College Physics Quiz: Scalars And Vectors

19 questions · exam conditions

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Scalars And VectorsQuestion 1 of 19

Three forces act on an object: $\vec{F_1} = (8.0\hat{i} + 6.0\hat{j})$ N, $\vec{F_2} = (-3.0\hat{i} + 4.0\hat{j})$ N, and $\vec{F_3} = (-2.0\hat{i} - 8.0\hat{j})$ N. What is the magnitude of the net force?

3.6 N

5.0 N

7.0 N

13.0 N

21.0 N

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College Physics Quiz

College Physics Quiz: Scalars And Vectors

Practice Scalars And Vectors in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Scalars And Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

3.6 N (correct answer)
5.0 N
7.0 N
13.0 N
21.0 N

Explanation: When you encounter multiple forces acting on an object, you need to find the net force by vector addition, then calculate its magnitude. This tests your understanding of vector components and the Pythagorean theorem. To find the net force, add the corresponding components of all three forces:

\vec{F_{net}} = \vec{F_1} + \vec{F_2} + \vec{F_3}

For the x-component:

F_{net,x} = 8.0 + (-3.0) + (-2.0) = 3.0

N For the y-component:

F_{net,y} = 6.0 + 4.0 + (-8.0) = 2.0

N So

\vec{F_{net}} = (3.0\hat{i} + 2.0\hat{j})

N The magnitude is found using:

|\vec{F_{net}}| = \sqrt{F_{net,x}^2 + F_{net,y}^2} = \sqrt{(3.0)^2 + (2.0)^2} = \sqrt{9.0 + 4.0} = \sqrt{13} = 3.6

N This confirms answer A is correct. Answer B (5.0 N) might result from incorrectly calculating

\sqrt{3^2 + 4^2} = 5

, confusing this with a 3-4-5 right triangle but using wrong components. Answer C (7.0 N) could come from simply adding the magnitudes of the net force components (3.0 + 2.0 + 2.0 = 7.0), which is incorrect vector addition. Answer D (13.0 N) results from forgetting to take the square root: using

3^2 + 2^2 = 13

as the final answer. Remember: always add vector components first, then find the magnitude of the resultant vector using the Pythagorean theorem. Never add magnitudes directly unless vectors are collinear.

Question 2

A projectile is launched at an angle above the horizontal. At the highest point of its trajectory, which statement about the velocity and acceleration vectors is correct?

Both velocity and acceleration are zero at the highest point
Velocity is horizontal and acceleration is vertical downward (correct answer)
Velocity is vertical upward and acceleration is horizontal
Both velocity and acceleration are vertical downward
Velocity is zero and acceleration is vertical downward

Explanation: When analyzing projectile motion, you need to separately consider the horizontal and vertical components of velocity and acceleration throughout the flight. At the highest point of a projectile's trajectory, the vertical component of velocity becomes zero because the projectile momentarily stops rising before beginning to fall. However, the horizontal component of velocity remains constant throughout the entire flight (assuming no air resistance). This means the velocity vector at the highest point is purely horizontal. The acceleration due to gravity acts downward with magnitude

g = 9.8 \text{ m/s}^2

throughout the entire trajectory, regardless of the projectile's position. Gravity doesn't "turn off" at the highest point—it's what causes the projectile to change direction from upward to downward motion. Looking at the incorrect options: (A) suggests both velocity and acceleration are zero, but gravity never stops acting, and the projectile maintains horizontal motion. (C) incorrectly states velocity is vertical upward—this would mean the projectile is still rising, which contradicts the definition of the highest point. (D) claims velocity is vertical downward, but at the precise highest point, the projectile hasn't started falling yet, and it still has horizontal velocity. The correct answer is (B): velocity is horizontal (only the horizontal component remains) and acceleration is vertical downward (gravity continues acting). Remember this key insight: at the highest point of any projectile's path, the vertical velocity component is always zero, but horizontal velocity and downward gravitational acceleration persist.

Question 3

A car travels around a circular track at constant speed. Which of the following statements about the car's velocity and acceleration is correct?

