2021 AP Exam Administration Scoring Guidelines AP Physics C Mechanics Set 2 AP ® Physics C Mechanics Scoring Guidelines Set 2 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 evaluating Newton’s second law equations for the system at rest m2 g − f =( m1 + m2 ) a = ∴ m2 g = f point For correctly substituting for static friction into above equation: m2 g= f= µ s FN= µ s m1 g point = µs m2 ( 0.20 kg ) = = 0.45 m1 ( 0.44 kg ) Total for part (a) (b) points For correctly drawing and labeling the horizontal forces of friction and tension on block of mass m1 point For correctly drawing and labeling the vertical forces of weight and normal force on block of mass m1 point For correctly drawing and labeling forces of weight and tension on block of mass m2 point For indicating that the gravitational forces on each block are different Example responses for part (b) point Scoring note: Examples of appropriate labels for the force due to gravity include: FG , Fg , Fgrav , W , mg , Mg , “grav force,” “F Earth on block,” “F on block by Earth,” FEarth on block , FE,Block , FBlock,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 (b) points © 2021 College Board AP® Physics C: Mechanics 2021 Scoring Guidelines (c) For correctly evaluating Newton’s second law equation for block 1: T − f = m1a point For correctly evaluating Newton’s second law equation for block 2: m2 g − T = m2 a point Combining the two equations m2 g − f = ( m1 + m2 ) a ∴ f = m2 g − ( m1 + m2 ) a Scoring note: Both points are earned for a single correct Newton’s second law equation for the two-block system For correctly substituting for kinetic friction into above equation m2 g − ( m1 + m2 ) a f = µk FN = µk m1 g = m2 g − ( m1 + m2 ) a ∴ µk = m1 g µk (d) point ( 0.20 kg ) ( 9.8 m s ) − ( 0.44 kg + 0.20 kg ) ( 2.3 m s ) 0.11 = ( 0.44 kg ) 9.8 m s ( ) Total for part (c) points For selecting “Yes” and attempting a relevant justification For a correct justification Example response for part (d) If the track is not level, the angle of the track must be incorporated into the equation for acceleration, and this could account for the larger coefficient of kinetic friction Total for part (d) point point points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines (e) i For drawing an appropriate best-fit line point ii For calculating slope using two points from the best-fit line point = slope ( ) ( − 2) m s ∆y = = 13.3 m kg s ∆x ( 0.45 − 0.15 )( kg ) For correctly using an expression that relates the slope to the acceleration due to gravity From= y mx + b = a (slope)m2 + (y -intercept) m2 g g = a ∴ slope = ( m1 + m2 ) ( m1 + m2 ) g = slope × ( m1 + m2 ) = (13.3 m kgs2 ) ( 0.44 kg + 0.20 kg ) = 8.5 m s Total for part (e) (f) point points For a correct justification point Example response for part (f) The acceleration would be greater because there would be a component of the gravitational force on block along the surface, which would be in the same direction as the tension force Total for question 15 points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines Question 2: Free-Response Question (a) 15 points point For integrating using the correct limits or constant of integration = I r =2L = λ r dr ∫ r =0 r =2L r r =0 λ = λ (( 2L )3 − 0) For correctly relating λ to M and L m M M λ= = ∴I = L3 = 2L 2L ( )( ) ( ) point ML2 Total for part (a) (b) i ii point For correctly substituting into an equation for the center of mass of an object in the horizontal direction ML ML M L + M ( L ) + mi xi 2 X CM L = = = = M M M mi + 2 ∑ ∑ ( )( ) ( ) ( ) ( ) For correctly substituting into an equation for the center of mass of an object in the vertical direction ML ML M ( L) + M L + mi yi 2 YCM L = = = = M M M mi + 2 ∑ ∑ ( ) ( )( ) ( ( ) point ) Total for part (b) (c) points For selecting “Less than” and attempting a relevant justification For a correct justification Example response for part (c) Because object B has more of its mass closer to the pivot than object A, the rotational inertia of object B must be less than that of object A Total for part (c) points point point points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines (d) For an acceleration graph that is concave down and begins horizontally point For an angular speed graph that is concave down and ends horizontally For consistency between the angular acceleration and angular speed graphs point point Example responses for part (d) (e) (f) Total for part (d) points For selecting “Decreasing” and attempting