L 2. AP Physics Free Response Practice Oscillations ANSWERS 1975B7. (a) F T2. (b) F NET(Y) = 0

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1 AP Physics Free Response Practice Oscillations ANSWERS 1975B7. (a) 60 F 1 F g (b) F NE(Y) = 0 F1 F1 = g / cos(60) = g (c) When the string is cut it swings fro top to botto, siilar to the diagra for 1974B1 fro wor-energy probles on the opposite side as shown below P U top = K bot gh = ½ v hen apply F NE(C) = v v v v g ( cos 60) g ( ) g (F 1 g) = (g) / / r F 1 = g. Since it s the sae force as before, it will be possible. (d) his otion is not siple haronic because the restoring force, (F gx exactly, but in this 315

2 1983B. a) Apply oentu conservation perfect inelastic. p before = p after Mv o = (3M)v f v f = /3 v o 4Mv o b) Apply energy conservation. K = U sp ½ (3M)(/3 v o ) = ½ 3 c) Period is given by B1. a) i) p = v = (0.)(3) = 0.6 g /s ii) K = ½ v = ½ (0.)(3) = 0.9 J b) i.) Apply oentu conservation p before = p after = 0.6 g /s ii) First find the velocity after, using the oentu above 0.6 = (1.3+0.) vf v f = 0.4 /s, then find K, K = ½ ( 1 + ) v f = ½ (1.3+0.)(0.4) = 0.1 J c) Apply energy conservation K = U sp 0.1 J = ½ = ½ (100) x = d) Period is given by 0. 77s 1996B. a) Use a ruler and nown ass. Hang the nown ass on the spring and easure the stretch distance pulling the spring F sp is equal to the weight (g). Plug into F sp = b) First find the period s 500 then the frequency is given by f = 1/ =.5 Hz c) Put the spring and ass on an incline and tilt it until it slips and easure the angle. Use this to find the coefficient of static friction on the incline u s. hen put the spring and ass on a horizontal surface and pull it until it slips. Based on F net = 0, we have F spring s g, Giving g = F spring only less than 1 this will allow an g value to be registered larger than the spring force. A sipler solution would be to put the bloc and spring in water. he upwards buoyant force will allow for a weight to be larger than the spring force. his will be covered in the fluid echanics unit. 316

3 005B. a) FBD b) Apply F net(x) = 0 F net(y) = 0 P cos 30 = g P = 0.37 N P sin 30 = H = N H c) Conservation of energy Diagra siilar to 1975B7. U top = K botto gh = ½ v g( (10) (.3.3 cos 30) = ½ v v botto =.5 /s d) he bob will reach the lowest position in ¼ of the period s 4 g B005B. FBD i) ii) b) Apply energy conservation? U top = K botto gh = ½ v (9.8)(.08) ½ v v = 1.3 /s c) F net(c) = v /r F t g = v /r F t = v /r + g (0.085)(1.3) /(1.5) + (0.085)(9.8) F t = 0.93 N d) g and are the two factors that deterine the pendulu period based on g o double the value of, should be increased by 4x or g should be decreased by ¼. he easiest odification would be siply to increase the length by 4 x 317

4 006B1. a) FBD b) Siply isolating the 4 g ass at rest. F net = 0 F t g = 0 F t = 39 N c) ension in the string is unifor throughout, now looing at the 8 g ass, F sp = F t = 39 = (0.05) = 780 N/ d) 4 g ass is in free fall. D = v i t + ½ g t 0.7 = 0 + ½ ( 9.8)t t = 0.38 sec e) First find the period s then the frequency is given by f = 1/ = 1.6 Hz f) he 8 g bloc will be pulled towards the wall and will reach a axiu speed when it passes the relaxed length of the spring. At this point all of the initial stored potential energy is converted to inetic energy U sp = K ½ = ½ v ½ (780) (0.05) = ½ (8) v v = 0.49 /s C1989M3. a) Apply energy conservation fro top to end of spring using h=0 as end of spring. U = K gh = ½ v (9.8)(0.45) = ½ v v = 3 /s b) At equilibriu the forces are balanced F net = 0 F sp = g =()(9.8) = 19.6 N c) Using the force fro part b, F sp = 19.6 = 00 x = d) Apply energy conservation using the equilibriu position as h = 0. (Note that the height at the top position is now increased by the aount of ound in part c h new = h+ U top = U sp + K (at equil) ghnew = ½ + ½ v ()(9.8)(0.548) = ½ (00)(0.098) + ½ ()(v ) v = 3.13 /s e) Use the turn horizontal tric. Set equilibriu position as zero spring energy then solve it as a horizontal proble where K equil = U sp(at ax ap.) ½ v = ½ ½ ()(3.13) = ½ (00)(A ) A = f) his is the axiu speed because this was the point when the spring force and weight were equal to each other and the acceleration was zero. Past this point, the spring force will increase above the value of gravity causing an upwards acceleration which will slow the box down until it reaches its axiu copression and stops oentarily. 00 g) 0. 63s 318