Both velocity and acceleration have constant magnitude and direction
Velocity has constant magnitude but changing direction; acceleration is zero
Velocity has constant magnitude but changing direction; acceleration has constant magnitude but changing direction (correct answer)
Velocity has changing magnitude and direction; acceleration has constant magnitude and direction
Both velocity and acceleration have changing magnitude but constant direction

Explanation: When analyzing circular motion problems, you need to carefully distinguish between velocity and acceleration as vector quantities that have both magnitude and direction. For a car moving at constant speed around a circular track, the velocity has constant magnitude (the speed remains the same) but continuously changing direction as the car follows the curved path. Since velocity is a vector, any change in direction means the velocity vector is changing, even when speed stays constant. This changing velocity requires acceleration. In circular motion, this is called centripetal acceleration, which always points toward the center of the circle. The magnitude of this acceleration remains constant (

a_c = \frac{v^2}{r}

), but its direction continuously changes to keep pointing inward as the car moves around the track. Option A is wrong because while acceleration magnitude is constant, its direction changes continuously. Option B makes the common error of thinking constant speed means zero acceleration—this ignores that acceleration occurs whenever velocity changes, including directional changes. Option D incorrectly states that velocity magnitude changes; the problem specifies constant speed, so only the direction of velocity changes. The correct answer is C: velocity has constant magnitude but changing direction, and acceleration has constant magnitude but changing direction. Study tip: Remember that in uniform circular motion, "constant speed" doesn't mean "constant velocity." Speed is scalar (magnitude only), while velocity is a vector. Any curved path requires acceleration, even at constant speed, because the direction is always changing.

Question 4

A particle's acceleration vector is $\vec{a} = (2.0\hat{i} + 4.0\hat{j})$ m/s². If the particle starts from rest at the origin, what is the angle that the velocity vector makes with the x-axis after 3.0 seconds?

26.6°
45.0°
63.4° (correct answer)
90.0°
The angle changes continuously with time

Explanation: This problem tests your understanding of kinematic motion with constant acceleration in two dimensions. When you see a constant acceleration vector, remember that you can find velocity by integrating acceleration over time, and the direction comes from the vector components. Since the particle starts from rest with constant acceleration

\vec{a} = (2.0\hat{i} + 4.0\hat{j})

m/s², you can find the velocity after 3.0 seconds using

\vec{v} = \vec{v_0} + \vec{a}t

. With

\vec{v_0} = 0

, this gives

\vec{v} = (2.0\hat{i} + 4.0\hat{j}) \times 3.0 = (6.0\hat{i} + 12.0\hat{j})

m/s. The angle with the x-axis is found using

\theta = \arctan\left(\frac{v_y}{v_x}\right) = \arctan\left(\frac{12.0}{6.0}\right) = \arctan(2.0) = 63.4°

. This confirms answer C is correct. Let's examine the wrong answers: A) 26.6° would result from incorrectly calculating

\arctan\left(\frac{v_x}{v_y}\right) = \arctan(0.5)

, essentially swapping the components. B) 45.0° occurs when

v_x = v_y

, but that's not the case here since the acceleration components are different. D) 90.0° would mean the velocity is purely in the y-direction, which would only happen if

a_x = 0

. Remember: when finding the angle of a vector from its components, always use

\theta = \arctan\left(\frac{y\text{-component}}{x\text{-component}}\right)

. The ratio of acceleration components determines the direction of motion for constant acceleration problems.

Question 5

The position of a particle is given by $\vec{r} = (5.0t^2)\hat{i} + (3.0t)\hat{j}$ where position is in meters and time is in seconds. What is the direction of the acceleration vector?

Along the positive x-axis only (correct answer)
Along the positive y-axis only
At 30.9° above the positive x-axis
At 59.0° above the positive x-axis
The direction changes with time

Explanation: When you encounter position as a function of time, remember that acceleration is the second derivative of position. This question tests your ability to work through the kinematic chain: position → velocity → acceleration. Starting with the given position vector

\vec{r} = (5.0t^2)\hat{i} + (3.0t)\hat{j}

, you find velocity by taking the first derivative:

\vec{v} = \frac{d\vec{r}}{dt} = (10.0t)\hat{i} + (3.0)\hat{j}

. Then find acceleration by taking the derivative of velocity:

