AP Physics 1 FRQs

▸ earns credit   ⚠︎ common error, partial credit   ✗ common error, no credit

Part C — Model Answer

The gravitational potential energy of the Earth–payload system is U = −GM_E m/r. With GM_E m/R_E = 2.50 × 10¹⁰ J,

U = −(2.50 × 10¹⁰ J)(R_E/r)

r = R_E: U = −2.50 × 10¹⁰ J

r = 2R_E: U = −1.25 × 10¹⁰ J

r = 4R_E: U = −0.625 × 10¹⁰ J

U is negative at every finite r because its zero is at infinite separation and gravity is attractive: moving the payload outward toward that zero takes energy. Doubling r halves the magnitude of U, so the curve rises steeply near Earth and flattens as r increases, approaching zero without reaching it (−0.417 × 10¹⁰ J at r = 6R_E). U depends only on the separation, so it is defined at every r in the range.

The system’s mechanical energy is constant (part B) and is fixed at release, where K = 0:

E = K + U = 0 + (−0.625 × 10¹⁰ J) = −0.625 × 10¹⁰ J

so E is a horizontal line at −0.625 × 10¹⁰ J. The kinetic energy at each r is K = E − U, the vertical gap from the U curve up to the E line: zero at r = 4R_E, 0.625 × 10¹⁰ J at r = 2R_E, which is ½mv_2² from part B, and 1.875 × 10¹⁰ J at r = R_E.

The kinetic energy at each r is the gap between the E line and the U curve, so the payload can only be where the E line lies on or above the U curve. For r > 4R_E, U is greater than E, and K = E − U would be negative, which is impossible: the payload, released from rest and falling inward, never reaches those separations, so K is not defined there.

Scoring (4 points):

▸C1 — 1 point: For sketching a line or curve, labelled as the gravitational potential energy, that is negative from r = R_E to r = 6R_E, increases as r increases, and becomes less steep as r increases without reaching zero.

▸C2 — 1 point: For a U curve at −2.50 × 10¹⁰ J at r = R_E, −1.25 × 10¹⁰ J at r = 2R_E and −0.625 × 10¹⁰ J at r = 4R_E.

▸C3 — 1 point: For sketching a horizontal line, labelled as the total mechanical energy, at −0.625 × 10¹⁰ J or at the value of the response’s own U curve at r = 4R_E, continuous from r = R_E to at least r = 4R_E.

▸C4 — 1 point: For sketching a line or curve, labelled as the kinetic energy, that is zero at r = 4R_E, increases as r decreases, and is not drawn for r > 4R_E, with values of 0.625 × 10¹⁰ J at r = 2R_E and 1.875 × 10¹⁰ J at r = R_E, or with values equal at each r to the vertical distance from the response’s own U curve up to its own E line.

⚠︎Common error (partial credit): Joins −2.50 × 10¹⁰ J at r = R_E and −0.625 × 10¹⁰ J at r = 4R_E with a straight line, as if U changed by the same amount for each metre of the fall, with E and K consistent with it — earns 2 of 4 points, C3 and C4. That line gives −1.875 × 10¹⁰ J at r = 2R_E, not −1.25 × 10¹⁰ J, and reaches zero at r = 5R_E; the field weakens with distance from Earth.

A disabled, untethered Sentinel payload of mass m = 4.00 × 10² kg is released from rest at r = 4R_E, where r is the payload’s distance from Earth’s centre and R_E = 6.37 × 10⁶ m is Earth’s radius, as shown in Figure 1. Its thrusters are inoperative, and it falls directly toward Earth along a radius. Once released, it touches nothing; air resistance is negligible at every point in the fall, and so are the gravitational forces exerted on the payload by the Moon, the Sun and every other body except Earth. The axes shown in Figure 2 are used in part C.

Take Earth’s mass to be M_E = 5.97 × 10²⁴ kg and use GM_E = 3.98 × 10¹⁴ N·m²/kg. Earth’s own motion may be neglected: use a reference frame in which Earth’s centre is at rest and which does not rotate with Earth, and treat it as inertial. All motion is one-dimensional along a radius from Earth’s centre; take radially outward as the positive direction, and take the gravitational potential energy of the Earth–payload system to be zero at infinite separation.

Part A

For this payload, the quantity GM_E m/R_E² has the value 3.92 × 10³ N. The side of each grid square in Figure 3 represents a force of magnitude 245 N.

