AP Physics C Mechanics Quiz: Scalars And Vectors
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Scalars And VectorsQuestion 1 of 20

Which of the following lists contains only vector quantities?

Velocity, acceleration, displacement, and force.
Speed, force, momentum, and electric charge.
Distance, work, power, and kinetic energy.
Temperature, mass, time, and electric potential.
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Scalars And Vectors

Practice Scalars And Vectors in AP Physics C Mechanics 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 AP Physics C Mechanics.

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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

Which of the following lists contains only vector quantities?

  1. Velocity, acceleration, displacement, and force. (correct answer)
  2. Speed, force, momentum, and electric charge.
  3. Distance, work, power, and kinetic energy.
  4. Temperature, mass, time, and electric potential.
Explanation: Vector quantities are defined as physical quantities that have both magnitude and direction. Velocity, acceleration, displacement, and force all fit this definition. The other choices contain scalar quantities: speed, distance, work, power, kinetic energy, electric charge, temperature, mass, time, and electric potential are all described by magnitude only.

Question 2

Two vectors P⃗\vec{P} and Q⃗\vec{Q} are added to form a resultant vector R⃗=P⃗+Q⃗\vec{R} = \vec{P} + \vec{Q}. If the magnitudes satisfy the relation ∣P⃗∣+∣Q⃗∣=∣R⃗∣|\vec{P}| + |\vec{Q}| = |\vec{R}|, what must be true about the vectors P⃗\vec{P} and Q⃗\vec{Q}?

  1. They are perpendicular to each other.
  2. They are in opposite directions.
  3. They are in the same direction. (correct answer)
  4. One of the vectors must be the zero vector.
Explanation: The magnitude of the sum of two vectors is equal to the sum of their individual magnitudes only in the specific case where the vectors are parallel and point in the same direction (an angle of 0∘0^\circ between them). In all other cases, due to the triangle inequality, the magnitude of the resultant vector will be less than the sum of the individual magnitudes.

Question 3

A particle's velocity vector is v⃗=(−3.0i^+4.0j^)\vec{v} = (-3.0 \hat{i} + 4.0 \hat{j}) m/s. What is the approximate angle of the velocity vector, measured counterclockwise from the positive x-axis?

  1. 53.1°
  2. 126.9° (correct answer)
  3. 233.1°
  4. 306.9°
Explanation: The vector has a negative x-component and a positive y-component, placing it in the second quadrant. The reference angle relative to the negative x-axis can be found using the arctangent: α=arctan⁡∣vyvx∣=arctan⁡∣4.0−3.0∣≈53.1∘\alpha = \arctan{\left|\frac{v_y}{v_x}\right|} = \arctan{\left|\frac{4.0}{-3.0}\right|} \approx 53.1^\circ. The angle measured counterclockwise from the positive x-axis is 180∘−α=180∘−53.1∘=126.9∘180^\circ - \alpha = 180^\circ - 53.1^\circ = 126.9^\circ. Distractor A is only the reference angle. The other distractors correspond to angles in the third and fourth quadrants.

Question 4

A particle's initial position is r⃗i=(2i^−3j^+5k^)\vec{r}_i = (2\hat{i} - 3\hat{j} + 5\hat{k}) m. Its final position is r⃗f=(−4i^+6j^+2k^)\vec{r}_f = (-4\hat{i} + 6\hat{j} + 2\hat{k}) m. What is the displacement vector Δr⃗\Delta\vec{r} of the particle?

  1. (−2i^+3j^+7k^)(-2\hat{i} + 3\hat{j} + 7\hat{k}) m
  2. (6i^−9j^+3k^)(6\hat{i} - 9\hat{j} + 3\hat{k}) m
  3. (−6i^+9j^−3k^)(-6\hat{i} + 9\hat{j} - 3\hat{k}) m (correct answer)
  4. (−6i^+3j^+7k^)(-6\hat{i} + 3\hat{j} + 7\hat{k}) m
Explanation: The displacement vector is the change in position, calculated as Δr⃗=r⃗f−r⃗i\Delta\vec{r} = \vec{r}_f - \vec{r}_i. Subtracting the components: x:(−4)−(2)=−6x: (-4) - (2) = -6. y:(6)−(−3)=9y: (6) - (-3) = 9. z:(2)−(5)=−3z: (2) - (5) = -3. Therefore, Δr⃗=(−6i^+9j^−3k^)\Delta\vec{r} = (-6\hat{i} + 9\hat{j} - 3\hat{k}) m. Distractor A represents the sum r⃗i+r⃗f\vec{r}_i + \vec{r}_f. Distractor B represents the vector r⃗i−r⃗f\vec{r}_i - \vec{r}_f. Distractor D contains calculation errors.

