Scalars and Vectors - AP Physics C Mechanics

1. A particle's velocity vector is $\vec{v} = (-3.0 \hat{i} + 4.0 \hat{j})$ m/s. What is the approximate angle of the velocity vector, measured counterclockwise from the positive x-axis?
   Answer: 126.9° 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: $\alpha = \arctan{\left|\frac{v_y}{v_x}\right|} = \arctan{\left|\frac{4.0}{-3.0}\right|} \approx 53.1^\circ$. The angle measured counterclockwise from the positive x-axis is $180^\circ - \alpha = 180^\circ - 53.1^\circ = 126.9^\circ$. Distractor A is only the reference angle. The other distractors correspond to angles in the third and fourth quadrants.
2. An athlete runs exactly one lap around a circular track of radius $R$. Which of the following statements about the athlete's motion is correct?
   Answer: The magnitude of the displacement is zero and the distance traveled is $2\pi R$. 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, $2\pi R$.
3. A particle has an initial velocity $\vec{v}_i = (4\hat{i} + 2\hat{j})$ m/s and a final velocity $\vec{v}_f = (-1\hat{i} + 5\hat{j})$ m/s. What is the change in velocity, $\Delta\vec{v} = \vec{v}_f - \vec{v}_i$?
   Answer: $(-5\hat{i} + 3\hat{j})$ m/s The change in velocity is found by subtracting the initial velocity vector from the final velocity vector component by component. $\Delta\vec{v} = \vec{v}_f - \vec{v}_i = ((-1) - 4)\hat{i} + (5 - 2)\hat{j} = (-5\hat{i} + 3\hat{j})$ m/s. Distractor B represents $\vec{v}_i - \vec{v}_f$. Distractor C represents $\vec{v}_i + \vec{v}_f$. Distractor D contains calculation errors.
4. The torque $\vec{\tau}$ about a pivot point is given by the vector product $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is the position vector from the pivot to the point of force application and $\vec{F}$ is the force vector. If $\vec{r} = (2\hat{j})$ m and $\vec{F} = (3\hat{i})$ N, what is the torque $\vec{\tau}$?
   Answer: $-6\hat{k}$ N·m The vector (cross) product is calculated using the properties of unit vectors: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$, and the anti-commutative property ($\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$). Thus, $\vec{\tau} = (2\hat{j}) \times (3\hat{i}) = 6 (\hat{j} \times \hat{i}) = 6(-\hat{k}) = -6\hat{k}$ N·m. Distractor A is the scalar product. Distractor C has the incorrect sign.
5. Vector $\vec{A}$ has a magnitude of 8 units, and vector $\vec{B}$ has a magnitude of 5 units. The vector $\vec{C}$ is defined as $\vec{C} = \vec{A} - \vec{B}$. Which of the following is a possible magnitude for vector $\vec{C}$?
   Answer: 12 When working with vector subtraction, you need to consider that the magnitude of the resultant vector depends on both the magnitudes of the individual vectors and the angle between them. For $\vec{C} = \vec{A} - \vec{B}$, the magnitude of $\vec{C}$ can range from the absolute difference to the sum of the magnitudes. The key insight is that vector subtraction $\vec{A} - \vec{B}$ is equivalent to $\vec{A} + (-\vec{B})$, where $-\vec{B}$ has the same magnitude as $\vec{B}$ but points in the opposite direction. Using the triangle inequality, the magnitude of $\vec{C}$ must satisfy: $|\vec{A}| - |\vec{B}| \leq |\vec{C}| \leq |\vec{A}| + |\vec{B}|$ With $|\vec{A}| = 8$ and $|\vec{B}| = 5$, this gives us: $3 \leq |\vec{C}| \leq 13$ The minimum occurs when vectors $\vec{A}$ and $\vec{B}$ point in the same direction (making $\vec{A}$ and $-\vec{B}$ opposite), and the maximum occurs when they point in opposite directions (making $\vec{A}$ and $-\vec{B}$ parallel). Answer B (12) falls within this valid range of 3 to 13 units, making it correct. Answer A (2) is too small—it's less than the minimum possible value of 3. Answer C (14) exceeds the maximum possible value of 13. Answer D (0) is impossible since the minimum magnitude is 3 when the vectors have different magnitudes. Remember: For any vector operation involving magnitudes, always check whether your answer falls within the physically possible range determined by the triangle inequality.
