Representing Motion AP Physics C Mechanics FRQ – Exam Style Questions etc.
Representing Motion AP Physics C Mechanics FRQ
Unit: 1. Kinematics
Weightage : 10-15%
Question
Two 5-kg masses are connected by a light string over two massless, frictionless pulleys. Each block sits on a frictionless inclined plane, as shown above. The blocks are released from rest.
(a) Determine the magnitude of the acceleration of the blocks. Now assume that the \(30^{\circ}\) incline provides a resistive force which depends on speed v. This resistive force causes the entire system’s acceleration to be given by the expression a = 1.8 − 0.03v where a speed v in m/s gives an acceleration in m/\(s^{2}\). The blocks are again released from rest.
(b) i. On the axes below, sketch a graph of the speed of the 5 kg block as a function of time. Label important values, including any asymptotes and intercepts.
ii. Explain how the expression for acceleration leads to the graph you drew.
(c) Explain how to figure out the terminal speed of the 5 kg block.
(d) The terminal speed is 60 m/s – that’s a typical speed in automobile racing. Explain briefly why this result is physically reasonable, even though the blocks are on a track in a physics laboratory.
Answer/Explanation
Ans:
(a) 1 pt: Write Newton’s second law for the direction along the plane for each block. For the right block, T − mg sin 30° = ma For the left block mg sin 60° − T = ma
1 pt: for including consistent directions for both equations
1 pt: for including a trig function in both equations, even if it’s the wrong function
1 pt: solve these equations simultaneously to get the acceleration
1 pt: answer is a = 1.8 m/\(s^{2}\). (An answer of a =−1.8 m/\(s^{2}\) is incorrect because the magnitude of a vector cannot be negative.) (Alternatively you can just recognize that mg sin 60° pulls left, while mg sin 30° pulls right, and use Newton’s second law directly on the combined system. Be careful, though, because the mass of the ENTIRE system is 10 kg, not 5 kg!)
(b) i. 
1 pt: for a graph that starts at v = 0, has exponential form, and sort of levels off to a maximum speed.
1 pt: for indicating that the asymptote represents the terminal speed, or 60 m/s.
ii. 1 pt: since acceleration is dv/dt, we have a differential equation: dv/dt = 1.8 − 0.03v. We need to find the expression for v.
1 pt: In a differential equation where the derivative is proportional to the function itself, the solution has an exponential term.
1 pt: In the solution e is raised to the −0.03t power because 0.03 is what multiplies the v.
1 pt: Since the blocks start at rest and speed up, the solution includes \((1-e^{-0.03})\) not just \(e^{-0.03t}\). [Those previous three points could be earned for a direct algorithmic solution for v, using separation of variables and integration.]
1 pt: For large t, the exponential term goes to zero, and so \((1-e^{-0.03t})\) should be multiplied by the terminal velocity. That’s why there’s an asymptote.
(c) 1 pt: The terminal velocity is when acceleration equals zero.
1 pt: solve the equation 0 = 1.8 − 0.03v. (You get 60 m/s.)
(d) 1 pt: The blocks only reach this terminal speed after a long time. An incline in a physics lab can’t be more than a couple of meters long, meaning the blocks don’t have the time and space to reach 60 m/s.