5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A robot vacuum moves 1.1 m east and then 1.6 m north, each along a straight line. What is the magnitude of its displacement from its starting point?
Answer and reasoning
A2.7 m A student who adds the two distances as ordinary numbers, 1.1 m + 1.6 m, picks this. That is the distance traveled; the displacement is the resultant of two perpendicular components and must be found with a² + b² = c².
B3.8 m A student who stops at 1.1² + 1.6² = 3.77 and forgets the square root picks this. That number is c² in m²; the displacement is its square root, 1.9 m.
C1.9 mCorrect The east and north displacements are perpendicular components of the total displacement, which is the hypotenuse of the right triangle they form: √((1.1 m)² + (1.6 m)²) = √(3.77 m²) = 1.9 m.
D1.2 m A student who treats the longer displacement, 1.6 m, as the hypotenuse and calculates √(1.6² − 1.1²) picks this. The resultant is the hypotenuse; both given displacements are the perpendicular sides.
Working The two displacements are perpendicular components of the total displacement. |Δr| = √(1.1² + 1.6²) m = √(1.21 + 2.56) m = √3.77 m = 1.9 m (1.94 m).
A cart moves up a straight ramp inclined at 30° to the horizontal. Student 1 resolves the cart's velocity into horizontal and vertical components. Student 2 resolves the same velocity using an x-axis along the ramp and a y-axis perpendicular to the ramp. Which statement about their results is correct?
Answer and reasoning
ATheir components must match, since both pairs describe one velocity. A student who thinks components are fixed properties of the vector picks this. The components depend on the chosen axes; the same velocity has components (v cos 30°, v sin 30°) in one system and (v, 0) in the other.
BTheir components differ, but each pair combines to give the same velocity.Correct Components depend on the axes chosen. Student 1 finds two nonzero components, v cos 30° and v sin 30°; Student 2, with an axis along the motion, finds v and 0. Each pair is a correct description: combined, both give the same vector, with the same speed v up the ramp.
CStudent 2's are not proper components, as components are horizontal and vertical. A student who thinks components must be horizontal and vertical picks this. Any pair of perpendicular axes can be used; axes along the ramp are often the most convenient choice here.
DTheir components differ, so the two students find different speeds. A student who thinks the magnitude depends on the axes picks this. Both pairs give the same magnitude: (v cos 30°)² + (v sin 30°)² = v², and v² + 0² = v². The speed does not depend on the coordinate system.
Working Speed v up the ramp. Student 1: vx = v cos 30°, vy = v sin 30° (both nonzero). Student 2: vx′ = v, vy′ = 0. The pairs differ, but √((v cos 30°)² + (v sin 30°)²) = v = √(v² + 0²): same vector, same speed.
The diagram shows the velocity of a ball just after it is kicked, drawn with its tail at the origin of an x-y coordinate system with x horizontal and y vertical. The velocity has magnitude v and makes an angle θ with the vertical y-axis. Which expression gives the horizontal component vx of the velocity?
Answer and reasoning
Av sin θCorrect θ is measured from the vertical, so the vertical component is adjacent to θ and the horizontal component is opposite it: sin θ = vx/v, giving vx = v sin θ. Check: if θ = 0 the ball moves straight up, and v sin 0 = 0, as it should be.
Bv cos θ A student who always uses cosine for the horizontal component picks this. That rule works only when the angle is measured from the horizontal; here θ is measured from the vertical, so v cos θ is the vertical component.
Cv/sin θ A student who rearranges sin θ = vx/v by dividing instead of multiplying picks this. Since sin θ ≤ 1, v/sin θ is larger than v, and a component can never be larger than the vector itself.
Dv tan θ A student who treats the velocity as one of the shorter sides of the triangle, using tan θ = vx/v, picks this. The velocity is the hypotenuse, so the ratio to use is sin θ = vx/v.
Working The angle θ is between the velocity and the y-axis, so the y-component is adjacent to θ and the x-component is opposite it. sin θ = vx/v, so vx = v sin θ (and vy = v cos θ). Check: θ = 0 gives a vertical velocity with vx = 0.
Ball A, of mass 2m, is launched horizontally from the edge of a table at the same instant that ball B, of mass m, is released from rest at the same height. Air resistance is negligible. Which statement correctly compares the times at which the two balls reach the floor?