\vec{a} = \frac{d\vec{v}}{dt} = (10.0)\hat{i} + (0)\hat{j} = 10.0\hat{i}

. The acceleration vector has only an x-component (10.0 m/s²) and no y-component. This means it points entirely along the positive x-axis. Choice A is correct because the acceleration vector

\vec{a} = 10.0\hat{i}

points only in the positive x-direction. Choice B incorrectly suggests the acceleration is along the y-axis, but the y-component of acceleration is zero since the y-component of velocity is constant. Choices C and D represent angles you might calculate if you mistakenly used velocity components instead of acceleration components, or confused the relationship between the x and y terms in the original position function. Key strategy: Always remember the derivative chain for kinematics. When finding acceleration from position, you must take two derivatives, not one. The direction of acceleration depends only on the acceleration vector's components, not the original position or velocity vectors.

Question 6

A particle moves in a circle of radius 5.0 m with constant angular velocity. If the particle completes one full revolution in 4.0 s, what is the magnitude of its centripetal acceleration?

7.9 m/s²
12.3 m/s² (correct answer)
15.7 m/s²
31.4 m/s²
2.5 m/s²

Explanation: When you encounter circular motion problems, you're dealing with objects that change direction continuously even at constant speed, which requires centripetal acceleration toward the center of the circle. To find centripetal acceleration, you can use

a_c = \omega^2 r

, where

\omega

is angular velocity and

r

is radius. First, calculate the angular velocity:

\omega = \frac{2\pi}{T} = \frac{2\pi}{4.0 \text{ s}} = \frac{\pi}{2} \text{ rad/s}

Now substitute into the centripetal acceleration formula:

a_c = \omega^2 r = \left(\frac{\pi}{2}\right)^2 \times 5.0 = \frac{\pi^2}{4} \times 5.0 = \frac{5\pi^2}{4} \approx 12.3 \text{ m/s}^2

Looking at the wrong answers: Choice A (7.9 m/s²) likely results from using

a_c = \frac{2\pi r}{T^2}

, which incorrectly applies linear motion formulas to circular motion. Choice C (15.7 m/s²) comes from calculating

\frac{2\pi r}{T} = \frac{2\pi \times 5.0}{4.0} \approx 7.85

and then doubling it, confusing speed with acceleration. Choice D (31.4 m/s²) represents

2\pi^2 r/T = 2\pi^2 \times 5.0/4.0

, mixing up the correct centripetal acceleration formula. For circular motion problems, always start by identifying what you know (radius, period, or frequency) and remember that centripetal acceleration depends on the square of angular velocity. The key formulas are

\omega = 2\pi/T

and

a_c = \omega^2 r

Question 7

A velocity vector $\vec{v}$ has components $v_x = 8.0$ m/s and $v_y = -6.0$ m/s. After rotating this vector 90° counterclockwise, what are the new components?

$v_x = 6.0$ m/s, $v_y = 8.0$ m/s (correct answer)
$v_x = -6.0$ m/s, $v_y = -8.0$ m/s
$v_x = -8.0$ m/s, $v_y = 6.0$ m/s
$v_x = 6.0$ m/s, $v_y = -8.0$ m/s
$v_x = 8.0$ m/s, $v_y = 6.0$ m/s

Explanation: When you encounter vector rotation problems, you're dealing with coordinate transformations that preserve the vector's magnitude while changing its direction. The key is understanding how the components transform under rotation. To rotate a vector 90° counterclockwise, there's a systematic transformation rule: the new x-component becomes the negative of the original y-component, and the new y-component becomes the original x-component. Mathematically:

v_x' = -v_y

and

v_y' = v_x

. Applying this to your vector with

v_x = 8.0

m/s and

v_y = -6.0

m/s:

New x-component: $v_x' = -v_y = -(-6.0) = 6.0$ m/s
New y-component: $v_y' = v_x = 8.0$ m/s

This confirms answer A is correct. Looking at the wrong answers: Answer B gives

v_x = -6.0

m/s,

v_y = -8.0

m/s, which corresponds to a 90° clockwise rotation (the opposite direction). Answer C provides

v_x = -8.0

m/s,

v_y = 6.0

m/s, which would result from a 180° rotation. Answer D gives

v_x = 6.0

m/s,

v_y = -8.0

m/s, which represents a 270° counterclockwise rotation. Remember this pattern for 90° counterclockwise rotations: "flip and negate the first." The x and y components swap positions, and the new x-component gets a sign flip. This rule works because rotation matrices have this specific structure, making it a reliable shortcut for this common transformation.

