Difference between revisions of "Skill Set 03.1"

Latest revision as of 10:24, 12 February 2009

Main Page > IB Physics Skill Sets > Skill Set 03.1

1. Find the components of the vector.

Since the angle given is the same as the trig angle, we can use 17^o in our equations.

32cos(17) = 30.60 m x
32sin(17) = 9.356 m y

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2. Find the components of the vector.

Converting the 23^o angle to a trig angle is done by subtracting it by 360, since the angle is just below the 0 or 360^o line. Using our new angle, 337^o, we can solve for the components.

15cos(337) = 13.81 m x
15sin(337) = -5.861 m y

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3. Convert 13.2 m/s x + 5.70 m/s y into an angle magnitude vector.

To get the angle:

tan^-1(5.7/13.2) = 23.4^o

To get the magnitude:

a² + b² = c²
5.7² + 13.2 ² = c²
c = 14.4 m

14.4 m @ 23.4^o

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4. Add and subtract the vector components.

All that's necessary to remember when doing these is to do the x and y components separately. For D+E:

16.2 + -13.7 = 2.5 m x
-3.5 + -4.2 = -7.7 m y

For E-D:

-13.7 - 16.2 = -29.9 m x
-4.2 - -3.5 = -.7 m y

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5. Calculate the sum of vectors A and B (From problem 1 and 2). Draw the resultant vector as an angle magnitude vector. (Label your angle and magnitude clearly – use an arrow) Do your work on the back of this sheet, and label and describe the three steps for doing so.

In order to add vectors, we first have to convert them into vector components rather than angle magnitude. Therefore we will use the answers of questions #1 and #2 in our addition.

30.60 m x + 5.861 m x = 36.46 m x
9.356 m y + -13.81 m y = -4.45 m y

Since we have our new vector components, we can convert them into an angle magnitude vector like we did in problem #3.

tan^-1(-4.45/36.46) = -6.95^o
-4.45² + 36.46 ² = c²
c = 36.7 m
36.7 m @ -6.95^o

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