Question 5

A particle undergoes a displacement of 6.06.0 m at an angle of 30∘30^\circ north of east, followed by a second displacement of 4.04.0 m at an angle of 60∘60^\circ west of north. What is the approximate magnitude of the particle's resultant displacement?

  1. 2.0 m
  2. 5.3 m (correct answer)
  3. 7.2 m
  4. 10.0 m
Explanation: First, resolve each displacement vector into its x (east) and y (north) components. For the first displacement: d1x=6.0cos⁡(30∘)≈5.20d_{1x} = 6.0 \cos(30^\circ) \approx 5.20 m, d1y=6.0sin⁡(30∘)=3.00d_{1y} = 6.0 \sin(30^\circ) = 3.00 m. For the second displacement: d2x=−4.0sin⁡(60∘)≈−3.46d_{2x} = -4.0 \sin(60^\circ) \approx -3.46 m, d2y=4.0cos⁡(60∘)=2.00d_{2y} = 4.0 \cos(60^\circ) = 2.00 m. The resultant components are Rx=5.20−3.46=1.74R_x = 5.20 - 3.46 = 1.74 m and Ry=3.00+2.00=5.00R_y = 3.00 + 2.00 = 5.00 m. The magnitude of the resultant displacement is found using the Pythagorean theorem: ∣R⃗∣=Rx2+Ry2=(1.74)2+(5.00)2≈5.3|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(1.74)^2 + (5.00)^2} \approx 5.3 m. Distractor D is the scalar sum of the magnitudes.

Question 6

An athlete runs exactly one lap around a circular track of radius RR. Which of the following statements about the athlete's motion is correct?

  1. The magnitude of the displacement is 2πR2\pi R and the distance traveled is zero.
  2. The magnitude of the displacement is zero and the distance traveled is 2πR2\pi R. (correct answer)
  3. Both the magnitude of the displacement and the distance traveled are equal to 2πR2\pi R.
  4. Both the magnitude of the displacement and the distance traveled are zero.
Explanation: Displacement is a vector quantity representing the change in position from the start point to the end point. Since the athlete finishes at the same point they started, the displacement vector is zero. Distance is a scalar quantity representing the total path length covered. For one lap around a circular track, the distance is the circumference, 2πR2\pi R.

Question 7

The position of a particle as a function of time is given by the vector r⃗(t)=(3t2−4t)i^+(2t3)j^−(5t)k^\vec{r}(t) = (3t^2 - 4t)\hat{i} + (2t^3)\hat{j} - (5t)\hat{k}, where tt is in seconds and r⃗\vec{r} is in meters. What is the particle's velocity vector v⃗(t)\vec{v}(t)?

  1. v⃗(t)=(6t−4)i^+6t2j^−5k^\vec{v}(t) = (6t - 4)\hat{i} + 6t^2\hat{j} - 5\hat{k} (correct answer)
  2. v⃗(t)=(t3−2t2)i^+(12t4)j^−(52t2)k^\vec{v}(t) = (t^3 - 2t^2)\hat{i} + (\frac{1}{2}t^4)\hat{j} - (\frac{5}{2}t^2)\hat{k}
  3. v⃗(t)=6i^+12tj^\vec{v}(t) = 6\hat{i} + 12t\hat{j}
  4. v⃗(t)=(3t2−4t)i^+(2t3)j^\vec{v}(t) = (3t^2 - 4t)\hat{i} + (2t^3)\hat{j}
Explanation: Velocity is the time derivative of the position vector, v⃗(t)=dr⃗dt\vec{v}(t) = \frac{d\vec{r}}{dt}. We must differentiate each component of r⃗(t)\vec{r}(t) with respect to time. ddt(3t2−4t)=6t−4\frac{d}{dt}(3t^2 - 4t) = 6t - 4. ddt(2t3)=6t2\frac{d}{dt}(2t^3) = 6t^2. ddt(−5t)=−5\frac{d}{dt}(-5t) = -5. Combining these gives the correct velocity vector. Distractor B is the integral of the position vector. Distractor C is the acceleration vector (the derivative of velocity). Distractor D omits the k-component and does not perform the differentiation.

