Solution - American Association of Physics Teachers
Solution - American Association of Physics Teachers Solution - American Association of Physics Teachers
2013 F = ma Exam 1618. Two point particles, each of mass 1 kg, begin in the state shown below.3y (m)210−1−21.0 m/s1.0 m/s−3−3 −2 −1 0 1 2 3x (m)The system evolves through internal forces only. Which of the following could be the state after some time haspassed?33(A)y (m)210−11.0 m/s1.0 m/s(B)y (m)210−12.0 m/s−2−2−3−3 −2 −1 0 1 2 332x (m)−3−3 −2 −1 0 1 2 332x (m)1.0 m/s(C)y (m)10−1−21.0 m/s1.0 m/s(D)y (m)10−1−21.0 m/s−3−3 −2 −1 0 1 2 33x (m)−3−3 −2 −1 0 1 2 3x (m)211.0 m/s(E)y (m)0−1−21.0 m/s−3−3 −2 −1 0 1 2 3x (m)SolutionFirst, linear momentum is conserved; as it happens, the system begins with zero linear momentum. Thiseliminates choice (B).Copyright c○2013 American Association of Physics Teachers
2013 F = ma Exam 17Angular momentum about any point is also conserved. (It is specified that the particles are point particlesto rule out their having spin angular momentum.) As it happens, the system has zero linear momentum, sothe angular momentum is the same measured about any point; otherwise it would suffice to consider any onepoint. The total angular momentum must be 2 kg m 2 /s counterclockwise. This eliminates choices (C) and(D).Finally, because the linear momentum of the system is zero, the center of mass does not move. This eliminateschoice (A).The following information applies to questions 19, 20, and 21.A simple pendulum experiment is constructed from a point mass m attached to a pivot by a massless rod of lengthL in a constant gravitational field. The rod is released from an angle θ 0 < π/2 at rest and the period of motion isfound to be T 0 . Ignore air resistance and friction.19. At what angle θ g during the swing is the tension in the rod the greatest?(A) The tension is the greatest at the point θ g = θ 0 .(B) The tension is the greatest at the point θ g = 0.(C) The tension is the greatest at an angle θ g with 0 < θ g < θ 0 .(D) The tension is constant.(E) None of the above is true for all values of θ 0 with 0 < θ 0 < π/2.20. What is the maximum value of the tension in the rod?(A) mg(B) 2mg(C) mLθ 0 /T 02(D) mg sin θ 0(E) mg(3 − 2 cos θ 0 )21. The experiment is repeated with a new pendulum with a rod of length 4L, using the same angle θ 0 , and the periodof motion is found to be T . Which of the following statements is correct?(A) T = 2T 0 regardless of the value of θ 0 .(B) T > 2T 0 with T ≈ 2T 0 if θ 0 ≪ 1.(C) T < 2T 0 with T ≈ 2T 0 if θ 0 ≪ 1.(D) T > 2T 0 for some values of θ 0 and T < 2T 0 for other values of θ 0 .(E) T 0 and T are undefined because the motion is not periodic unless θ 0 ≪ 1.SolutionThe mass accelerates towards the pivot due to centripetal acceleration; if its speed is v, this acceleration isa c = v2LTwo forces act in the radial direction; the tension in the rod, inward, and the radial component of gravity,outward. At an angle θ to the vertical the radial component of gravity isF g,rad = mg cos θFrom Newton’s second law we thus have for the tension FF − F g,rad = ma cCopyright c○2013 American Association of Physics Teachers
- Page 2 and 3: 2013 F = ma Exam 21. An observer st
- Page 4: 2013 F = ma Exam 43. Tom throws a f
- Page 8 and 9: 2013 F = ma Exam 8The following inf
- Page 10 and 11: 2013 F = ma Exam 1011. A right-tria
- Page 12 and 13: 2013 F = ma Exam 1213. There is a r
- Page 14 and 15: 2013 F = ma Exam 1416. Inspired by
- Page 18 and 19: 2013 F = ma Exam 18F = mg cos θ +
- Page 20: 2013 F = ma Exam 2025. A box with w
2013 F = ma Exam 1618. Two point particles, each <strong>of</strong> mass 1 kg, begin in the state shown below.3y (m)210−1−21.0 m/s1.0 m/s−3−3 −2 −1 0 1 2 3x (m)The system evolves through internal forces only. Which <strong>of</strong> the following could be the state after some time haspassed?33(A)y (m)210−11.0 m/s1.0 m/s(B)y (m)210−12.0 m/s−2−2−3−3 −2 −1 0 1 2 332x (m)−3−3 −2 −1 0 1 2 332x (m)1.0 m/s(C)y (m)10−1−21.0 m/s1.0 m/s(D)y (m)10−1−21.0 m/s−3−3 −2 −1 0 1 2 33x (m)−3−3 −2 −1 0 1 2 3x (m)211.0 m/s(E)y (m)0−1−21.0 m/s−3−3 −2 −1 0 1 2 3x (m)<strong>Solution</strong>First, linear momentum is conserved; as it happens, the system begins with zero linear momentum. Thiseliminates choice (B).Copyright c○2013 <strong>American</strong> <strong>Association</strong> <strong>of</strong> <strong>Physics</strong> <strong>Teachers</strong>
Change language