2021 AP Exam Administration Scoring Guidelines AP Physics C Mechanics Set 1 AP ® Physics C Mechanics Scoring Guidelines Set 1 2021 © 2021 College Board College Board, Advanced Placement, AP, AP Centra[.]
2021 AP Physics C: Mechanics ® Scoring Guidelines Set © 2021 College Board College Board, Advanced Placement, AP, AP Central, and the acorn logo are registered trademarks of College Board Visit College Board on the web: collegeboard.org AP Central is the official online home for the AP Program: apcentral.collegeboard.org AP® Physics C: Mechanics 2021 Scoring Guidelines Question 1: Free-Response Question (a) 15 points For correctly drawing and labeling all the forces on the cart on the flat surface point For correctly drawing and labeling the weight of the cart on the incline point For correctly drawing and labeling the normal force on the cart on the incline point For correctly drawing and labeling the force of the fan on the cart on the incline point Scoring note: A maximum of three points can be earned if there are any extraneous vectors Example responses for part (a) Scoring note: Examples of appropriate labels for the force due to gravity include: FG , Fg , Fgrav , W , mg , Mg , “grav force,” “F Earth on cart,” “F on cart by Earth,” FEarth on cart , FE,Cart , FCart,E The labels G or g are not appropriate labels for the force due to gravity Fn , FN , N , “normal force,” “ground force,” or similar labels may be used for the normal force Total for part (a) points (b) For correctly applying Newton’s second law for the cart on the flat surface point Ffan = mcart a1 For the correct answer with units ( point ) = Ffan (= 0.50 kg ) 0.8 m s 0.40 N Total for part (b) points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines (c) For including the correct component of weight in a Newton’s second law equation for the cart point Ffan + mg sin θ = ma2 point For a correct Newton’s second law equation Ffan + mg sin θ = ma2 point For correct substitutions consistent with part (b) into the above equation = θ ma2 − Ffan mg sin ma2 − Ffan mg θ = sin −1 ( ) ( ) ( 0.50 kg ) 2.4 m s − ( 0.50 kg ) 0.8 m s 0.50 kg 9.8 m s ( ) θ = sin −1 ( ) = θ 9.4° Total for part (c) (d) points For selecting “No” and an attempted justification point For a correct justification point Example response for part (d) The mass of the cart cancels out in the equation used to find the angle of the incline Total for part (d) points © 2021 College Board AP® Physics C: Mechanics 2021 Scoring Guidelines (e) i For drawing an appropriate best-fit line ii point point For correctly calculating slope using two points from the best-fit line slope = ( ) ∆y ( − 1) m s = = 10.52 m s ∆x ( 0.42 − 0.04 ) For correctly using an expression that relates the slope to the acceleration due to gravity point Fnet = Ffan + mg sin θ = ma F ∴= a g sin θ + fan m from= y mx + b = a ( slope ) sin θ + ( y -intercept ) ∴ slope == g 10.52 m s (f) Total for part (e) points For a correct explanation Example response for part (f) The mass of the cart is in the denominator of the y-intercept, so increasing the mass decreases the y-intercept without changing the rest of the graph So the new line of data is predicted to be parallel to and below the original line Total for question 1 point 15 points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines Question 2: Free-Response Question (a) 15 points For a single acceleration vector pointing down and to the left and attempting a justification point For a correct justification point Example response for part (a) The block is changing direction with a centripetal component of the acceleration toward the center of the circle and a gravitational acceleration downward Therefore, the acceleration of the block will be down and to the left Total for part (a) (b) i For correctly using conservation of energy for the block at point B points point K A + U gA =K B + U gB mv + mghA = mvB2 + mghB A point For correctly substituting into the above equation + mgh = = vB mv + mgR B 2g ( h − R ) ii For correctly substituting the expression for vB from part (b)(i) into an expression for point centripetal force Fc = mv = r mvB2 = R m ( 2g ( h − R ) ) R For correctly substituting into an equation for vector addition to derive an expression for the net force at point B F = net Fc2 + ( mg )2 F= net mvB2 )2 R + ( mg= point 2mg ( h − R ) + ( mg )2 R Total for part (b) points © 2021 College Board AP® Physics C: Mechanics 2021 Scoring Guidelines (c) point For correctly using conservation of energy for the speed of the block at point C K A + U gA =K C + U gC mvA + mghA = mvC2 + mghC 2 mv + mg (2 R ) + mgh = C = vC g ( h − 2R ) point For correctly applying Newton’s second law at point C Fc = mvC2 r