Question 8

A boat travels 4.0 km due north, then 3.0 km due east, then 2.0 km due south. What is the magnitude of the boat's total displacement from its starting point?

9.0 km
5.0 km
3.6 km (correct answer)
4.2 km
7.0 km

Explanation: When you encounter displacement problems involving multiple movements in different directions, you're dealing with vector addition. Unlike distance (which adds up all the lengths traveled), displacement is the straight-line distance from start to finish, requiring you to consider direction. To solve this, set up a coordinate system and track the boat's position. Starting at the origin (0,0), the boat moves:

4.0 km north: position becomes (0, 4.0)
3.0 km east: position becomes (3.0, 4.0)
2.0 km south: final position is (3.0, 2.0)

The displacement is the straight-line distance from the starting point (0,0) to the final position (3.0, 2.0). Using the Pythagorean theorem:

d = \sqrt{(3.0)^2 + (2.0)^2} = \sqrt{9 + 4} = \sqrt{13} = 3.6 \text{ km}

Answer A (9.0 km) represents the total distance traveled, not displacement—this ignores that displacement is a vector quantity. Answer B (5.0 km) comes from incorrectly adding the final coordinates (3.0 + 2.0), forgetting to use the Pythagorean theorem. Answer D (4.2 km) might result from calculation errors or incorrectly handling the vector components. Remember: displacement problems require vector addition, not simple arithmetic. Always break movements into x and y components, find the final position, then calculate the straight-line distance back to the starting point using the Pythagorean theorem.

Question 9

Two vectors $\vec{M}$ and $\vec{N}$ lie in the xy-plane. Vector $\vec{M}$ has components (6.0, 8.0) and vector $\vec{N}$ has components (-3.0, 4.0). What is the cosine of the angle between vectors $\vec{M}$ and $\vec{N}$ ?

0.28 (correct answer)
0.56
0.80
0.14
-0.28

Explanation: When you encounter questions about the angle between vectors, you're dealing with the dot product formula:

\vec{M} \cdot \vec{N} = |\vec{M}||\vec{N}|\cos\theta

, where

\theta

is the angle between them. First, calculate the dot product:

\vec{M} \cdot \vec{N} = (6.0)(-3.0) + (8.0)(4.0) = -18.0 + 32.0 = 14.0

Next, find the magnitudes:

|\vec{M}| = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = 10.0

and

|\vec{N}| = \sqrt{(-3.0)^2 + 4.0^2} = \sqrt{9 + 16} = 5.0

Now solve for cosine:

\cos\theta = \frac{\vec{M} \cdot \vec{N}}{|\vec{M}||\vec{N}|} = \frac{14.0}{10.0 \times 5.0} = \frac{14.0}{50.0} = 0.28

This confirms answer (A) is correct. Looking at the wrong answers: (B) 0.56 is exactly double the correct answer, suggesting you might have made an error in the denominator calculation. (C) 0.80 could result from incorrectly using only the positive components in your dot product calculation (6×3 + 8×4 = 50, then 40/50 = 0.80). (D) 0.14 appears to be the raw dot product value without dividing by the product of magnitudes. Remember this three-step process: dot product, then magnitudes, then divide. Also, pay careful attention to negative components—they're crucial for getting the correct dot product value. The cosine will tell you if the angle is acute (positive) or obtuse (negative).

Question 10

A particle moves along a curved path from point A to point B. The displacement vector from A to B has magnitude 15 m and points due east. The particle travels a total distance of 25 m along the curved path. What is the magnitude of the average velocity vector if the trip takes 5.0 s?