Question 8

Two force vectors, F⃗1\vec{F}_1 and F⃗2\vec{F}_2, have magnitudes of 8 N and 6 N, respectively. They are applied to an object at the same point. Which of the following is NOT a possible magnitude for the resultant force F⃗res=F⃗1+F⃗2\vec{F}_{res} = \vec{F}_1 + \vec{F}_2?

  1. 1 N (correct answer)
  2. 5 N
  3. 10 N
  4. 14 N
Explanation: The magnitude of the resultant of two vectors depends on the angle between them. The maximum possible magnitude occurs when the vectors are parallel (angle 0∘0^\circ), resulting in a magnitude of 8+6=148 + 6 = 14 N. The minimum possible magnitude occurs when the vectors are anti-parallel (angle 180∘180^\circ), resulting in a magnitude of ∣8−6∣=2|8 - 6| = 2 N. All other possible magnitudes lie between these two extremes. Therefore, a magnitude of 1 N is not possible.

Question 9

A particle has an initial velocity v⃗i=(4i^+2j^)\vec{v}_i = (4\hat{i} + 2\hat{j}) m/s and a final velocity v⃗f=(−1i^+5j^)\vec{v}_f = (-1\hat{i} + 5\hat{j}) m/s. What is the change in velocity, Δv⃗=v⃗f−v⃗i\Delta\vec{v} = \vec{v}_f - \vec{v}_i?

  1. (−5i^+3j^)(-5\hat{i} + 3\hat{j}) m/s (correct answer)
  2. (5i^−3j^)(5\hat{i} - 3\hat{j}) m/s
  3. (3i^+7j^)(3\hat{i} + 7\hat{j}) m/s
  4. (5i^+3j^)(5\hat{i} + 3\hat{j}) m/s
Explanation: The change in velocity is found by subtracting the initial velocity vector from the final velocity vector component by component. Δv⃗=v⃗f−v⃗i=((−1)−4)i^+(5−2)j^=(−5i^+3j^)\Delta\vec{v} = \vec{v}_f - \vec{v}_i = ((-1) - 4)\hat{i} + (5 - 2)\hat{j} = (-5\hat{i} + 3\hat{j}) m/s. Distractor B represents v⃗i−v⃗f\vec{v}_i - \vec{v}_f. Distractor C represents v⃗i+v⃗f\vec{v}_i + \vec{v}_f. Distractor D contains calculation errors.

Question 10

A constant force F⃗=(2i^+4j^)\vec{F} = (2\hat{i} + 4\hat{j}) N acts on an object that undergoes a displacement Δr⃗=(3i^−1j^)\Delta\vec{r} = (3\hat{i} - 1\hat{j}) m. The work done by the force is given by the scalar product W=F⃗⋅Δr⃗W = \vec{F} \cdot \Delta\vec{r}. What is the work done?

  1. 2 J (correct answer)
  2. 8 J
  3. 14 J
  4. A vector quantity with magnitude 2 N·m
Explanation: The scalar (dot) product is calculated as W=FxΔx+FyΔyW = F_x \Delta x + F_y \Delta y. Substituting the given components: W=(2)(3)+(4)(−1)=6−4=2W = (2)(3) + (4)(-1) = 6 - 4 = 2 J. The result of a scalar product is a scalar quantity, so work is a scalar, making choice D incorrect. Other distractors result from incorrect vector operations.

Question 11

The torque τ⃗\vec{\tau} about a pivot point is given by the vector product τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}, where r⃗\vec{r} is the position vector from the pivot to the point of force application and F⃗\vec{F} is the force vector. If r⃗=(2j^)\vec{r} = (2\hat{j}) m and F⃗=(3i^)\vec{F} = (3\hat{i}) N, what is the torque τ⃗\vec{\tau}?