mvC2 FN + mg = r Set the normal force equal to zero mvC2 ∴ vC = R + mg = Rg point For combining the two equations above Rg = g ( h − R ) ∴ R = ( h − R ) ∴ R = h − R h = 2.5 R Total for part (c) (d) points For correctly using conservation of energy for the block compressing the spring point For correctly substituting the gravitational and elastic potential energies into an equation for conservation of energy point mgh = kx MAX point For correctly substituting for the spring constant k into the equation above x= MAX 2mgh = k 2mgh = mg ( R ) = 4hR = m ) 0.35 m ( )( 0.30 m )( 0.10 Total for part (d) points © 2021 College Board AP® Physics C: Mechanics 2021 Scoring Guidelines (e) i point For a correct justification Example response for part (e)(i) Because the block does not make it through the loop at this height, it will not compress the spring ii For indicating that as the height increases, the compression of the spring increases point For indicating that the height is proportional to the square of the compression of the spring point Example response for part (e)(ii) From the equation in part (c), the compression of the spring is directly proportional to the square root of the height that the block is released Thus, the graph would be an x-axis parabola, as shown Total for part (e) points Total for question 15 points © 2021 College Board AP® Physics C: Mechanics 2021 Scoring Guidelines Question 3: Free-Response Question (a) 15 points point For using integral calculus to calculate the rotational inertia of the rod I = ∫r dm dm λ= dr γ x dx = point For correctly substituting γ x into the above equation = I dx ) ∫ x (γ x= 2 x=L x=L x5 3M L = γ = x dx γ = − 0 ∫ 5 L x =0 x =0 ML2 Total for part (a) (b) point For using integral calculus to determine the center of mass of the rod = X CM points ∑i mi xi ∫ xdm = ∑i mi ∫ dm For correctly substituting γ x into the numerator of the above equation point For correctly substituting M into the denominator of the above equation OR evaluating the integral ∫ dm to find the mass of the rod point x=L x=L γ x4 3M L x dx γ ∫ xλdx ∫ x γ x dx x ∫= = X CM L = = x 0= = = L= M M M M M ( ) OR x=L x=L γ x4 γ L4 γ x dx ∫ x γ x dx x ∫ xλdx = ∫= x 0= X= L = = = = CM = 2 x L= x L γ γ L L λ dx γ x ∫ ∫x = γ x dx 3 x =0 ( ) Total for part (b) points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines (c) For selecting “Greater than” with an attempted justification point For a correct justification point Example responses for part (c) Because more of the mass of the rod is at the end of the rod opposite point P, more mass is concentrated away from the axis of rotation; thus, the rotational inertia of the rod would be greater around point P than around its center of mass OR According to the parallel axis theorem, I=Icm + md2, if the axis is at a position away from the center of mass, the rotational inertia is larger than if the axis were at the center of mass Total for part (c) points (d) For a concave down curve that decreases to zero for the graph of τ as a function of t point For a concave down curve that approaches horizontal for the graph of ω as a function of t point For consistency between the two graphs point Total for part (d) points (e) For selecting “Decreases” with an attempted justification point For a correct justification point Example responses for part (e) As the rod rotates downward, the angle θ in the torque equation τ=rFsinθ decreases Thus, the torque on the rod decreases OR As the rod rotates downward, the lever arm between point P and the rod’s center of mass continues to decrease; thus, the torque on the rod decreases Total for part (e) points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines (f) point For using conservation of energy to calculate the speed of the rotating rod U i + Ki = U f + K f Ui + = + K f Ui = K f point For correctly substituting into the above equation mghi = Iω f point For correctly solving for the linear speed of point S Mg ( ) ( 3 L = ML2 )( ) 3 MgL = Mv ∴ = v 10 v L = gL ( 9.8 m s2 ) (1.0 m=) 4.9 m s Total for part (f) points Total for question 15 points © 2021 College Board ... the graph So the new line of data is predicted to be parallel to and below the original line Total for question 1 point 15 points â 20 21 College Board AP? ? Physics C: Mechanics 20 21 Scoring Guidelines. .. ) 0.35 m ( )( 0.30 m )( 0 .10 Total for part (d) points © 20 21 College Board AP? ? Physics C: Mechanics 20 21 Scoring Guidelines (e) i point For a correct justification Example response for part... released Thus, the graph would be an x-axis parabola, as shown Total for part (e) points Total for question 15 points © 20 21 College Board AP? ? Physics C: Mechanics 20 21 Scoring Guidelines Question