6. A vector $\vec{F}$ can be written as $\vec{F} = F_x\hat{i} + F_y\hat{j}$ where $F_x = 12$ N and $F_y = -5$ N. If this vector is rotated counterclockwise by $90°$ about the origin, what are the components of the resulting vector $\vec{F}'$?
   Answer: $F'_x = 5$ N, $F'_y = 12$ N When a vector is rotated counterclockwise by 90°, the transformation is: $\vec{F}' = -F_y\hat{i} + F_x\hat{j}$. With $F_x = 12$ N and $F_y = -5$ N, we get: $F'_x = -F_y = -(-5) = 5$ N and $F'_y = F_x = 12$ N. Choice B results from incorrectly applying $F'_x = F_y$ and $F'_y = -F_x$. Choice C comes from using the clockwise rotation formula. Choice D results from simply changing the sign of $F_y$ without proper rotation.
7. A position vector $\vec{r}$ makes an angle of $\theta$ with the positive x-axis and has magnitude $r$. The vector is then rotated clockwise by angle $\phi$ to create a new vector $\vec{r}'$. Which expression correctly represents the x-component of $\vec{r}'$?
   Answer: $r\cos(\theta - \phi)$ When a vector is rotated clockwise by angle $\phi$, the new angle with respect to the positive x-axis becomes $\theta - \phi$ (since clockwise rotation decreases the angle). The x-component of the rotated vector is $r'_x = r\cos(\theta - \phi)$. Choice B represents counterclockwise rotation. Choices C and D incorrectly use sine instead of cosine for the x-component.
8. A particle's velocity changes from $\vec{v}_i = (3.0\hat{i} - 4.0\hat{j})$ m/s to $\vec{v}_f = (5.0\hat{i} + 4.0\hat{j})$ m/s. The average acceleration is in the same direction as the change in velocity, $\Delta\vec{v}$. What is the unit vector $\hat{u}$ in the direction of the average acceleration?
   Answer: $\frac{1}{\sqrt{17}}\hat{i} + \frac{4}{\sqrt{17}}\hat{j}$ When you encounter vector problems involving velocity changes, remember that average acceleration points in the same direction as the change in velocity $\Delta\vec{v}$. To find a unit vector in this direction, you need to calculate $\Delta\vec{v}$ and then normalize it. First, find the change in velocity: $\Delta\vec{v} = \vec{v}_f - \vec{v}_i = (5.0\hat{i} + 4.0\hat{j}) - (3.0\hat{i} - 4.0\hat{j}) = 2.0\hat{i} + 8.0\hat{j}$ m/s. Next, calculate the magnitude: $|\Delta\vec{v}| = \sqrt{(2.0)^2 + (8.0)^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}$ m/s. The unit vector is: $\hat{u} = \frac{\Delta\vec{v}}{|\Delta\vec{v}|} = \frac{2.0\hat{i} + 8.0\hat{j}}{2\sqrt{17}} = \frac{1}{\sqrt{17}}\hat{i} + \frac{4}{\sqrt{17}}\hat{j}$ This confirms answer A is correct. Let's examine the wrong answers: Answer B has the wrong sign on the $\hat{i}$ component—this would result from incorrectly calculating $\vec{v}_i - \vec{v}_f$ instead of $\vec{v}_f - \vec{v}_i$. Answer C suggests the acceleration is purely in the $\hat{i}$ direction, ignoring the significant $\hat{j}$ component. Answer D uses the magnitude of $\vec{v}_f$ ($\sqrt{41}$) instead of the magnitude of $\Delta\vec{v}$, representing a fundamental misunderstanding of what quantity needs to be normalized. Always remember: when finding unit vectors, normalize the actual vector you're interested in—here, the change in velocity, not the final velocity itself.
9. A particle's position is described by the vector $\vec{r}(t) = (R\cos(\omega t))\hat{i} + (2R\sin(\omega t))\hat{j}$, where $R$ and $\omega$ are positive constants. At which of the following times $t > 0$ are the particle's velocity and acceleration vectors first perpendicular to each other?