5 C1990M3. a) Equilibriu so F net = 0, F sp = g x = g (0.0) = (8)(9.8) = 39 N/ b) First deterine the speed of the 3 g bloc prior to ipact using energy conservation U = K gh = ½ v (9.8)(0.50) = ½ v v = 3.13 /s hen solve perfect inelastic collision. p before = p after 1 v 1i = ( 1 + ) v f (3)(3.13)=(8)v f v f = 1.17 /s c) Since we do not now the speed at equilibriu nor do we now the aplitude the turn horizontal tric would not wor initially. If you first solve for the speed at equilibriu as was done in 1989M3 first, you could then use the turn horizontal tric. However, since this question is siply looing for an equation to be solved, we will use energy conservation fro the top position to the lowest position where the ax aplitude is reached. For these two positions, the total distance traveled is equal to the distance traveled to equilibriu + the distance traveled to the ax copression ( 1 + ) = (0.0 + ) which will serve as both the initial height as well as the total copression distance. We separate it this way because the distance traveled to the axiu copression fro equilibriu is the resulting aplitude x that the question is asing for. Apply energy conservation U top + K top = U sp(ax-cop) gh + ½ v = ½ (8)(9.8)(0.0 + ) + ½ (8)(1.17) = ½ (39)(0.0 + ) he solution of this quadratic would lead to the answer for which is the aplitude d) First find period 0. 90s hen find frequency f = 1/ = 1.11 Hz e) he axiu speed will occur at equilibriu because the net force is zero here and the blocs stop accelerating in the direction of otion oentarily. Past this point, an upwards net force begins to exist which will slow the blocs down as they approach axiu copressions and begin to oscillate. f) his otion is siple haronic because the force acting on the asses is given by F= directly proportional to the displaceent eeting the definition of siple haronic otion 319

6 C000M1. a) (c) t 10 (s) (s) (s ) b) c) We want a linear equation of the for y = x. Based on g g 4 g y = x his fits our graph with y being and x being. Finding the slope of the line will give us a value that we can equate to the slope ter above and solve it for g. Since the points don t fall on the line we pic rando points / g and solve for g = 8.69 /s d) A +/- 4% deviation of the answer (8.69) puts its possible range in between so this result does not agree with the given value 9.8 e) Since the value of g is less than it would norally be (you feel lighter) the elevator oving down would also need to be accelerating down to create a lighter feeling and saller F n. Using down as the positive direction we have the following relationship, F net = a g F n = a F n = g a For Fn the be saller than usual, a would have to be + which we defined as down. 30

7 C003M. a) Apply energy conservation U top = K bot gh = ½ v v gh b) Apply oentu conservation perfect inelastic p before = p after Mv ai = (M+M) v f 1 M ( gh ) = Mv f v f = gh c) Again we cannot use the turn horizontal tric because we do not now inforation at the equilibriu position. While the tray was initially at its equilibriu position, its collision with the clay changed where this location would be. Even though the initial current rest position iediately after the collision has an unnown initial stretch to begin with due to the weight of the tray and contains spring energy, we can set this as the zero spring energy position and use the additional stretch distance H/ given to equate the conversion of inetic and gravitational energy after the collision into the additional spring energy gained at the end of stretch. Apply energy conservation K + U = U sp (gained) ½ v + gh = ½ Plug in ass (), h = H/ and ½ ()v + ()g(h/) = ½ (H/) plug in vf fro part b (gh/4) + gh = H /8. Both sides * (1/H) g/ + g = H/8 3/ g = H/8 = 1g / H d) Based on M 1Mg H H 6g 31

8 C008M3 (a) (b) he slope of the line is F / Slope = 4 N/ (c) Apply energy conservation. U top = U sp (botto). Note that the spring stretch is the final distance the initial length of the spring = 0.90 gh = ½ (9.8)(1.5) = ½ (4)(0.9) = 0.66 g (d) i) At equilibriu, the net force on the ass is zero so F sp = g F sp = (0.66)(9.8) F sp = 6.5 N ii) F sp = 6.5 = (4) x = 0.7 iii) Measured fro the starting position of the ass, the equilibriu position would be located at the location ared by the unstretched cord length + the stretch found above = Set this as the h=0 location and equate the U top to the U sp + K here. gh = ½ +1/ v (0.66)(9.8)(0.87) = ½ (4)(0.7) + ½ (0.66) v v = 3.8 /s iv) his is the axiu speed because this is the point when the spring force and weight were equal to each other and the acceleration was zero. Past this point, the spring force will increase above the value of gravity causing an upwards acceleration which will slow the ass down until it reaches its axiu copression and stops oentarily. 3

9 Suppleental. (a) (b) F net = 0 F t = F sp = x = F t / (c) Using energy conservation U sp = U sp + K note that the second postion has both K and U sp since the spring still has stretch to it. ½ = ½ + ½ v ( = ( + Mv ¾ ( = Mv, plug in t /) = Mv v F t 3 M (d) o reach the position fro the far left will tae ½ of a period of oscillation. 1 M t (e) he forces acting on the bloc in the x direction are the spring force and the friction force. Using left as + we get F net = a F sp f = a Fro (b) we now that the initial value of F sp is equal to F t which is an acceptable variable so we siply plug in F t for F sp to get F t µ g = a a = F t / µ g M 33