Answer and reasoning
ABall A lands together with B, as their vertical motions match.Correct Separate each motion into components. Vertically, both balls start with zero vertical velocity, fall through the same height and accelerate at g whatever their mass, so they reach the floor together. A's horizontal velocity only carries it farther sideways.
BBall A lands later, as its horizontal motion helps to hold it up. A student who thinks sideways motion keeps a projectile aloft picks this. With air resistance negligible, A's vertical acceleration is g, the same as B's, whatever its horizontal velocity.
CBall A lands first, as its greater mass makes it fall faster. A student who thinks heavier objects fall faster picks this. With air resistance negligible, both balls have the same downward acceleration, g, whatever their masses.
DBall A lands later, as it starts to fall once its forward motion dies. A student who thinks a launched ball moves forward until its 'impetus' is used up, and then falls, picks this. A starts to fall at the instant it leaves the table, and its horizontal velocity does not die away.
Working Vertical motion of each ball: vy0 = 0, ay = −g (independent of mass and of horizontal velocity), same drop h, so both take t = √(2h/g).
A ball is launched at an angle above the horizontal from level ground. Air resistance is negligible. Which statement describes the ball's acceleration at the highest point of its path?
Answer and reasoning
AIt is g, directed downward, the same as at every other point of the path.Correct A projectile's acceleration is g downward at every point of its path. At the highest point vy is zero for an instant but is still changing, from upward to downward, at the same rate as everywhere else, while vx does not change at all.
BIt is zero, since the ball stops rising and moves horizontally there. A student who thinks the acceleration is zero at the highest point picks this. vy is zero there for only an instant and is still changing by g each second, so the acceleration is still g downward.
CIt is horizontal, pointing the way the ball is moving at that point. A student who thinks acceleration points along the direction of motion picks this. The ball's velocity is horizontal at the top, but only its vertical component is changing, so the acceleration is vertical, g downward.
DIt is less than g, since the ball's horizontal motion helps hold it up. A student who thinks sideways motion keeps a projectile aloft picks this. With air resistance negligible the vertical acceleration is g whatever the horizontal velocity.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.5.A.1 Vector component Fix
Vector component
The part of a vector along one axis of a coordinate system, written as a signed number (for example vx = −3.0 m/s). A vector can be modeled as the resultant of two perpendicular components, and the components have the same unit as the vector.
Resultant of perpendicular components
The single vector equivalent to two perpendicular components placed tip to tail. Its magnitude is found with the Pythagorean relationship, √(Ax² + Ay²), which is never more than the sum of the component magnitudes and never less than the larger of them.
Students often think Vector magnitudes add like ordinary numbers, so a vector's magnitude is the sum of its components' magnitudes (and one component is the magnitude minus the other). In fact No. Perpendicular components combine by the Pythagorean relationship, |A| = √(Ax² + Ay²). Adding the component magnitudes gives a larger value, correct only if both components point the same way along one line.
Students often think When a relationship contains a squared quantity, the result can be read off before taking the square root, so c = a² + b² and t = 2h/g. In fact No. If the relationship gives c² (as in a² + b² = c²) or t² (as in Δy = (1/2)g t²), the quantity itself is the square root. Leaving the root out gives a value with the wrong units and the wrong dependence on the data.
1.5.A.2 Coordinate system for resolving a vector Fix
Coordinate system for resolving a vector
A chosen pair of perpendicular axes with positive directions. The same vector has different components in different coordinate systems (for example horizontal–vertical axes or axes along and perpendicular to a ramp), but its magnitude and direction do not change; the axes are chosen for convenience.
Students often think A vector's components are fixed properties of the vector, so every correct coordinate system must give the same pair of components. In fact No. The components depend on the chosen axes. The same velocity has two nonzero components with horizontal and vertical axes, but only one nonzero component if one axis is chosen along the velocity. The vector itself, and so its magnitude and direction, is unchanged.
Students often think Components are by definition horizontal and vertical, so resolving a vector along tilted axes does not give true components. In fact No. Any pair of perpendicular axes can be chosen, for example along and perpendicular to a ramp. The choice is made for convenience, and every choice describes the same vector correctly.