  1. 0 N·m
  2. 6 N·m
  3. 6k^6\hat{k} N·m
  4. −6k^-6\hat{k} N·m (correct answer)
Explanation: The vector (cross) product is calculated using the properties of unit vectors: i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j}, and the anti-commutative property (A⃗×B⃗=−B⃗×A⃗\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}). Thus, τ⃗=(2j^)×(3i^)=6(j^×i^)=6(−k^)=−6k^\vec{\tau} = (2\hat{j}) \times (3\hat{i}) = 6 (\hat{j} \times \hat{i}) = 6(-\hat{k}) = -6\hat{k} N·m. Distractor A is the scalar product. Distractor C has the incorrect sign.

Question 12

Vector A⃗\vec{A} has magnitude 5 and vector B⃗\vec{B} has magnitude 3. The magnitude of their sum, ∣A⃗+B⃗∣|\vec{A} + \vec{B}|, is 4. What is the approximate angle between vectors A⃗\vec{A} and B⃗\vec{B} when they are placed tail-to-tail?

  1. 0°
  2. 90°
  3. 127° (correct answer)
  4. 180°
Explanation: The law of cosines for vector addition states ∣R⃗∣2=∣A⃗∣2+∣B⃗∣2+2∣A⃗∣∣B⃗∣cos⁡θ|\vec{R}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta, where θ\theta is the angle between the vectors when placed tail-to-tail. Substituting the given values: 42=52+32+2(5)(3)cos⁡θ4^2 = 5^2 + 3^2 + 2(5)(3)\cos\theta. This simplifies to 16=25+9+30cos⁡θ16 = 25 + 9 + 30\cos\theta, so 16=34+30cos⁡θ16 = 34 + 30\cos\theta. Solving for cos⁡θ\cos\theta gives cos⁡θ=(16−34)/30=−18/30=−0.6\cos\theta = (16 - 34) / 30 = -18 / 30 = -0.6. The angle is θ=arccos⁡(−0.6)≈127∘\theta = \arccos(-0.6) \approx 127^\circ.

Question 13

A car travels east at 20 m/s for 10 s, then turns and travels north at 20 m/s for 10 s. Which statement correctly compares the average speed and the magnitude of the average velocity for the entire 20 s trip?

  1. The average speed is equal to the magnitude of the average velocity because the speed was constant.
  2. The average speed is greater than the magnitude of the average velocity because the distance traveled is greater than the displacement. (correct answer)
  3. The average speed is less than the magnitude of the average velocity because the turn adds to the overall motion.
  4. The relationship cannot be determined without knowing the path of the turn.
Explanation: The total distance traveled is (20 m/s)(10 s)+(20 m/s)(10 s)=400(20 \text{ m/s})(10 \text{ s}) + (20 \text{ m/s})(10 \text{ s}) = 400 m. The average speed is distance/time = 400 m / 20 s = 20 m/s. The displacement is the vector sum of 200 m east and 200 m north. The magnitude of the displacement is 2002+2002≈283\sqrt{200^2 + 200^2} \approx 283 m. The magnitude of the average velocity is |displacement|/time = 283 m / 20 s ≈14.1 \approx 14.1 m/s. Since the path was not a straight line, the distance is greater than the magnitude of displacement, making the average speed greater than the magnitude of average velocity.

Question 14

A block of mass mm rests on an inclined plane that makes an angle θ\theta with the horizontal. What is the component of the gravitational force vector, F⃗g\vec{F}_g, that is parallel to the surface of the plane?

  1. mgsin⁡θmg \sin\theta (correct answer)
  2. mgcos⁡θmg \cos\theta
  3. mgmg
  4. mgtan⁡θmg \tan\theta
Explanation: The gravitational force vector F⃗g\vec{F}_g has a magnitude of mgmg and is directed vertically downward. When resolving this vector into components parallel and perpendicular to the inclined plane, the angle between the vertical gravitational force and the perpendicular to the plane is θ\theta. Therefore, the component parallel to the plane is given by mgsin⁡θmg \sin\theta, and the component perpendicular to the plane is mgcos⁡θmg \cos\theta.