   Answer: $t = \frac{\pi}{2\omega}$ When you encounter parametric motion problems involving perpendicular vectors, you need to find when their dot product equals zero. This tests your understanding of vector calculus and the geometric relationship between velocity and acceleration. To find when velocity and acceleration are perpendicular, first calculate these vectors by taking derivatives of the position vector. The velocity is: $\vec{v}(t) = \frac{d\vec{r}}{dt} = -R\omega\sin(\omega t)\hat{i} + 2R\omega\cos(\omega t)\hat{j}$ The acceleration is: $\vec{a}(t) = \frac{d\vec{v}}{dt} = -R\omega^2\cos(\omega t)\hat{i} - 2R\omega^2\sin(\omega t)\hat{j}$ For perpendicular vectors, their dot product must equal zero: $\vec{v} \cdot \vec{a} = (-R\omega\sin(\omega t))(-R\omega^2\cos(\omega t)) + (2R\omega\cos(\omega t))(-2R\omega^2\sin(\omega t)) = 0$ Simplifying: $R^2\omega^3\sin(\omega t)\cos(\omega t) - 4R^2\omega^3\sin(\omega t)\cos(\omega t) = 0$ This gives us: $-3R^2\omega^3\sin(\omega t)\cos(\omega t) = 0$ Since $R$ and $\omega$ are positive constants, either $\sin(\omega t) = 0$ or $\cos(\omega t) = 0$. The first occurrence for $t > 0$ is when $\cos(\omega t) = 0$, which happens at $\omega t = \frac{\pi}{2}$, giving $t = \frac{\pi}{2\omega}$. Answer A is correct. Answer B gives $t = \frac{\pi}{4\omega}$, where neither sine nor cosine is zero. Answer C gives $t = \frac{\pi}{\omega}$, which corresponds to the second occurrence. Answer D is wrong since we found specific times when the vectors are perpendicular. Remember: perpendicular vectors have zero dot product, so always set up the dot product equation and solve systematically.
10. A particle moves along a curved path such that its position vector is $\vec{r}(t) = (3t^2 - 2t)\hat{i} + (4t - t^2)\hat{j}$ meters. At $t = 2$ seconds, what is the magnitude of the particle's displacement from its position at $t = 0$?
    Answer: $\sqrt{52}$ m First find the position at t = 0: $\vec{r}(0) = 0\hat{i} + 0\hat{j} = \vec{0}$. Then find position at t = 2: $\vec{r}(2) = (3(4) - 2(2))\hat{i} + (4(2) - 4)\hat{j} = 8\hat{i} + 4\hat{j}$. The displacement vector is $\vec{r}(2) - \vec{r}(0) = 8\hat{i} + 4\hat{j}$. The magnitude is $\sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} = \sqrt{52}$ m.
11. Consider three physical quantities that can be represented by fields in a region of space: temperature $T(x,y,z)$, gravitational potential $V(x,y,z)$, and fluid velocity $\vec{v}(x,y,z)$. Which of the following mathematical operations on these fields results in a physically meaningful vector quantity?
    Answer: The cross product of the gradient of temperature and the gradient of potential, $\nabla T \times \nabla V$ When analyzing mathematical operations on physical fields, you need to consider both the mathematical validity of the operations and whether the result has physical meaning as a vector quantity. The correct answer is D because $\nabla T \times \nabla V$ produces a meaningful vector. Both $\nabla T$ and $\nabla V$ are vector fields (since gradients of scalar fields yield vectors), and their cross product gives a vector perpendicular to both gradients. This vector points along lines where both temperature and gravitational potential are constant simultaneously, which has physical significance in describing the intersection of isothermal and equipotential surfaces. Option A is wrong because $T + |\vec{v}|$ adds a scalar (temperature) to another scalar (speed), yielding only a scalar result, not a vector. Option B fails because $V(\nabla \cdot \vec{v})$ multiplies two scalars: the gravitational potential $V$ and the divergence of velocity $\nabla \cdot \vec{v}$ (which measures compression/expansion of fluid flow). The product of scalars remains scalar. Option C produces a scalar because $\vec{v} \cdot \nabla T$ is a dot product between velocity vector and temperature gradient vector. This actually represents the rate of temperature change along the flow direction—physically meaningful but scalar, not vector. Study tip: Remember that gradients of scalars create vectors, divergences and dot products create scalars, and cross products of vectors create vectors. Always check both the mathematical dimension of your result and whether it has physical meaning in the given context.