3.0 m/s (correct answer)
5.0 m/s
8.0 m/s
4.0 m/s
Cannot be determined without knowing the path shape

Explanation: When you encounter problems involving curved motion, the key distinction is between displacement and distance traveled. These are fundamentally different quantities that lead to different velocity calculations. Average velocity is defined as displacement divided by time, not distance divided by time. Displacement is a vector quantity representing the straight-line path from start to finish, regardless of the actual route taken. Here, the displacement vector has magnitude 15 m pointing due east, and the time is 5.0 s. Therefore:

\text{Average velocity magnitude} = \frac{\text{displacement magnitude}}{\text{time}} = \frac{15 \text{ m}}{5.0 \text{ s}} = 3.0 \text{ m/s}

This confirms answer choice A is correct. The wrong answers stem from common misconceptions. Choice B (5.0 m/s) incorrectly uses distance instead of displacement:

\frac{25 \text{ m}}{5.0 \text{ s}} = 5.0 \text{ m/s}

. This actually gives you the average speed, not average velocity. Choice C (8.0 m/s) appears to result from calculation errors, possibly confusing the given values. Choice D (4.0 m/s) might come from incorrectly manipulating the displacement and distance values. Remember this crucial distinction: average velocity uses displacement (straight-line distance between start and end points), while average speed uses total distance traveled. The magnitude of average velocity will always be less than or equal to average speed when the path is curved, since displacement is the shortest possible distance between two points.

Question 11

Two vectors $\vec{A}$ and $\vec{B}$ have magnitudes of 8.0 and 6.0 units respectively. If their vector sum $\vec{A} + \vec{B}$ has a magnitude of 10.0 units, what is the magnitude of $\vec{A} - \vec{B}$ ?

2.0 units
6.0 units
10.0 units (correct answer)
14.0 units
Cannot be determined without knowing the angle between the vectors

Explanation: When you encounter vector problems involving magnitudes of sums and differences, think about the geometric relationship between the vectors. The key insight is that vector addition and subtraction follow specific mathematical relationships based on the angle between the vectors. Given that

|\vec{A}| = 8.0

|\vec{B}| = 6.0

, and

|\vec{A} + \vec{B}| = 10.0

, you can use the law of cosines. For vector addition:

|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta

, where

\theta

is the angle between the vectors. Substituting:

10^2 = 8^2 + 6^2 + 2(8)(6)\cos\theta

100 = 64 + 36 + 96\cos\theta

0 = 96\cos\theta

Therefore,

\cos\theta = 0

, meaning

\theta = 90°

. For vector subtraction with perpendicular vectors:

|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta

|\vec{A} - \vec{B}|^2 = 8^2 + 6^2 - 2(8)(6)(0) = 64 + 36 = 100

|\vec{A} - \vec{B}| = 10.0

units. Answer A (2.0) incorrectly assumes simple arithmetic subtraction of magnitudes. Answer B (6.0) might come from mistakenly using just one vector's magnitude. Answer D (14.0) incorrectly adds the magnitudes arithmetically. Remember: when

|\vec{A} + \vec{B}|^2 + |\vec{A} - \vec{B}|^2 = 2(|\vec{A}|^2 + |\vec{B}|^2)

, this relationship often helps verify your vector calculations quickly.

Question 12

Two displacement vectors $\vec{A}$ and $\vec{B}$ are added graphically using the head-to-tail method. If the angle between $\vec{A}$ and $\vec{B}$ is 120°, and both vectors have the same magnitude of 10.0 units, what is the magnitude of the resultant vector $\vec{A} + \vec{B}$ ?

20.0 units
17.3 units
10.0 units (correct answer)
14.1 units
5.0 units

Explanation: When you encounter vector addition problems, especially with specific angles given, you're dealing with the geometric nature of vectors where direction matters as much as magnitude. To find the resultant of two vectors using the head-to-tail method, you can use the law of cosines. When vectors

\vec{A}

and

\vec{B}

have magnitudes

A

and

B

with angle

\theta

between them, the magnitude of their sum is:

|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}

Here, both vectors have magnitude 10.0 units and the angle between them is 120°. Substituting:

|\vec{A} + \vec{B}| = \sqrt{10^2 + 10^2 + 2(10)(10)\cos(120°)}

Since

\cos(120°) = -0.5

|\vec{A} + \vec{B}| = \sqrt{100 + 100 + 200(-0.5)} = \sqrt{200 - 100} = \sqrt{100} = 10.0

Answer A (20.0 units) represents simple scalar addition, ignoring that vectors have direction—this would only be correct if the vectors were parallel (0° angle). Answer B (17.3 units) would result from using

\cos(30°)

instead of

\cos(120°)

, a common mistake when students confuse supplementary angles. Answer D (14.1 units) comes from using

\cos(45°)

, perhaps from misremembering the given angle. The correct answer is C (10.0 units). Remember: when two equal-magnitude vectors meet at 120°, they form an equilateral triangle with the resultant, making all sides equal. This geometric insight can help you check your algebra quickly.

