125 Physics Final Optional
December 12, 1998 This is a closed book exam. You have 3 hours to do the exam. IT IS OPTIONAL. This exam will be used to replace one of the in class tests. Your grade will be based on the four highest grades. You may use a calculator. Where appropriate, draw neat diagrams, labeling your axes and clearly indicating what variables you are using. Use Cartesian coordinates. Give solutions to 2 or 3 significant figures. Grading is based on what appears on your paper. In general, the answer to a problem by itself is not sufficient to obtain credit. The grader should be able to determine, without guessing, the steps you took to arrive at a solution. REGRADES can go either way. That is, errors made in grading do not always reduce your grade; they might have incorrectly increased your grade. (Most requests for regrades yield no change. Usually the student just did not understand how the grade was arrived at.) If more than one answer is given only one can be graded. Please indicate clearly which one you want to be graded. Please remember to put your name on the test booklet and SIGN THE PLEDGE.
1. (15 points) Trajectories You are riding in an elevator rising with a constant velocity, v Y = 4.0 m/s. At a height of y = 10 m above the ground, you throw a ball horizontally, with a initial velocity of v X = 10 m/s. Use g = 9.8 m/s2. From the frame of reference of a person on the ground, a). (7 points) What is the highest point above the ground the ball reaches?
2 v F = v 0 + 2a x 2
where a = -g, vF = 0.0, and v0 = 4.0. hence y = 16/(2 x 9.8) y = 0.816 m y = 10.816 m b). (8 points) Where does the ball hit the ground, if it was released at x = 0 m?
x - x 0 = v0 t + 1 2 at 2
where x x0 = -10.0, v0 = 4.0 and a = -g thus 4.9t2 4.0t 10.0 = 0.0 using x =
- b b 2 - 4ac 2a
we get t = 1.89 s thus x = vt x = 10 x 1.89 x = 18.9 m
2. (20 points) Springs and Collisions A mass M1 = 0.5 kg is attached to a massless ideal spring with spring constant k = 200 N/m. It is initially at rest and compressed to a distance x = 25 cm. When it is released it performs Simple Harmonic motion. a). (5 points) What is the angular frequency of the SHO motion? 2 = k/m = 20 rad/s or s-1 b). (5 points) What is v1 at x = 0.0? x = xM cos(t) v = -xM sin(t) v = 0.25 x 20 v = 5.0 m/s (did not ask for sign) Now a mass M2 = 0.75 kg is placed (at rest) at x = 0.0 cm. Then M1 is again compresses to 25 cm and released. M1 and M2 make a perfectly inelastic collision, (i.e. they stick together). c) (5 points) What is the new amplitude, x M of the combined masses? For a perfectly inelastic collision, m1v1 + m2v2 = (m1+m2)VF v1 = 5.0, v2 = 0.0 VF = 0.5 x 5.0/(0.5+0.75) VF = 2.0 m/s ET = 1/2kxM2 = 1/2MV2 xM2 = MV2/k xM2 = 1.25 x 2 x 2/200 xM = 16 cm d). (5 points) What is the angular frequency of this new SHO? 2 = k/M 2 = 200/1.25 = 12.6 rad/s or s-1
3. (15 points) Rolling and Energy A uniform solid sphere of mass M and radius R is at the top of a ramp that flattens out to a level surface. The sphere starts a rest at the top of the ramp with its CM at y = H. When it reaches the level surface at point B, its CM is at y = 0. Assume there is no energy lost during the motion. a). (5 points) What is the velocity of the sphere at point B if it is sliding without rolling on a frictionless ramp? MgH = 1/2Mv2 v = (2gH)1/2 b). (5 points) What is the velocity of the sphere at point B if it rolls without slipping on the ramp? Mgh = 1/2Mv2 + 1/2I2 I = 2/5MR2 v = R MgH = 1/2Mv2 + 1/2 x 2/5x MR2 x (v/R)2 MgH = 7/10Mv2 v = (10/7gH)1/2 c). (5 points) If the sphere is replaced by a hoop of mass M and R radius what is the answer to part b). I = MR2 MgH = 1/2Mv2 + 1/2MR2(v/R)2 MgH = Mv2 v = (gH)1/2 y=H M