1.5.A.3 Right-triangle relationships Fix
Right-triangle relationships
For a right triangle with hypotenuse c, side a opposite the angle θ and side b adjacent to θ: sin θ = a/c, cos θ = b/c, tan θ = a/b and a² + b² = c². When a vector is resolved, the vector is the hypotenuse and its components are the two perpendicular sides.
Resolving a vector with trigonometry
A vector of magnitude A making angle θ with an axis has component A cos θ along that axis and A sin θ along the perpendicular axis. Which function goes with which component depends on where the angle is marked, not on whether the axis is horizontal or vertical. The angle of a vector follows from its components, for example tan θ = Ay/Ax.
Students often think The vector, or the longer of two given lengths, can be treated as one of the two perpendicular sides of the triangle (for example using tan θ = component/vector, or c² = (longer side)² − (shorter side)²). In fact No. When a vector is resolved into perpendicular components, the vector is the hypotenuse and the components are the two shorter sides, so neither component can be larger than the vector.
Students often think Sine and cosine are chosen by habit rather than by the position of the angle: the x (horizontal) component is always v cos θ and the y (vertical) component is always v sin θ. In fact No. The component adjacent to the marked angle uses cosine and the opposite component uses sine. If the angle is measured from the vertical, the horizontal component is v sin θ.
1.5.B.1 Independence of perpendicular components of motion Fix
Independence of perpendicular components of motion
Motion in two dimensions can be separated into motion along two perpendicular axes. Each component obeys the one-dimensional kinematic relationships with its own initial velocity and acceleration, and the two are linked only by the shared time t.
Students often think Horizontal motion helps hold a projectile up, so an object moving sideways falls more slowly, or with a smaller acceleration, than one dropped from rest. In fact No. With air resistance negligible, the vertical motion is the same whatever the horizontal velocity: the vertical acceleration is g downward in every case, so an object launched horizontally reaches the ground at the same time as one dropped from the same height.
Students often think Heavier objects fall with a greater acceleration than lighter ones, so a heavier projectile reaches the ground sooner. In fact No. Near Earth's surface every object in free fall has the same downward acceleration, g, whatever its mass; this holds for the vertical motion of a projectile as well.
1.5.B.2 Projectile motion Fix
Projectile motion
Motion of an object moving only under the influence of gravity near Earth's surface (air resistance negligible). With x horizontal and y vertical (up positive): ax = 0, so vx is constant; ay = −g, so the vertical motion is the same as that of an object thrown straight up or dropped. The path is a parabola.
Horizontal launch
A projectile launched horizontally has vy0 = 0. Its time of flight is set by the vertical motion alone, Δy = (1/2)g t² for a drop of height h, so t = √(2h/g); its horizontal distance is then vx0 t.
Highest point of a projectile's path
At the highest point vy = 0 for an instant, but vx keeps its constant value, so the velocity is horizontal and nonzero and the speed is least there. The acceleration is still g downward.
Students often think A launched object carries a forward 'impetus' that is gradually used up, so its horizontal velocity decreases during the flight, and it moves forward and only drops once the forward motion has died away. In fact No. With air resistance negligible, nothing acts on a projectile horizontally, so its horizontal acceleration is zero and vx keeps the value it had at launch until it lands.
Students often think The kinematic equation for a fall can be used without its numerical factor, as h = g t², giving t = √(h/g). In fact No. The (1/2) comes from averaging a steadily changing velocity. For a drop of height h from rest, h = (1/2)g t², so t = √(2h/g); dropping the factor gives √(h/g), a time too short by a factor of √2.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
A ball is kicked with a velocity of magnitude 10 m/s directed at an angle above the horizontal. The vertical component of this velocity is 6.0 m/s. What angle does the velocity make with the horizontal?
Answer and reasoning
A53° A student who swaps sine and cosine, using cos θ = 6.0/10 for the vertical component, picks this. The vertical component is opposite the angle to the horizontal, so sin θ = 6.0/10; 53° is the angle the velocity makes with the vertical.
B31° A student who uses tan θ = 6.0/10, treating the velocity as one of the perpendicular sides, picks this. The velocity is the hypotenuse; tan θ needs the two components, 6.0/8.0.