Question 13

A projectile is launched horizontally from a cliff. After 2.0 seconds, its velocity vector makes an angle of 45° below the horizontal. What was the initial horizontal velocity of the projectile?

9.8 m/s
19.6 m/s (correct answer)
13.9 m/s
27.7 m/s
39.2 m/s

Explanation: When you encounter projectile motion problems involving angles, remember that horizontal and vertical velocity components behave independently. The horizontal velocity remains constant throughout flight, while the vertical velocity increases due to gravity. At any moment, the angle of the velocity vector below horizontal is determined by the ratio of vertical to horizontal velocity components. Since the velocity makes a 45° angle below horizontal after 2.0 seconds, we know that

\tan(45°) = 1 = \frac{v_y}{v_x}

, meaning the vertical and horizontal components are equal at that instant. The vertical velocity after 2.0 seconds is

v_y = gt = 9.8 \times 2.0 = 19.6 \text{ m/s}

. Since

v_y = v_x

at this moment, and the horizontal velocity never changes in projectile motion, the initial horizontal velocity must be 19.6 m/s. Looking at the wrong answers: (A) 9.8 m/s represents just the acceleration due to gravity, not the velocity after 2 seconds. (C) 13.9 m/s might result from incorrectly calculating

\frac{19.6}{\sqrt{2}}

, perhaps confusing velocity components with the total velocity magnitude. (D) 27.7 m/s could come from mistakenly adding the horizontal and vertical components instead of recognizing they're equal. The key insight is recognizing that when the angle is 45°, the horizontal and vertical velocity components are equal. Always identify what the angle tells you about the velocity component ratio, then use kinematic equations to find the specific values.

Question 14

Vector $\vec{F}$ has magnitude 15.0 N and points at 30.0° above the positive x-axis. What are the x and y components of the vector $2\vec{F}$ ?

$F_x = 26.0$ N, $F_y = 15.0$ N (correct answer)
$F_x = 15.0$ N, $F_y = 26.0$ N
$F_x = 13.0$ N, $F_y = 7.5$ N
$F_x = 30.0$ N, $F_y = 15.0$ N
$F_x = 7.5$ N, $F_y = 13.0$ N

Explanation: When you encounter vector problems involving components and scalar multiplication, remember that you need to find the components first, then apply any scalar operations to those components. To find the components of vector

\vec{F}

, use the standard trigonometric relationships:

F_x = F\cos\theta

and

F_y = F\sin\theta

. With magnitude 15.0 N and angle 30.0°:

F_x = 15.0 \cos(30.0°) = 15.0 \times 0.866 = 13.0 \text{ N}

F_y = 15.0 \sin(30.0°) = 15.0 \times 0.500 = 7.5 \text{ N}

Since we need the components of

2\vec{F}

, we multiply each component by 2:

(2\vec{F})_x = 2 \times 13.0 = 26.0 \text{ N}

(2\vec{F})_y = 2 \times 15.0 = 15.0 \text{ N}

This confirms answer A is correct. Answer B (

F_x = 15.0

F_y = 26.0

N) incorrectly swaps the x and y components, likely confusing sine and cosine functions. Answer C (

F_x = 13.0

F_y = 7.5

N) gives the components of the original vector

\vec{F}

without applying the scalar multiplication by 2. Answer D (

F_x = 30.0

F_y = 15.0

N) appears to multiply the magnitude by 2 for the x-component but uses the original magnitude for the y-component, showing incomplete application of scalar multiplication. Remember: when multiplying a vector by a scalar, multiply each component individually. Always double-check that you're using cosine for the x-component and sine for the y-component.