a relevant justification For a justification that indicates the lever arm for the torque is decreasing Example response for part (e) Because the horizontal position of the center of mass for the object is moving closer to the pivot, the lever arm for the force of gravity is decreasing so the angular acceleration decreases Total for part (e) point point points For using conservation of energy U g1 = K point For correctly relating the change in rotational kinetic energy to the change in gravitational potential energy mgh = I ω 2 For correctly substituting for h into the equation above L mg = Iω 2 point For an expression for ω that uses only the allowed symbols and is algebraically consistent with the previous steps Mg ( L ) 2mgh = ω = I IB point ( ) ω= point MgL IB Total for part (f) points Total for question 15 points â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines Question 3: Free-Response Question (a) 15 points point point point For correctly drawing and labeling the weight of the block For correctly drawing and labeling the force exerted by the track on the block For a correct justification consistent with the diagram Example response for part (a) The force Fg represents the weight of the block and always points downward The force FN represents the force the track exerts on the block to keep it moving in a circular path and points perpendicular to the surface of the track Scoring Note: Examples of appropriate labels for the force due to gravity include: FG , Fg , Fgrav , W , mg , Mg , “grav force,” “F Earth on block,” “F on block by Earth,” FEarth on block , FE,Block , FBlock,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 Scoring Note: If extraneous forces are present, a maximum of points can be earned Total for part (a) (b) i points For using conservation of energy U1 + K1 = U + K ∴ U1 + = U + K point For correctly relating the elastic potential energy at maximum spring compression to the gravitational potential energy at point B point U s1 + 0= U g + K ∴ K = U s1 − U g For a correct substitution into the equation above mv = k ( ∆x )2 − mgh2 B vB2 = k ( ∆x )2 − g ( 3R ) ∴ vB = m point k ( ∆x )2 − gR m â 2021 College Board APđ Physics C: Mechanics 2021 Scoring Guidelines ii point For correctly relating the centripetal force to speed from part (b)(i) mvB2 mv = r R For an answer consistent with part (b)(i) = FC FC = point 2 k ( ∆x ) m k − 6mg ( ∆x )2 − gR= R m R Total for part (b) (c) For correctly relating the net force to the diagram from part (a) and setting the normal force equal to zero For correctly substituting into the equation above 2 k ( ∆x ) k ( ∆x ) − 6mg = mg ∴ = mg ∴ ∆= x R R point point mgR k Total for part (c) (d) points For correctly relating the height of fall to the time of fall ( )( R ) 1 y = y0 + voy t + a y t ∴ H = + + gt ∴ t = = 2 g points point 8R g For correctly substituting into the equation for constant velocity consistent with part (b)(i) point k 8R D = vx t = ( ∆x )2 − gR g m (e) Total for part (d) points i For a correct justification ii For indicating that as the maximum compression of the spring increases, the distance D increases For indicating that the minimum value is due to the minimum speed needed to get through the track Example responses for part (e) The block needs a minimum speed to make it through point B on the track; thus, the horizontal line segment represents compressions of the spring for which the block does not make it to point B OR From the equation in part (d), the compression of the spring is directly proportional to the horizontal distance traveled by the block; thus, the graph would be a straight line The minimum value is the distance traveled when the compression of the spring generates the minimum speed needed to reach point B on the track Total for part (e) point point point points Total for question 15 points © 2021 College Board ... the equation above mv = k ( ∆x )2 − mgh2 B vB2 = k ( ∆x )2 − g ( 3R ) ∴ vB = m point k ( ∆x )2 − gR m â 20 21 College Board AP? ? Physics C: Mechanics 20 21 Scoring Guidelines ii point For correctly... points â 20 21 College Board AP? ? Physics C: Mechanics 20 21 Scoring Guidelines (d) For an acceleration graph that is concave down and begins horizontally point For an angular speed graph that is... kinetic friction Total for part (d) point point points â 20 21 College Board AP? ? Physics C: Mechanics 20 21 Scoring Guidelines (e) i For drawing an appropriate best-fit line point ii For calculating