Physics 12 Vectors Worksheet featuring 9 problems on vector concepts, including vector addition, displacement, force, and motion.

Problem: Physics 12 Vectors Worksheet

Below, I will solve the problems step by step and explain the solutions.

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#### Problem 1: Label each quantity as being vector or scalar
- Distance: Scalar (magnitude only)
- Time: Scalar (magnitude only)
- Mass: Scalar (magnitude only)
- Area: Scalar (magnitude only)
- Energy: Scalar (magnitude only)
- Impulse: Vector (involves force and time, both of which are vectors)
- Temperature: Scalar (magnitude only)
- Displacement: Vector (has magnitude and direction)
- Volume: Scalar (magnitude only)
- Speed: Scalar (magnitude only)
- Acceleration: Vector (has magnitude and direction)
- Momentum: Vector (mass × velocity, both of which are vectors)
- Work: Scalar (force × displacement in the direction of force, but work is a scalar result)
- Velocity: Vector (has magnitude and direction)

Answer:
- Vector: Impulse, Displacement, Acceleration, Momentum, Velocity
- Scalar: Distance, Time, Mass, Area, Energy, Temperature, Volume, Speed, Work

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#### Problem 2: Sketch the following vectors on a separate piece of paper and draw the resultant
This problem requires graphical or analytical solutions for vector addition. Here’s how to approach it:

a) A + B
- Place vector A, then place vector B starting from the tip of A.
- The resultant is the vector from the tail of A to the tip of B.

b) B - A
- Subtracting A is equivalent to adding -A (reverse direction of A).
- Place vector B, then place -A starting from the tip of B.
- The resultant is the vector from the tail of B to the tip of -A.

c) A + D - C
- Add A and D first, then subtract C.
- Place A, then place D starting from the tip of A.
- From the resultant of A + D, place -C (reverse direction of C).
- The resultant is the vector from the tail of A to the tip of -C.

d) B - C + D
- Subtract C from B, then add D.
- Place B, then place -C starting from the tip of B.
- From the resultant of B - C, place D.
- The resultant is the vector from the tail of B to the tip of D.

e) C - 2D + A
- Subtract 2D from C, then add A.
- Place C, then place -2D (double the length of D in the opposite direction) starting from the tip of C.
- From the resultant of C - 2D, place A.
- The resultant is the vector from the tail of C to the tip of A.

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#### Problem 3: Jagger runs 300 m due west and then turns and runs 500 m due south
a) What is the total distance that she ran?
- Distance is the sum of the magnitudes of the paths traveled.
- Total distance = 300 m + 500 m = 800 m

b) What is her total displacement?
- Displacement is a vector quantity representing the straight-line distance from the starting point to the ending point.
- Use the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(300)^2 + (500)^2} = \sqrt{90000 + 250000} = \sqrt{340000} \approx 583.1 \, \text{m}
\]
- Direction: Use the tangent function to find the angle:
\[
\theta = \tan^{-1}\left(\frac{\text{south component}}{\text{west component}}\right) = \tan^{-1}\left(\frac{500}{300}\right) \approx 59.0^\circ \, \text{south of west}
\]

c) If it takes her 135 s to complete the route, calculate her speed and velocity.
- Speed: Total distance / time
\[
\text{Speed} = \frac{800 \, \text{m}}{135 \, \text{s}} \approx 5.93 \, \text{m/s}
\]
- Velocity: Displacement / time
\[
\text{Velocity} = \frac{583.1 \, \text{m}}{135 \, \text{s}} \approx 4.32 \, \text{m/s}, \, 59.0^\circ \, \text{south of west}
\]

Answer:
a) 800 m
b) 583.1 m, 59.0° south of west
c) Speed: 5.93 m/s, Velocity: 4.32 m/s, 59.0° south of west

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#### Problem 4: Two ropes attached to a heavy object
Each rope exerts a force of 1000 N. Determine the resultant force in different scenarios.

a) In the same direction east
- Resultant force = 1000 N + 1000 N = 2000 N, east

b) In opposite directions
- Resultant force = 1000 N - 1000 N = 0 N

c) At right angles, south and east
- Use the Pythagorean theorem:
\[
\text{Resultant force} = \sqrt{(1000)^2 + (1000)^2} = \sqrt{1000000 + 1000000} = \sqrt{2000000} \approx 1414.2 \, \text{N}
\]
- Direction: 45° southeast (since the forces are equal and perpendicular)

Answer:
a) 2000 N, east
b) 0 N
c) 1414.2 N, 45° southeast

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#### Problem 5: Forces acting on an object
Two forces act on an object: 200 N due north and 300 N due east.

