PHYSICS 12 VECTORS WORKSHEET 1. Label each ... - mbourget - Free Printable

1.
a) distance - scalar
b) time - scalar
c) mass - scalar
d) area - scalar
e) energy - scalar
f) impulse - vector
g) temperature - scalar
h) displacement - vector
i) volume - scalar
j) speed - scalar
k) acceleration - vector
l) momentum - vector
m) work - scalar
n) velocity - vector
o) force - vector

2.
[Note: Since drawing is not possible here, instructions for each operation are given.]
a) C – A: Draw vector A, then from its tip draw vector C in the opposite direction (or reverse A and add to C). Resultant starts at tail of C, ends at tip of reversed A.
b) D – B: Reverse vector B, then add to D. Resultant starts at tail of D, ends at tip of reversed B.
c) A + D – B: First add A and D head-to-tail, then subtract B by reversing it and adding. Resultant starts at tail of A, ends at tip of reversed B.
d) B – (C – D): First compute C – D (reverse D and add to C), then subtract that result from B (reverse the result and add to B).
e) C – 2B: Scale vector B by 2, reverse it, then add to C.
f) 3C – 2D + A: Scale C by 3, scale D by 2 and reverse it, then add A. Add all three scaled/reversed vectors head-to-tail. Resultant starts at tail of first, ends at tip of last.

3.
a) Total distance = 300 m + 500 m = 800 m
b) Displacement: Use Pythagoras. West and south are perpendicular.
Displacement magnitude = √(300² + 500²) = √(90000 + 250000) = √340000 ≈ 583.1 m
Direction: tan⁻¹(500/300) ≈ tan⁻¹(1.6667) ≈ 59.0° south of west
So displacement is 583.1 m at 59.0° south of west.
c) Speed = total distance / time = 800 m / 135 s ≈ 5.93 m/s
Velocity = displacement / time = 583.1 m / 135 s ≈ 4.32 m/s at 59.0° south of west

4.
a) Same direction: F_net = 1000 N + 1000 N = 2000 N
b) Opposite directions: F_net = 1000 N – 1000 N = 0 N
c) At right angles: F_net = √(1000² + 1000²) = √2,000,000 = 1000√2 ≈ 1414 N, direction 45° between the two pulls.

5.
a) Forces are perpendicular: South 200 N, East 300 N.
F_net = √(200² + 300²) = √(40000 + 90000) = √130000 ≈ 360.6 N
Direction: tan⁻¹(200/300) = tan⁻¹(2/3) ≈ 33.7° south of east
b) To make net force zero, third force must be equal in magnitude and opposite in direction to the resultant.
Magnitude: 360.6 N
Direction: 33.7° north of west

6.
Plane flies 10,000 km at 30° N of W.
a) North component: 10000 × sin(30°) = 10000 × 0.5 = 5000 km north
b) West component: 10000 × cos(30°) = 10000 × (√3/2) ≈ 10000 × 0.8660 = 8660 km west

7.
Force applied at 40° to horizontal. Only horizontal component pushes mower along ground.
F_horizontal = 250 N × cos(40°) ≈ 250 × 0.7660 ≈ 191.5 N

8.
a) Resultant velocity: Boat’s velocity is 30.0 m/s at 30° N of E.
East component: 30.0 × cos(30°) ≈ 30.0 × 0.8660 = 25.98 m/s
North component: 30.0 × sin(30°) = 30.0 × 0.5 = 15.00 m/s
Resultant velocity magnitude: √(25.98² + 15.00²) ≈ √(675.0 + 225.0) = √900.0 = 30.0 m/s
Direction: tan⁻¹(15.00/25.98) ≈ tan⁻¹(0.577) ≈ 30.0° N of E — same as boat’s heading since no wind or current mentioned.
b) Distance east = 5000 m. Eastward speed = 25.98 m/s.
Time = distance / speed = 5000 / 25.98 ≈ 192.5 seconds

9.
a) Boat velocity relative to water: 20.0 m/s due east. Stream velocity: 5.00 m/s due south.
Resultant velocity magnitude: √(20.0² + 5.00²) = √(400 + 25) = √425 ≈ 20.6 m/s
Direction: tan⁻¹(5.00/20.0) = tan⁻¹(0.25) ≈ 14.0° south of east
b) Width of stream = 100 m. Eastward speed = 20.0 m/s.
Time = 100 / 20.0 = 5.00 seconds
c) Downstream distance = southward speed × time = 5.00 m/s × 5.00 s = 25.0 m
d) To end up directly across, boat must head upstream at an angle such that its northward component cancels the stream’s southward flow.
Let θ be angle north of east. Then:
20.0 × sin(θ) = 5.00 → sin(θ) = 5.00 / 20.0 = 0.25 → θ = sin⁻¹(0.25) ≈ 14.5° north of east