Question 15

Vector $\vec{W}$ makes an angle of 60° with the positive x-axis and has a magnitude of 20.0 units. If this vector is reflected across the x-axis, what are the components of the reflected vector?

$W_x = 10.0$ , $W_y = -17.3$ (correct answer)
$W_x = 17.3$ , $W_y = -10.0$
$W_x = -10.0$ , $W_y = 17.3$
$W_x = 10.0$ , $W_y = 17.3$
$W_x = -17.3$ , $W_y = 10.0$

Explanation: When dealing with vector reflections, you need to understand how each component transforms. Reflecting a vector across the x-axis keeps the x-component unchanged but reverses the sign of the y-component. First, let's find the original vector's components. For a vector with magnitude 20.0 units at 60° above the positive x-axis:

$W_x = 20.0 \cos(60°) = 20.0 \times 0.5 = 10.0$
$W_y = 20.0 \sin(60°) = 20.0 \times 0.866 = 17.3$

When reflected across the x-axis, the x-component stays the same (10.0), but the y-component becomes negative (-17.3). This gives us the reflected vector components:

W_x = 10.0

W_y = -17.3

. Answer A is correct with these exact values. Answer B incorrectly swaps the component values, giving you

\cos(60°)

for the y-component and

\sin(60°)

for the x-component - this suggests confusion about which trigonometric function corresponds to which component. Answer C has the wrong signs for both components, as if the vector were reflected across the y-axis instead. Answer D gives the original vector's components before any reflection, ignoring the transformation entirely. Remember: reflection across the x-axis only changes the sign of the y-component. The x-component always stays the same. This is because the x-axis acts like a mirror - distances from the axis (y-values) flip signs, but positions along the axis (x-values) remain unchanged.

Question 16

A position vector $\vec{r}(t)$ has components $x(t) = 3t^2$ and $y(t) = 4t - 2$ where distances are in meters and time is in seconds. What is the magnitude of the velocity vector at $t = 2.0$ s?

4.0 m/s
6.0 m/s
10.0 m/s
12.6 m/s (correct answer)
14.0 m/s

Explanation: When you encounter a position vector problem asking for velocity magnitude, you need to find the velocity components first, then calculate the magnitude using the Pythagorean theorem. To find velocity, take the derivative of each position component with respect to time. Given

x(t) = 3t^2

and

y(t) = 4t - 2

, the velocity components are:

$v_x(t) = \frac{dx}{dt} = 6t$
$v_y(t) = \frac{dy}{dt} = 4$

t = 2.0

$v_x(2) = 6(2) = 12$ m/s
$v_y(2) = 4$ m/s

The magnitude of velocity is

|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} = 12.6

m/s. Looking at the wrong answers: Choice A (4.0 m/s) likely comes from using only the y-component of velocity, ignoring the x-component entirely. Choice B (6.0 m/s) might result from using the coefficient 6 from the derivative without substituting

t = 2

. Choice C (10.0 m/s) could come from incorrectly adding the velocity components (12 + 4 = 16, then perhaps making an arithmetic error) rather than using the proper vector magnitude formula. Remember: velocity is the derivative of position, and vector magnitude always requires

\sqrt{v_x^2 + v_y^2}

, never simple addition of components. Always substitute the specific time value after taking derivatives.

Question 17

Two vectors $\vec{P}$ and $\vec{Q}$ satisfy the equation $2\vec{P} - 3\vec{Q} = \vec{R}$ , where $\vec{R}$ has magnitude 15 units and points in the positive x-direction. If $\vec{P} = 6\hat{i} + 8\hat{j}$ units, what is the y-component of vector $\vec{Q}$ ?

$\frac{8}{3}$ units because the y-components must balance in the vector equation
$\frac{16}{3}$ units because substitution and algebraic manipulation of the constraint equation yields this value (correct answer)
$-\frac{8}{3}$ units because the y-component of $\vec{Q}$ must be negative to satisfy the given constraint
$\frac{4}{3}$ units because this value makes the magnitude relationships consistent with the problem constraints

Explanation: Given:

2\vec{P} - 3\vec{Q} = \vec{R}

, where

\vec{R} = 15\hat{i}

and

\vec{P} = 6\hat{i} + 8\hat{j}

. Substituting:

2(6\hat{i} + 8\hat{j}) - 3\vec{Q} = 15\hat{i}

, which gives

12\hat{i} + 16\hat{j} - 3\vec{Q} = 15\hat{i}

. Rearranging:

3\vec{Q} = 12\hat{i} + 16\hat{j} - 15\hat{i} = -3\hat{i} + 16\hat{j}

. Therefore:

\vec{Q} = -\hat{i} + \frac{16}{3}\hat{j}

. The y-component of

\vec{Q}

\frac{16}{3}

units. Choice A incorrectly assumes the y-components of

2\vec{P}

and

3\vec{Q}

are equal rather than that

3Q_y = 16

. Choice C gets the wrong sign from algebraic errors. Choice D results from incorrectly dividing 16 by 12 instead of by 3.

Question 18

Consider the vector equation $\vec{A} \times \vec{B} = \vec{C}$ where $|\vec{A}| = 4$ , $|\vec{B}| = 3$ , and $|\vec{C}| = 6$ . If vector $\vec{D}$ is defined as $\vec{D} = \vec{A} + \vec{B}$ , what is the minimum possible magnitude of $\vec{D}$ ?

1 unit because $\vec{A}$ and $\vec{B}$ can be antiparallel while satisfying the cross product constraint
$\sqrt{13}$ units because the cross product constraint forces a specific angle between $\vec{A}$ and $\vec{B}$ (correct answer)
5 units because this represents the case when the vectors are perpendicular as required by the cross product
2 units because the constraint allows partial cancellation but not complete cancellation of the vectors

Explanation: From the cross product magnitude formula:

|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin\theta = 4 \times 3 \times \sin\theta = 12\sin\theta = 6

. Therefore

\sin\theta = \frac{1}{2}

, which gives

\theta = 30°

\theta = 150°

. For the magnitude of

\vec{D} = \vec{A} + \vec{B}

|\vec{D}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta = 16 + 9 + 24\cos\theta = 25 + 24\cos\theta

. For

\theta = 30°

\cos(30°) = \frac{\sqrt{3}}{2}

, so

|\vec{D}|^2 = 25 + 24 \cdot \frac{\sqrt{3}}{2} = 25 + 12\sqrt{3} \approx 45.78

. For

\theta = 150°

\cos(150°) = -\frac{\sqrt{3}}{2}

, so

|\vec{D}|^2 = 25 - 12\sqrt{3} \approx 4.22

. The minimum occurs at

\theta = 150°

, giving

|\vec{D}| = \sqrt{25 - 12\sqrt{3}} = \sqrt{13}

. Choice A is wrong because antiparallel vectors would give

\sin\theta = 0

, violating the cross product constraint. Choice C assumes perpendicularity but this isn't the minimum. Choice D is approximate but lacks precision.

Question 19

A displacement vector $\vec{d}$ has components $d_x = 12.0$ m and $d_y = -5.0$ m. If this vector is rotated counterclockwise by 90° about the origin, what are the components of the resulting vector $\vec{d}'$ ?

$(d'_x, d'_y) = (5.0, 12.0)$ m because rotation swaps components and changes signs appropriately (correct answer)
$(d'_x, d'_y) = (-12.0, 5.0)$ m because the x-component becomes negative and y-component becomes positive
$(d'_x, d'_y) = (5.0, -12.0)$ m because counterclockwise rotation preserves the magnitude but changes orientation
$(d'_x, d'_y) = (-5.0, -12.0)$ m because both components change sign during the 90° rotation

Explanation: For a 90° counterclockwise rotation, the transformation is

(x,y) \rightarrow (-y,x)

. Applying this to the given vector:

d'_x = -d_y = -(-5.0) = 5.0

m and

d'_y = d_x = 12.0

m. Therefore

(d'_x, d'_y) = (5.0, 12.0)

m. We can verify this preserves magnitude: original magnitude is

\sqrt{12^2 + (-5)^2} = 13

m, and new magnitude is

\sqrt{5^2 + 12^2} = 13

m. Choice B applies the wrong transformation

(x,y) \rightarrow (-x,y)

, which would be a reflection. Choice C gets the x-component correct but applies the wrong sign to the y-component. Choice D applies a 180° rotation transformation instead of 90°.