Question 15

Two non-zero vectors A⃗\vec{A} and B⃗\vec{B} have equal magnitudes. If the magnitude of their difference, ∣A⃗−B⃗∣|\vec{A} - \vec{B}|, is also equal to the magnitude of A⃗\vec{A}, what is the angle between vectors A⃗\vec{A} and B⃗\vec{B} when placed tail-to-tail?

  1. 0°
  2. 60° (correct answer)
  3. 90°
  4. 120°
Explanation: Let the magnitude of each vector be xx. The law of cosines for vector subtraction is ∣A⃗−B⃗∣2=∣A⃗∣2+∣B⃗∣2−2∣A⃗∣∣B⃗∣cos⁡θ|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta. Substituting xx for all magnitudes gives x2=x2+x2−2(x)(x)cos⁡θx^2 = x^2 + x^2 - 2(x)(x)\cos\theta. This simplifies to 0=x2−2x2cos⁡θ0 = x^2 - 2x^2\cos\theta. Since xx is non-zero, we can divide by x2x^2 to get 0=1−2cos⁡θ0 = 1 - 2\cos\theta, which gives cos⁡θ=1/2\cos\theta = 1/2. The angle for which this is true is 60∘60^\circ.

Question 16

A vector is given by A⃗=6i^+8j^\vec{A} = 6\hat{i} + 8\hat{j}. A second vector is given by B⃗=2i^\vec{B} = 2\hat{i}. What is the magnitude of the component of vector A⃗\vec{A} that is parallel to the direction of vector B⃗\vec{B}?

  1. 6 (correct answer)
  2. 8
  3. 10
  4. 12
Explanation: The magnitude of the component of vector A⃗\vec{A} along the direction of vector B⃗\vec{B} is found by the scalar projection, which is calculated as the dot product of A⃗\vec{A} with the unit vector of B⃗\vec{B}. The unit vector for B⃗\vec{B} is B^=2i^∣2∣=i^\hat{B} = \frac{2\hat{i}}{|2|} = \hat{i}. The scalar projection is then A⃗⋅B^=(6i^+8j^)⋅(i^)=(6)(1)+(8)(0)=6\vec{A} \cdot \hat{B} = (6\hat{i} + 8\hat{j}) \cdot (\hat{i}) = (6)(1) + (8)(0) = 6. Distractor D is the dot product A⃗⋅B⃗\vec{A} \cdot \vec{B}, not the projection.

Question 17

Three displacement vectors are given by A⃗=2i^−3j^\vec{A} = 2\hat{i} - 3\hat{j}, B⃗=−4i^−j^\vec{B} = -4\hat{i} - \hat{j}, and C⃗=i^+5j^\vec{C} = \hat{i} + 5\hat{j}. What is the resultant displacement vector R⃗=A⃗+B⃗+C⃗\vec{R} = \vec{A} + \vec{B} + \vec{C}?

  1. R⃗=−i^+j^\vec{R} = -\hat{i} + \hat{j} (correct answer)
  2. R⃗=−i^+9j^\vec{R} = -\hat{i} + 9\hat{j}
  3. R⃗=7i^+j^\vec{R} = 7\hat{i} + \hat{j}
  4. R⃗=7i^+9j^\vec{R} = 7\hat{i} + 9\hat{j}
Explanation: To find the resultant vector, add the corresponding components of the individual vectors. For the x-components: Rx=Ax+Bx+Cx=2+(−4)+1=−1R_x = A_x + B_x + C_x = 2 + (-4) + 1 = -1. For the y-components: Ry=Ay+By+Cy=(−3)+(−1)+5=1R_y = A_y + B_y + C_y = (-3) + (-1) + 5 = 1. Therefore, the resultant vector is R⃗=−1i^+1j^=−i^+j^\vec{R} = -1\hat{i} + 1\hat{j} = -\hat{i} + \hat{j}.

Question 18

A particle moves along a curved path in three-dimensional space. Its velocity vector is v⃗\vec{v} and its acceleration vector is a⃗\vec{a}. Which of the following scalar quantities is equivalent to the instantaneous rate of change of the particle's kinetic energy with respect to time?