On the dots in Figure 3 that represent the payload at r = 4R_E and at r = 2R_E, draw and label the forces (not components) that are exerted on the payload.

• Each arrow must start on, and point away from, each dot.
• Draw the length of each arrow to represent the magnitude of each force, consistent with the scale used in Figure 3.

Part B

Starting with conservation of energy, derive an expression for the speed v_2 of the payload when it reaches r = 2R_E. Express your final answer in terms of m, M_E, R_E, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

For this payload, the quantity GM_E m/R_E has the value 2.50 × 10¹⁰ J.

On the axes shown in Figure 2, do the following for r = R_E to r = 6R_E.

• Sketch and label a line or curve that represents the gravitational potential energy U of the Earth–payload system as a function of r.
• Sketch and label a line or curve that represents the total mechanical energy E of the Earth–payload system as a function of r during the drop.
• Sketch and label a line or curve that represents the kinetic energy K of the payload as a function of r during the drop.

If a line or curve is not defined over part of this range, draw it only where it is defined.

Part D

Released from rest at r = 4R_E, the payload gains kinetic energy ΔK_4 as it falls from r = 2R_E to r = R_E. Suppose instead that the same payload is released from rest at r = 6R_E, and that it gains kinetic energy ΔK_6 as it falls from r = 2R_E to r = R_E. Use the lines and curves you drew on Figure 2 in part C to compare ΔK_6 and ΔK_4.

Indicate whether ΔK_6 is greater than, less than, or equal to ΔK_4 by writing one of the following.

• ΔK_6 > ΔK_4
• ΔK_6 < ΔK_4
• ΔK_6 = ΔK_4

Justify your response.

▸ earns credit   ⚠︎ common error, partial credit   ✗ common error, no credit

Part C — Model Answer

The speed on reaching roof level is the magnitude of the velocity vector there, not either component on its own. The horizontal component is unchanged throughout the flight, because nothing accelerates the motorcycle horizontally: v_x = v_0. The vertical component follows from the relation linking velocity to displacement:

v_y² = v_y0² + 2 · a_y · Δy = 0 + 2(−g)(−h) = 2gh

The two components are perpendicular, so they combine by the Pythagorean relationship:

v_land = √(v_x² + v_y²) = √(v_0² + 2gh)

Three features of that expression fix the shape of the sketch. At v_0 = 0 the motorcycle would simply fall, reaching roof level at v_land = √(2gh) = √240 = 15.5 m/s — the curve starts well above the origin and already below the 22 m/s limit. The curve then rises throughout, gradually at first because a small horizontal component adds little in quadrature, and more steeply as v_0 grows. For large v_0 the vertical contribution becomes negligible beside it and the curve approaches the line v_land = v_0 from above.

Setting v_land equal to the 22 m/s limit:

v_0,max = √(v_land² − 2gh) = √(22² − 2 × 10 × 12.0) = √(484 − 240) = √244 = 15.6 m/s

The limit binds where the motorcycle actually lands on the lower roof, and v_0,max = 15.6 m/s lies in that range.

The same relation follows from conservation of energy, ½mv_land² = ½mv_0² + mgh, which gives v_land² = v_0² + 2gh directly; either route is correct. The vertical contribution to the speed at roof level is the same for every launch speed, so only the horizontal contribution changes.

Scoring (3 points):

▸C1 — 1 point: For a curve of v_land against v_0 that begins at a clearly non-zero v_land at v_0 = 0, below the 22 m/s limit, rather than at the origin. The expected intercept is the free-fall value √(2gh) ≈ 15.5 m/s; the point is earned from the sketch alone, whether or not a value is stated.

▸C2 — 1 point: For a curve that increases over the whole range, rising gradually at small v_0 and more steeply at large v_0.

Scoring Note: C2 requires the increasing steepness, not a particular curvature elsewhere. The approach to the line v_land = v_0 at large v_0 need not be drawn or named.

▸C3 — 1 point: For marking on the sketched curve the launch speed at which v_land reaches the 22 m/s limit, and labelling it v_0,max. The expected value is v_0,max ≈ 15.6 m/s. Award the point wherever the marking is read correctly from the response’s own curve, whatever value that curve gives.

