{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Quiz5Solutions

# Quiz5Solutions - Physics 2 Quiz 5 Solutions November 4 2010...

This preview shows pages 1–3. Sign up to view the full content.

Physics 2: Quiz 5 Solutions November 4, 2010 1. Vector -→ A has length 4 and a polar angle of 140 degrees, vector -→ B = 3 ˆ i + 2 ˆ j , and vector -→ C has length 2 and a polar angle of 30 degrees. Find: a) the x component of -→ C - -→ A We need to start by getting vectors -→ A and -→ C into Cartesian coordinates which will allow us to take the di ff erence. We can use the formula: ( r, θ ) ( r cos θ , r sin θ ) to convert from polar to Cartesian coordinates. For -→ A we have: -→ A = (4 , 140 ) = (4 cos 140 , 4 sin 140 ) = ( - 3 . 57 , 2 . 57) = - 3 . 57 ˆ i + 2 . 57 ˆ j For -→ C we have: -→ C = (2 , 30 ) = (2 cos 30 , 2 sin 30 ) = 3 , 1 = 3 ˆ i + ˆ j For the x component of the di ff erence between the vectors, we just take the di ff erence between the x components of each vector: -→ C - -→ A x = C x - A x = 3 - ( - 3 . 57) = 5 . 3 b) the magnitude of 2 -→ B - 3 ˆ i To get the magnitude we will need both the x and y components of the di ff erence. We have: 2 -→ B - 3 ˆ i = 2 3 ˆ i + 2 ˆ j - 3 ˆ i = 6 ˆ i + 4 ˆ j - 3 ˆ i = 3 ˆ i + 4 ˆ j so the magnitude is given by: 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 -→ B - 3 ˆ i = 3 ˆ i + 4 ˆ j = 3 2 + 4 2 = 5 c) the polar angle of -→ A - -→ B + 2 -→ C We can get the polar angle from the Cartesian components. We have: -→ A - -→ B + 2 -→ C = - 3 . 57 ˆ i + 2 . 57 ˆ j - 3 ˆ i + 2 ˆ j + 2 3 ˆ i + ˆ j = - 3 . 57 ˆ i - 3 ˆ i + 2 3 ˆ i + 2 . 57 ˆ j - 2 ˆ j + 2 ˆ j = - 3 . 11 ˆ i + 2 . 57 ˆ j And we know that, since the vector is in the second quadrant: θ = arctan y x + 180 =
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Quiz5Solutions - Physics 2 Quiz 5 Solutions November 4 2010...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online