  1. m(v⃗⋅a⃗)m (\vec{v} \cdot \vec{a}) (correct answer)
  2. m∣v⃗×a⃗∣m |\vec{v} \times \vec{a}|
  3. md∣v⃗∣dtm \frac{d|\vec{v}|}{dt}
  4. 12m∣a⃗∣2\frac{1}{2}m|\vec{a}|^2
Explanation: When you encounter questions about the rate of change of kinetic energy, think about the fundamental relationship between energy and motion. The key insight is recognizing that kinetic energy KE=12m∣v⃗∣2KE = \frac{1}{2}m|\vec{v}|^2 depends on the magnitude of velocity squared. To find how kinetic energy changes with time, you need to take the derivative: dKEdt=ddt(12m∣v⃗∣2)=12m⋅2∣v⃗∣d∣v⃗∣dt\frac{dKE}{dt} = \frac{d}{dt}\left(\frac{1}{2}m|\vec{v}|^2\right) = \frac{1}{2}m \cdot 2|\vec{v}|\frac{d|\vec{v}|}{dt}. However, there's a more elegant approach using the velocity vector directly. Since v⃗⋅v⃗=∣v⃗∣2\vec{v} \cdot \vec{v} = |\vec{v}|^2, we can write KE=12m(v⃗⋅v⃗)KE = \frac{1}{2}m(\vec{v} \cdot \vec{v}). Taking the time derivative: dKEdt=12m(dv⃗dt⋅v⃗+v⃗⋅dv⃗dt)=ma⃗⋅v⃗=m(v⃗⋅a⃗)\frac{dKE}{dt} = \frac{1}{2}m\left(\frac{d\vec{v}}{dt} \cdot \vec{v} + \vec{v} \cdot \frac{d\vec{v}}{dt}\right) = m\vec{a} \cdot \vec{v} = m(\vec{v} \cdot \vec{a}), confirming answer A. Answer B, m∣v⃗×a⃗∣m|\vec{v} \times \vec{a}|, gives the magnitude of the cross product, which relates to the component of acceleration perpendicular to velocity—this doesn't affect kinetic energy. Answer C, md∣v⃗∣dtm\frac{d|\vec{v}|}{dt}, is missing the factor of ∣v⃗∣|\vec{v}| needed from the chain rule. Answer D, 12m∣a⃗∣2\frac{1}{2}m|\vec{a}|^2, has units of power but represents something entirely different—it's not derived from differentiating kinetic energy. Remember: the dot product v⃗⋅a⃗\vec{v} \cdot \vec{a} captures how much acceleration aligns with velocity, which directly determines whether kinetic energy increases or decreases. This connection appears frequently in mechanics problems involving energy and power.

Question 19

A particle's position vector is given by r⃗(t)=(2t)i^+(3t2−1)j^\vec{r}(t) = (2t)\hat{i} + (3t^2 - 1)\hat{j} meters. Its velocity vector is v⃗(t)\vec{v}(t). Let C⃗=r⃗×v⃗\vec{C} = \vec{r} \times \vec{v}. Which of the following best describes the vector C⃗\vec{C} at time t=1t=1 s?