⚠︎Common error (partial credit): Draws a straight line from the origin of slope 1, so that v_land = v_0. This leaves the vertical component out of the speed at roof level entirely and fails C1 and C2 together. The crossing at 22 m/s is still read correctly from the line drawn — earns 1 of 3 points, for C3.

✗Common error (no credit): Sketches the motorcycle’s trajectory — height against horizontal distance — on the axes supplied, rather than v_land against the launch speed. The axes are v_land against v_0, so the curve is not a curve of v_land against v_0 at all and meets none of C1, C2 or C3 whatever its intercept happens to be.

A vehicle-testing facility has two curves that share the same radius, R = 50 m. Curve A is level. On its dry surface the coefficient of static friction between the car’s tyres and the road is μ_s = 0.80. Curve B is banked at a constant angle θ = 30° to the horizontal. It is built so that a car travelling at the curve’s design speed can round it with no friction force exerted on the car.

The same car, of mass m = 1500 kg, is driven around each curve at constant speed; on Curve B it travels at the design speed. Air resistance and rolling resistance are negligible.

Model the car as a particle. Use an inertial reference frame fixed to the ground. Both circular paths lie in horizontal planes, and every force exerted on the car lies in the vertical plane that contains the car and the centre of its circular path. For horizontal forces take the direction toward the centre of the curve as positive, and for vertical forces take upward as positive. Use g = 10 m/s².

Part A

Part B

A sudden downpour leaves both road surfaces effectively frictionless. The car enters Curve B on its 50 m circle, moving horizontally along the circle at 12 m/s.

Indicate whether the car’s distance from the centre of the curve increases, decreases, or remains constant just after the car enters the curve.

______ Increases

______ Decreases

______ Remains constant

Justify your answer. In your justification, include qualitative reasoning beyond mathematical derivations or expressions.

▸ earns credit   ⚠︎ common error, partial credit   ✗ common error, no credit

Part D — Model Answer

The altimeter reports a height above the surface, but the r in Newton’s law of universal gravitation is the distance between the two centres of mass, so the height must be converted first:

r = R_T + h = R_T + 1.50R_T = 2.50R_T

With G·M_T eliminated by the surface value, as in part C, the dependence from part B and an inverse fall-off with the same surface value predict

inverse square: a_g = g_T·(R_T/r)² = (1.60 m/s²)(1/2.50)² = 0.256 m/s²

inverse: a_g = g_T·(R_T/r) = (1.60 m/s²)/2.50 = 0.640 m/s²

The dependence from part B

The reported 0.254 m/s² differs from the inverse-square prediction by about 0.8%, inside the 2% spread of the repeated runs, while the inverse prediction is two and a half times the reported value. The measurement therefore supports the dependence from part B, the inverse square, and rules out the inverse one.

Both curves start from g_T at r = R_T. Only a reading at a second, independently measured distance can separate two dependences that share a surface value.

Scoring (3 points):

▸D1 — 1 point: For converting the altimeter height into a distance from Theron’s centre, r = R_T + h = 2.50R_T.

▸D2 — 1 point: For predicting a_g from the part B expression with its value at r = R_T set equal to g_T — 0.256 m/s² at r = 2.50R_T — or the value that follows correctly at the response’s own r from its own part B expression, so anchored.

▸D3 — 1 point: For predicting the inverse value g_T·(R_T/r) — 0.640 m/s² at r = 2.50R_T, or the value at the response’s own r — and indicating the dependence the measurement supports, justified by comparing each prediction with the reported 0.254 m/s². The indication is “The dependence from part B” for the values above (0.254 m/s² is within 1% of 0.256 m/s², far from 0.640 m/s²), and otherwise the one the response’s own two predictions support.

⚠︎Common error (partial credit): Uses the altimeter height as the distance from Theron’s centre, r = 1.50R_T, then predicts correctly at that distance, 0.711 m/s² from the part B expression and 1.07 m/s² for the inverse, and indicates “Neither” because both lie far from 0.254 m/s² — earns 2 of 3 points, D2 and D3. The r in the law is measured between the centres of mass, so r = R_T + h.

✗Common error (no credit): With a correct inverse-square expression in part B, ignores it: takes the altimeter height as the radius and divides the surface value by the ratio of the distances, not its square, to get 1.07 m/s², then indicates “Neither” and reports the tracking data as faulty. The radius is wrong, 1.07 m/s² does not follow from the part B expression, and only one prediction is compared with the data, so no point is earned; the data, not a model, are rejected.