101e2sol01 - Physics 171.101 Exam 2 Oct 23 2001 Prof Barnett s a 59 T Q v.5 Do all FOUR problems You may use anything written on the front and back

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Unformatted text preview: Physics 171.101 Exam 2 Oct. 23, 2001 Prof. Barnett s a 59 T: Q v.5 Do all FOUR problems. You may use anything written on the front and back of ONE 3”x5” index card. Everything should be written in ink and no “snow paint” should be used. Use = 10 m/sec2, 7r : 3.0, Sin(370):O.6, Cos(37"):0.8, Tan(370):0.75. 1. Two cars A and B slide on an icy Baltimore road as they attempt to stop at a traffic light. The mass of A is 1000 kg and the mass of B is 1500 kg. The coefficient of kinetic friction between the icy road and the car tires is 0.10. Car A succeeds in stopping but car B does not and rear—ends car A. After the collision both cars slid through the intersection with their brakes locked. Car A stops 8 meters ahead and car B stops 2 meters ahead of their positions at impact. A) (10 pts) How much work does friction do on each car while bringing them to a halt? lFfl at W! = Mlmgl W = F - R WA 2 —(0.1)(1000 kg)(10 m/s2)(8 m) : —8000 Joules W3 : —(0.1)(1500 kg)(10 m/s2)(2 m) : —3000 Joules B) (10 pts) What was the speed of each car immediately after impact? %mAl7[§ 2 —WA V3 2 2(8000 J)/(1000 kg) = 16(m/s)2 17f : 4 m/s left. 17,; = 2 m/s left. C) (5 pts) What was the speed of car B immediately before impact? Use momentum conservationjsi : 15f. ’ffLAl/Xc + m1;ng 4000 kg m/s + 3000 kg m/s : 7000 kg m/s 2 mBVg V13: 4.67 m/s left D) (5 pts) Was this an elastic collision between the cars? If not, how much mechanical energy was gained or lost? The final kinetic enregy is |WA| + IWB| : 11, 000.]. The initial kinetic energy is %mB (V13)? émB(Vl§)2 = 16,333 J 5, 333 Joules lost. 2. Three masses are connected by a massless rOpe over a perfect pulley in Bloomberg Center as shown in the adjacent figure. The inclined surface is frictionless. The mass on the incline is m1 = 5 kg and the masses hanging from the rope are m2 = 2 kg and m3 2 3 kg. The incline makes an angle of 6 : 37° with respect to the horizontal surface of the earth. A) (5 pts) \Vhat is the value of the “normal” force acting on m1? |j\7| : mlg 003(37) : 40 newtons. Direction is up, J. to incline. B) (10 pts) What is the acceleration of mg? (5 kg)g— (5kg)g sin(37) : (m1+mg +m3)a, where “a>0” means the mass m3 accelerate towards the earth. a = 2 m/s downward, toward earth. C) (10 pts) What are the tensions T1 and T2 in the two ropes, R1 and R2, as defined in the figure? Find T2 by looking at all the forces on 7713. Gravity pulls down with force = mgg = 30 Newtons. The rope, R2, pulls up with tension T2. The sum of these equals mga. Thus, T2 2 24 Newtons. Find T1 by looking at all the forces on m2. Rope 1 pulls up while rope 2 and gravity pull down. The sums of these three forces equals mga. Thus, T1 : 40 Newtons. 3. NASA proposes to put a large spring on the surface of an asteroid and use it to shoot rocks with masses of 2 kg into space with escape velocity for use in assembling a space station. The mass of the asteroid is 1018 kg. The radius of the asteroid is 1.334 X 106 meters and G = 6.67 X 10—11 N mZ/kgz. The spring is to have a spring constant of 50 N/m. A) (10 pts) What speed must be imparted to the rocks by the spring? The total mechanical energy for escape velocity is “zero”. So at the surface of the asteroid K.E. + PE. : 0. §mV2—GMm/R = 0 ——> V2 = 2GM/R = 2(6.67><10‘“1018/(1.334>< 106) 2102(m/s)2. V : 10 m/s. ' B) (10 pts) How much must the spring be compressed in order for the rocks to achieve this speed? The energy stored in the Spring when it is compressed must equal the kinetic energy of the mass immediately after it is shot. éml/2 : ékXZ‘ X2 = 4 meters2 —% X : 2 meters. 4. Loretta Lockhorn is exercising in New Jersey by jumping rope. Her mass is 60 kg. She makes 30 jumps per minute and jumps 20 cm each time. A) (10 pts) How much work does she do for each jump? W : mgh = (60 kg)(10 m/s2)(0.2 m) = 120 Joules. B) (5 pts) What power is she producing? P = dW/dt = (120 Joules)/(2 seconds) = 60 watts. C) (10 pts) What impulse does the floor give to her during each impact with the floor? %ml/2 : 120 Joules —> IV] = 2 m/sec. I 2 AP. [II = 2mV : 240 kg m/sec. Direction : upward. ...
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This test prep was uploaded on 04/15/2008 for the course PHYSICS 171.101 taught by Professor Professorhenry during the Fall '08 term at Johns Hopkins.

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101e2sol01 - Physics 171.101 Exam 2 Oct 23 2001 Prof Barnett s a 59 T Q v.5 Do all FOUR problems You may use anything written on the front and back

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