a) Determine the resultant net force.
- Use the Pythagorean theorem:
\[
\text{Resultant force} = \sqrt{(200)^2 + (300)^2} = \sqrt{40000 + 90000} = \sqrt{130000} \approx 360.6 \, \text{N}
\]
- Direction: Use the tangent function:
\[
\theta = \tan^{-1}\left(\frac{\text{east component}}{\text{north component}}\right) = \tan^{-1}\left(\frac{300}{200}\right) = \tan^{-1}(1.5) \approx 56.3^\circ \, \text{east of north}
\]

b) A third force now acts on the object so that the net force is 0. Determine its magnitude and direction.
- The third force must be equal in magnitude to the resultant force but in the opposite direction.
- Magnitude: 360.6 N
- Direction: 180° + 56.3° = 236.3° (measured counterclockwise from east, or 56.3° south of west)

Answer:
a) 360.6 N, 56.3° east of north
b) 360.6 N, 56.3° south of west

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#### Problem 6: Pilot flies a plane 10,000 km in a direction 30° N of W
To find how much farther north and west the pilot has gone:

- North component: \( 10000 \times \sin(30^\circ) = 10000 \times 0.5 = 5000 \, \text{km} \)
- West component: \( 10000 \times \cos(30^\circ) = 10000 \times \frac{\sqrt{3}}{2} \approx 8660 \, \text{km} \)

Answer:
5000 km north, 8660 km west

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#### Problem 7: Student mows lawn with a push mower
The student exerts a force of 250 N at an angle of 40° below the horizontal.

- Horizontal component (pushing force along the ground):
\[
F_{\text{horizontal}} = 250 \times \cos(40^\circ) \approx 250 \times 0.766 \approx 191.5 \, \text{N}
\]

Answer:
191.5 N

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#### Problem 8: Phaedo drives his turbo scooter
Phaedo travels 5000 m due west at 30.0 m/s.

- Time taken:
\[
t = \frac{\text{distance}}{\text{speed}} = \frac{5000 \, \text{m}}{30.0 \, \text{m/s}} \approx 166.7 \, \text{s}
\]

Answer:
166.7 s

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#### Problem 9: Boat heads directly across a river
The boat travels at 20.0 m/s relative to the water, and the river flows at 5.0 m/s downstream.

a) What is the resultant velocity of the boat?
- Use the Pythagorean theorem:
\[
v_{\text{resultant}} = \sqrt{(20.0)^2 + (5.0)^2} = \sqrt{400 + 25} = \sqrt{425} \approx 20.6 \, \text{m/s}
\]
- Direction: Use the tangent function:
\[
\theta = \tan^{-1}\left(\frac{\text{downstream component}}{\text{cross-stream component}}\right) = \tan^{-1}\left(\frac{5.0}{20.0}\right) = \tan^{-1}(0.25) \approx 14.0^\circ \, \text{downstream}
\]

b) How long does it take the boat to reach the other side?
- Time = width of river / cross-stream velocity:
\[
t = \frac{100 \, \text{m}}{20.0 \, \text{m/s}} = 5.0 \, \text{s}
\]

c) How far downstream is the boat when it reaches the other side?
- Distance downstream = downstream velocity × time:
\[
d_{\text{downstream}} = 5.0 \, \text{m/s} \times 5.0 \, \text{s} = 25.0 \, \text{m}
\]

d) In which direction should the boat head to end up directly across?
- The boat must compensate for the downstream flow by heading upstream at an angle.
- Let \( \theta \) be the angle upstream from the perpendicular direction.
- The cross-stream component of the boat's velocity must still be 20.0 m/s, and the upstream component must cancel the 5.0 m/s downstream flow:
\[
v_{\text{upstream}} = 20.0 \times \sin(\theta) = 5.0 \implies \sin(\theta) = \frac{5.0}{20.0} = 0.25 \implies \theta = \sin^{-1}(0.25) \approx 14.5^\circ
\]
- The boat should head 14.5° upstream from the perpendicular direction.

Answer:
a) 20.6 m/s, 14.0° downstream
b) 5.0 s
c) 25.0 m
d) 14.5° upstream from the perpendicular direction

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Final Answers:

1. Vectors: Impulse, Displacement, Acceleration, Momentum, Velocity
Scalars: Distance, Time, Mass, Area, Energy, Temperature, Volume, Speed, Work
2. (Graphical/Analytical solution provided above)
3. a) 800 m, b) 583.1 m, 59.0° south of west, c) Speed: 5.93 m/s, Velocity: 4.32 m/s, 59.0° south of west
4. a) 2000 N, east, b) 0 N, c) 1414.2 N, 45° southeast
5. a) 360.6 N, 56.3° east of north, b) 360.6 N, 56.3° south of west
6. 5000 km north, 8660 km west
7. 191.5 N
8. 166.7 s
9. a) 20.6 m/s, 14.0° downstream, b) 5.0 s, c) 25.0 m, d) 14.5° upstream from the perpendicular direction

Final Boxed Answer:
\[
\boxed{14.5^\circ}
\]