C56° A student who finds the horizontal component as 10 − 6.0 = 4.0 m/s, as if the components added up to the magnitude, and then uses tan θ = 6.0/4.0, picks this. Perpendicular components combine by the Pythagorean relationship, so the horizontal component is √(10² − 6.0²) = 8.0 m/s.
D37°Correct The velocity is the hypotenuse and the vertical component is the side opposite the angle to the horizontal: sin θ = 6.0 m/s ÷ 10 m/s = 0.60, so θ = 37°. (The horizontal component is then 8.0 m/s, and tan θ = 6.0/8.0 gives the same angle.)
Working The velocity (10 m/s) is the hypotenuse and the vertical component (6.0 m/s) is opposite the angle to the horizontal: sin θ = 6.0/10 = 0.60, so θ = 37° (the horizontal component is √(10² − 6.0²) = 8.0 m/s, and tan θ = 6.0/8.0 = 0.75 agrees).
A ball rolls off the edge of a horizontal ledge 1.8 m above the floor with a horizontal speed of 4.0 m/s. Air resistance is negligible. Use g = 10 m/s². What is the horizontal distance from the edge of the ledge to the point where the ball lands on the floor?
Answer and reasoning
A1.7 m A student who drops the 1/2 from Δy = (1/2)g t², writing 1.8 = 10t², gets t = 0.42 s and picks this. The fall takes 0.60 s, √2 times longer.
B1.4 m A student who solves 1.8 = (1/2)(10)t² as t = 2(1.8)/10 = 0.36, forgetting the square root, picks this. 0.36 is t² in s²; the time is 0.60 s.
C2.4 mCorrect Vertically the ball starts with zero velocity and falls 1.8 m with acceleration g: 1.8 m = (1/2)(10 m/s²)t², so t = 0.60 s. Horizontally its acceleration is zero, so it moves at a constant 4.0 m/s for that time: x = (4.0 m/s)(0.60 s) = 2.4 m.
D4.2 m A student who applies g to the horizontal motion as well, writing x = v t + (1/2)g t² = 2.4 m + 1.8 m, picks this. A projectile's horizontal acceleration is zero, so x = v t = 2.4 m.
Working Vertical: vy0 = 0, ay = −g, drop 1.8 m: 1.8 m = (1/2)(10 m/s²)t², so t² = 0.36 s² and t = 0.60 s. Horizontal: ax = 0, x = vx0 t = (4.0 m/s)(0.60 s) = 2.4 m.
The diagram shows the path of a ball launched from level ground and moving from left to right. Air resistance is negligible. Points P and R are at the same height, and Q is the highest point of the path. Which ranking of the horizontal component of the ball's velocity, vx, at the three points is correct?
Answer and reasoning
AP > Q > R A student who thinks the ball's forward motion is gradually used up picks this. With air resistance negligible nothing acts horizontally on the ball, so vx stays constant along the whole path.
BP = R > Q A student who thinks the ball stops at the highest point picks this. At Q only the vertical component is zero; the ball is still moving sideways with the same vx as at P and R.
CQ > P = R A student who thinks the speed stays the same and becomes all horizontal at the top picks this. The speed is actually least at Q; vx is the same at all three points, and at Q it is simply the whole velocity.
DP = Q = RCorrect A projectile's acceleration is vertical (g downward), so its horizontal acceleration is zero and vx never changes: it has the same value at P, at Q and at R. Only the vertical component changes along the path.
Working Projectile: ax = 0, so vx is constant: vx(P) = vx(Q) = vx(R). (Speed ranking would be P = R > Q, because vy = 0 at Q.)
A ball is launched into the air. The horizontal and vertical components of its velocity, vx and vy, are shown as functions of time t in the graphs, with upward positive. Air resistance is negligible. Which claim about the ball's motion is supported by the graphs?
Answer and reasoning
AIt is momentarily at rest at t = 2 s, where vy is zero. A student who thinks a projectile stops at its highest point picks this. At t = 2 s only vy is zero; the upper graph shows vx = 6 m/s at that time, so the ball is moving horizontally.