  1. A zero vector.
  2. A non-zero vector in the xyxy-plane.
  3. A non-zero vector perpendicular to the xyxy-plane in the negative k^\hat{k} direction.
  4. A non-zero vector perpendicular to the xyxy-plane in the positive k^\hat{k} direction. (correct answer)
Explanation: When you encounter cross products involving position and velocity vectors, you're dealing with angular momentum concepts. The cross product r⃗×v⃗\vec{r} \times \vec{v} gives a vector perpendicular to both r⃗\vec{r} and v⃗\vec{v}, and since both vectors lie in the xyxy-plane, their cross product must point along the zz-axis. First, find the velocity by differentiating the position vector: v⃗(t)=dr⃗dt=2i^+6tj^\vec{v}(t) = \frac{d\vec{r}}{dt} = 2\hat{i} + 6t\hat{j}. At t=1t = 1 s, you have r⃗(1)=2i^+2j^\vec{r}(1) = 2\hat{i} + 2\hat{j} and v⃗(1)=2i^+6j^\vec{v}(1) = 2\hat{i} + 6\hat{j}. Now calculate the cross product: C⃗=r⃗×v⃗=(2i^+2j^)×(2i^+6j^)\vec{C} = \vec{r} \times \vec{v} = (2\hat{i} + 2\hat{j}) \times (2\hat{i} + 6\hat{j}). Using the distributive property: C⃗=2⋅2(i^×i^)+2⋅6(i^×j^)+2⋅2(j^×i^)+2⋅6(j^×j^)\vec{C} = 2 \cdot 2(\hat{i} \times \hat{i}) + 2 \cdot 6(\hat{i} \times \hat{j}) + 2 \cdot 2(\hat{j} \times \hat{i}) + 2 \cdot 6(\hat{j} \times \hat{j}). Since i^×i^=0\hat{i} \times \hat{i} = 0, j^×j^=0\hat{j} \times \hat{j} = 0, i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, and j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}, you get: C⃗=12k^−4k^=8k^\vec{C} = 12\hat{k} - 4\hat{k} = 8\hat{k}. This confirms answer D is correct. Answer A is wrong because the cross product equals 8k^8\hat{k}, not zero. Answer B is incorrect since cross products of vectors in the xyxy-plane cannot lie in that same plane. Answer C has the wrong direction—the result is positive k^\hat{k}, not negative. Remember: when computing r⃗×v⃗\vec{r} \times \vec{v} for motion in a plane, use the right-hand rule or systematically apply cross product properties to determine the direction of the resulting vector.

Question 20

Two vectors are defined as A⃗=3i^+4j^\vec{A} = 3\hat{i} + 4\hat{j} and B⃗=−8i^+6j^\vec{B} = -8\hat{i} + 6\hat{j} in a standard Cartesian coordinate system S. The coordinate system is then rotated counterclockwise by 30 degrees to form a new system S', with new basis vectors i^′\hat{i}' and j^′\hat{j}'. Which of the following quantities has a different value when calculated in system S' compared to its value in system S?

  1. The magnitude of vector A⃗\vec{A}
  2. The dot product A⃗⋅B⃗\vec{A} \cdot \vec{B}
  3. The angle between vectors A⃗\vec{A} and B⃗\vec{B}
  4. The component of vector B⃗\vec{B} along the i^′\hat{i}' direction (correct answer)
Explanation: When you encounter coordinate system rotations in physics, remember that some quantities are invariant (unchanged) while others depend on your choice of coordinate system. This distinction is crucial for understanding vectors in different reference frames. Let's examine what happens when we rotate the coordinate system by 30°. The key insight is that scalar quantities and relationships between vectors remain the same, while individual vector components change. The correct answer is D because the component of B⃗\vec{B} along i^′\hat{i}' represents a projection onto a new axis direction. In system S, B⃗\vec{B} has an x-component of -8. After rotation, the i^′\hat{i}' direction points differently in space, so B⃗\vec{B}'s component along this new direction will be different. You'd calculate this using: Bi′=Bxcos⁡(30°)+Bysin⁡(30°)=−8(32)+6(12)=−43+3≠−8B_{i'} = B_x \cos(30°) + B_y \sin(30°) = -8(\frac{\sqrt{3}}{2}) + 6(\frac{1}{2}) = -4\sqrt{3} + 3 \neq -8. Choice A is wrong because vector magnitude is invariant under rotation—∣A⃗∣=32+42=5|\vec{A}| = \sqrt{3^2 + 4^2} = 5 regardless of coordinate system. Choice B is incorrect since the dot product A⃗⋅B⃗=(3)(−8)+(4)(6)=0\vec{A} \cdot \vec{B} = (3)(-8) + (4)(6) = 0 remains unchanged because it represents the intrinsic relationship between vectors. Choice C is wrong because the angle between vectors is a geometric property independent of how you orient your axes. Study tip: Remember that coordinate rotations change components but preserve magnitudes, dot products, and angles. When you see rotation problems, ask yourself: "Is this quantity about the vectors themselves, or about how we measure them?"