12. Two vectors $\vec{P}$ and $\vec{Q}$ are added to form a resultant vector $\vec{R} = \vec{P} + \vec{Q}$. If the magnitudes satisfy the relation $|\vec{P}| + |\vec{Q}| = |\vec{R}|$, what must be true about the vectors $\vec{P}$ and $\vec{Q}$?
    Answer: They are in the same direction. 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 $0^\circ$ 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.
13. A vector is given by $\vec{A} = 6\hat{i} + 8\hat{j}$. A second vector is given by $\vec{B} = 2\hat{i}$. What is the magnitude of the component of vector $\vec{A}$ that is parallel to the direction of vector $\vec{B}$?
    Answer: 6 The magnitude of the component of vector $\vec{A}$ along the direction of vector $\vec{B}$ is found by the scalar projection, which is calculated as the dot product of $\vec{A}$ with the unit vector of $\vec{B}$. The unit vector for $\vec{B}$ is $\hat{B} = \frac{2\hat{i}}{|2|} = \hat{i}$. The scalar projection is then $\vec{A} \cdot \hat{B} = (6\hat{i} + 8\hat{j}) \cdot (\hat{i}) = (6)(1) + (8)(0) = 6$. Distractor D is the dot product $\vec{A} \cdot \vec{B}$, not the projection.
14. Two vectors are defined as $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = -8\hat{i} + 6\hat{j}$ 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 $\hat{i}'$ and $\hat{j}'$. Which of the following quantities has a different value when calculated in system S' compared to its value in system S?
    Answer: The component of vector $\vec{B}$ along the $\hat{i}'$ direction 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 $\vec{B}$ along $\hat{i}'$ represents a projection onto a new axis direction. In system S, $\vec{B}$ has an x-component of -8. After rotation, the $\hat{i}'$ direction points differently in space, so $\vec{B}$'s component along this new direction will be different. You'd calculate this using: $B_{i'} = B_x \cos(30°) + B_y \sin(30°) = -8(\frac{\sqrt{3}}{2}) + 6(\frac{1}{2}) = -4\sqrt{3} + 3 \neq -8$. Choice A is wrong because vector magnitude is invariant under rotation—$|\vec{A}| = \sqrt{3^2 + 4^2} = 5$ regardless of coordinate system. Choice B is incorrect since the dot product $\vec{A} \cdot \vec{B} = (3)(-8) + (4)(6) = 0$ 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?"
15. 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?
    Answer: The average speed is greater than the magnitude of the average velocity because the distance traveled is greater than the displacement. The total distance traveled is $(20 \text{ m/s})(10 \text{ s}) + (20 \text{ m/s})(10 \text{ s}) = 400$ 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 $\sqrt{200^2 + 200^2} \approx 283$ m. The magnitude of the average velocity is |displacement|/time = 283 m / 20 s $\approx 14.1$ 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.
16. Three displacement vectors are given by $\vec{A} = 2\hat{i} - 3\hat{j}$, $\vec{B} = -4\hat{i} - \hat{j}$, and $\vec{C} = \hat{i} + 5\hat{j}$. What is the resultant displacement vector $\vec{R} = \vec{A} + \vec{B} + \vec{C}$?
    Answer: $\vec{R} = -\hat{i} + \hat{j}$ To find the resultant vector, add the corresponding components of the individual vectors. For the x-components: $R_x = A_x + B_x + C_x = 2 + (-4) + 1 = -1$. For the y-components: $R_y = A_y + B_y + C_y = (-3) + (-1) + 5 = 1$. Therefore, the resultant vector is $\vec{R} = -1\hat{i} + 1\hat{j} = -\hat{i} + \hat{j}$.
17. A particle undergoes a displacement of $6.0$ m at an angle of $30^\circ$ north of east, followed by a second displacement of $4.0$ m at an angle of $60^\circ$ west of north. What is the approximate magnitude of the particle's resultant displacement?