y=0
4. (15 points) Work, Kinetic Energy, Potential Energy and Friction A body of mass, m = 10 kg, is sliding down a plane. (See figure). The plane has a coefficient of friction of = 0.25 and makes an angle of = 30 with respect to the horizontal. The mass moves down the plane a distance s = 10 m and with an initial kinetic energy of 100 J. Take the potential energy to be zero at the lowest point it reaches. Use g = 9.8 m/s2. a). (5 points) What is the change in potential energy, U? U = mg(hF-hI) h = -s sin 30 U = -10 x 9.8 x 10 x 0.5 U = -490 J b). (5 points) What is the energy lost to the frictional forces? Wf = fK s fK = KN where N = mg cos30 Wf = 0.25 x 10 x 9.8 x 0.866 x 10 Wf = 212.5 J c). (5 points) What is the velocity of the mass after it has gone 10 m? KEF = KEI -U Wf 1/2mv2 = 100 + 490 212.5 v = 8.7 m/s FK
mg
30
5. (15 points) Waves and resonance. a). (5 points) A pulse is propagating along a string with velocity, v = 20 m/s. Which of the following equations could be a possible wavefunction for the pulse? (1) y = At2/(x2 + 40) (2) y = A/(C + (x 10t)2) (3) y = A/(C + (x + 20t2)) (4) y = A/(x2 - 40xt + 400t2) The answer is (4). The wave must be of the form f(x-vt) with v = 20 m/s. Number 4 has the form A/(x-20.0t)2. b). (5 points) For the following refer to the diagram below. A string is stretched between two points a distance, L = 3.0 m. One end is driven by an oscillator. The wave speed on the string is v = 120 m/s. Assume that the amplitude of the oscillator is small. What frequency results in the pattern shown? = 3.0 m v = 120 m/s f = v/ f = 40 Hz c). (5 points) Now assume that the right hand side of the string can slide up and down freely, (not shown). The wave speed on the string is still v = 120 m/s. What frequency of the oscillator will drive the fundamental mode, f0. Now the fundamental is 1/4 of a wave length. = 12 m f = v/ f = 10 Hz
L
Osc.
6. (20 points) SHO (Advice: This is the hardest problem on the exam. Do the others first). A uniform solid sphere of radius r = 5.0 cm and mass M = 0.50 kg is rolling back and forth in the bottom of a spherical bowl. It rolls without slipping and performs Simple Harmonic motion. The bowl has a radius of R = 2.0 m. Assume that R>>r, that is assume that the CM of the sphere is at a distance R from the center of the bowl. Use g = 9.80 m/s2 a). (10 points) What is the force on the mass due to the rolling friction? The rolling force is given by = rFR = I where = d2/dt2 and is angular variable describing the rotation of the sphere around its CM. I = 2/5Mr2 putting all this together we get FR = 2/5Mrd2/dt2 b). (10 points) What is the period of this SHO? again = I but now I = MR2 (CM at distance R) and = d2/dt2 also = RF where F is the sum of all forces acting on the sphere perpendicular to the radius. These are the rolling force FR from part a) above and the component of gravity, F = -mgsin (See figure below) MR2d2/dt2 = R(-mgsin + FR) we can plug in the equation for FR from part a). but it is in terms of the variable and we need it in terms of the variable . We can relate the two using the rolling condition. R = -r (The minus sign is really tricky. When the sphere is moving clockwise in the bowl, it is rotating counterclockwise!). Thus d2/dt2 = -d2/dt2(R/r) Putting it all together we get MR2d2/dt2 = -R(mgsin - 2/5Mrd2/dt2(R/r)) using the small angle approximation and doing the algebra d2/dt2(1+2/5) = -g/R 2 = 5g/7R T = 2(7R/5g)1/2 T = 3.36 s
R
r -mgsin
FR N = mgcos
mg