BIts speed is least at t = 2 s, where it equals 6 m/s.Correct The speed is √(vx² + vy²). vx is 6 m/s throughout, and vy is zero at t = 2 s and larger in size at every other time, so the speed is least at 2 s, where it is just vx = 6 m/s.
CIts speed at t = 0 is 26 m/s, the sum of 6 m/s and 20 m/s. A student who adds the component magnitudes as ordinary numbers picks this. Perpendicular components combine by the Pythagorean relationship: √((6 m/s)² + (20 m/s)²) ≈ 21 m/s.
DIts acceleration is zero at t = 2 s, where vy is zero. A student who thinks the acceleration is zero at the highest point picks this. The vy graph is a straight line with slope −40 m/s ÷ 4 s = −10 m/s² at every time, including t = 2 s.
Working vx = 6 m/s constant; vy = 20 − 10t (m/s), zero at t = 2 s. Speed = √(6² + vy²): at t = 0 it is √(36 + 400) = 20.9 m/s; least at t = 2 s, where it is 6 m/s. Slope of vy–t = −40/4 = −10 m/s² at every time, including t = 2 s.
A ball is launched horizontally from the top of a wall and lands a horizontal distance d from the base of the wall. Air resistance is negligible. The ball is then launched with the same horizontal speed from the top of a wall twice as high. The new horizontal distance is approximately how many times d?
Answer and reasoning
A1.0 A student who reasons that the horizontal motion is independent of the vertical, so the height cannot matter, picks this. The two motions share the same time, and a higher launch gives a longer time of fall.
B2.0 A student who treats the time of fall as proportional to the height, as if t = 2h/g without the square root, picks this. t is proportional to √h, so doubling h multiplies the time, and the distance, by √2.
C1.4Correct The time of fall comes from the vertical motion: h = (1/2)g t², so t = √(2h/g) is proportional to √h. The horizontal distance is vx0 t, also proportional to √h. Doubling the height multiplies the distance by √2 ≈ 1.4.
D4.0 A student who inverts h = (1/2)g t² as if t were proportional to h² picks this. The square belongs to t: t² ∝ h, so t ∝ √h.
Working Vertical: h = (1/2)g t², so t = √(2h/g) ∝ √h. Horizontal: x = vx0 t ∝ √h. Height × 2 → t × √2 → distance × √2 ≈ 1.4.
A ball is launched horizontally with speed vx0 from the edge of a cliff of height h above level ground. Air resistance is negligible. Just before the ball lands, its velocity makes an angle φ below the horizontal. Which expression gives tan φ?
Answer and reasoning
A√(2gh)/vx0Correct Horizontally, vx = vx0 throughout. Vertically, the ball falls h from rest, so vy² = 2gh and |vy| = √(2gh) at landing. The components are perpendicular, so tan φ = |vy|/vx = √(2gh)/vx0.
B√(gh)/vx0 A student who uses h = gt², without the factor 1/2, gets t = √(h/g) and |vy| = gt = √(gh), and picks this. For a fall from rest, h = (1/2)gt², so t = √(2h/g) and |vy| = √(2gh).
C2gh/vx0 A student who writes vy² = 2gh and then uses 2gh as vy, without taking the square root, picks this. The expression is not even a ratio of two speeds: 2gh has units m²/s². Take the square root: |vy| = √(2gh).
D√(gh/2)/vx0 A student who takes the velocity to point along the straight line from the launch point to the landing point uses tan φ = h/R, with R = vx0√(2h/g), and picks this. The velocity is tangent to the path, not along the displacement; its direction comes from its components, and its tangent is twice as large.
Working Separate the motion into components. Vertical: the ball starts with vy0 = 0 and falls a height h with acceleration of magnitude g, so vy² = vy0² + 2gh = 2gh and |vy| = √(2gh) at landing. Horizontal: no acceleration, so vx = vx0 throughout. The components are perpendicular, so tan φ = |vy|/vx = √(2gh)/vx0. (Using h = gt² instead of h = (1/2)gt²: t = √(h/g), |vy| = gt = √(gh). Omitting the square root in vy² = 2gh: 2gh/vx0, which has units m/s and so cannot be a tangent. Along the straight line from launch point to landing point: tan = h/R with R = vx0√(2h/g), which is √(gh/2)/vx0, half the key.)
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
