is not equal to 0 !
(that means they are not prependicular):
= (Bx)(Cx) + (By)(Cy) + (Bz)(Cz)
= (-2)(2) + (-1)(-5) + (3)(-3)
= -8 !
Great, thanks Ricky, good to have my thoughts confirmed :)