    Answer: 5.3 m First, resolve each displacement vector into its x (east) and y (north) components. For the first displacement: $d_{1x} = 6.0 \cos(30^\circ) \approx 5.20$ m, $d_{1y} = 6.0 \sin(30^\circ) = 3.00$ m. For the second displacement: $d_{2x} = -4.0 \sin(60^\circ) \approx -3.46$ m, $d_{2y} = 4.0 \cos(60^\circ) = 2.00$ m. The resultant components are $R_x = 5.20 - 3.46 = 1.74$ m and $R_y = 3.00 + 2.00 = 5.00$ m. The magnitude of the resultant displacement is found using the Pythagorean theorem: $|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(1.74)^2 + (5.00)^2} \approx 5.3$ m. Distractor D is the scalar sum of the magnitudes.
18. Two non-zero vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. If the magnitude of their difference, $|\vec{A} - \vec{B}|$, is also equal to the magnitude of $\vec{A}$, what is the angle between vectors $\vec{A}$ and $\vec{B}$ when placed tail-to-tail?
    Answer: 60° Let the magnitude of each vector be $x$. The law of cosines for vector subtraction is $|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta$. Substituting $x$ for all magnitudes gives $x^2 = x^2 + x^2 - 2(x)(x)\cos\theta$. This simplifies to $0 = x^2 - 2x^2\cos\theta$. Since $x$ is non-zero, we can divide by $x^2$ to get $0 = 1 - 2\cos\theta$, which gives $\cos\theta = 1/2$. The angle for which this is true is $60^\circ$.
19. A force $\vec{F} = (4\hat{i} + 6\hat{j})$ N acts on an object that undergoes a displacement $\vec{d} = (3\hat{i})$ m. A second force $\vec{G}$ also acts on the object. The work done by force $\vec{G}$ on the object is zero, and the net force $\vec{F}_{net}$ on the object is parallel to the displacement $\vec{d}$. What is the vector $\vec{G}$?
    Answer: $(-6\hat{j})$ N This problem combines work-energy concepts with vector addition, testing your understanding of when forces do zero work and how net forces relate to motion direction. Since force $\vec{G}$ does zero work on the object, $\vec{G}$ must be perpendicular to the displacement $\vec{d} = (3\hat{i})$ m. The work formula $W = \vec{F} \cdot \vec{d}$ equals zero only when the dot product is zero, which happens when vectors are perpendicular. Since $\vec{d}$ points purely in the x-direction, $\vec{G}$ can only have a y-component. The key insight is that the net force $\vec{F}_{net}$ is parallel to $\vec{d}$, meaning $\vec{F}_{net}$ has no y-component. Since $\vec{F}_{net} = \vec{F} + \vec{G}$, and $\vec{F} = (4\hat{i} + 6\hat{j})$ N has a 6 N component in the y-direction, $\vec{G}$ must cancel this y-component. Therefore, $\vec{G} = (-6\hat{j})$ N. Let's verify: $\vec{F}_{net} = (4\hat{i} + 6\hat{j}) + (-6\hat{j}) = (4\hat{i})$ N, which is indeed parallel to $\vec{d}$. Choice A) $(-4\hat{i})$ N would reduce the x-component of the net force but wouldn't eliminate the y-component. Choice C) $(-4\hat{i} - 6\hat{j})$ N would make the net force zero, not parallel to $\vec{d}$. Choice D) $(3\hat{i})$ N would do positive work since it's parallel to the displacement. Remember: when a force does zero work, look for perpendicularity to displacement. When forces must produce motion in a specific direction, focus on which components need to be eliminated through vector addition.
20. A particle moves in a plane such that its velocity vector is $\vec{v}(t) = (6t - 3)\hat{i} + (8 - 4t^2)\hat{j}$ m/s. At what time(s) during the first 3 seconds is the particle moving purely in the x-direction?
    Answer: $t = \sqrt{2}$ s only For the particle to move purely in the x-direction, the y-component of velocity must be zero: $v_y = 8 - 4t^2 = 0$. Solving: $4t^2 = 8$, so $t^2 = 2$, giving $t = \pm\sqrt{2}$. Since we need $0 \leq t \leq 3$, only $t = \sqrt{2} \approx 1.41$ s is valid. At this time, $v_x = 6\sqrt{2} - 3 \neq 0$, confirming motion is purely in x-direction. Choice B results from incorrectly solving $6t - 3 = 0$. Choice C includes the negative solution which is outside our time interval. Choice D incorrectly concludes no solution exists.

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