PHY 211 Laboratory for 10/14/03

Exam II

Exam II will take place during the lab period on Tuesday, Oct 14. As before, you will have the full three hours. The exam will consist of 6 to 8 problems and cover from Chapter 4 through Chapter 7.3.

Below are a few sample exam problems. I have included hints and answers for each.

You are responsible for all material covered in lecture and homework assignments. DO NOT ASSUME that a subject not covered in the sample exam below will not show up on Tuesday's exam! During the exam, you will be provided with an equation sheet. You will be allowed to use only the equations on the sheet, or others that you derive from those on the sheet. As before, I would suggest that you create your own equation sheet from the equations in the summary at the end of each chapter and use only that when trying the problems below. Resist the temptation to simply look for a similar problem in the text or notes (the standard "get the homework done" technique). That will better prepare you for the exam.

Sample Physics 211 Exam II

  1. A 70 kg running back is traveling east to west at 8 m/s when he is tackled. At 0.5 seconds later he is traveling west to east at 2 m/s. What average force was exerted on him?

    Use F = DP/Dt, OR use F = ma, finding a from the cosntant acceleration equations. Ans: 1400 N

  2. A 200 lb wrecking ball is suspended vertically by a 50 ft cable. You push horizontally until the ball moves 5 ft. What force is necessary to hold it there?

    First draw the FBD! The cable makes an angle of sin-15/50 = 5.7o from the vertical. SFy = Tcos5.7 - 200 = 0 and SFx = Fpush - Tsin5.7 = 0. Fpush = 200tan5.7 = 20 lb.

  3. A 50 kg skier starts from rest at the top of a frictionless 37o slope and slides100 m down the incline to level ground.
    a) What was her speed at the bottom?
    b) Once on the level, a constant frictional force stops her in a distance of 200 m. What is the coefficient of friction between her skis and the snow.

    You can use F = ma, but I suggest conservation of energy.
    a) PEtop = KEbottom --> mg(100sin37) = 1/2mv2, so v = 34 m/s.
    b) WNC = Ffriction(200m)cos180 = DKE + DPE = 0-1/2m(34)2 + 0-0, so Ffriction = 147 N. But Ffriction = mkN = mkmg, so mk = 0.3

  4. An elevator (with passengers) has a mass of 700 kg and the counterweight has mass 400 kg.
    a) How much work is done by the motor in raising the elevator (and thus lowering the counterweight) by 50 m?
    b) What minimum power must the motor generate in order to do this in 12 seconds?

    Wmotor = DKE + DPE = 0 - 0 + 700kg(9.8)(50-0) +400(9.8)(0-50) = 1.47 x 105 J
    Power= Work/time = 1.47 x 105 J / 12 = 12.3 kW.

  5. A 20 kg coyote jumps off a 30 m cliff with his new ACME Springy Shoes. At the bottom, the springs compress 20 cm in bringing him to rest.
    a) What is the spring constant for the shoes?
    b) What average force did the shoes exert on him over the 20 cm?

    a) Conservation of energy --> PEg at the top = PEspring at the bottom. mgh = 1/2kx2 --> k = 2.94 x 105 N/m.
    b) Work = Fave(.2 m) = either PEg at the top or PEspring at the bottom. --> Fave = 29,400 N ... a squashed coyote!

  6. Your 80 kg friend is running W to E at 5 m/s. You (70 kg) make a flying tackle with speed 10 m/s running S to N. You and your friend stick together.
    a) What is the velocity of you and your friend just after the tackle?
    b) Was KE conserved in this collision? If not, how much was lost or gained?

    a) 80kg(5m/s)i + 70kg(10m/s)j = (80+70)v'. --> v' = 2.7i + 4.7j m/s.
    b) KE is lost. Find KEf - Find KEi = 1/2(150)(2.72 + 4.72) - [1/2(80)52 + 1/2(70)102]

  7. A 200 gm cue ball traveling at 3 m/s makes a head-on elastic collision with a 150 gm billiard ball initially at rest. What is the final velocity of each?

    Use 200(3) + 150(0) = 200v'1 + 150v'2 and 3-0 = v'2 - v'1. Substitute to find v'1 = .43 m/s and v'2 = 3.43 m/s

  8. A bicycle accelerates uniformly from rest to a speed of 10 m/s over a distance of 20 m. The wheels have radius of 30 cm. Assume the wheels do not slip on the ground.
    a) Through how many radians have the wheels turned after the 20 m linear distance?
    b) What was the angular acceleration of the wheels?

    a) s = rq --> 20 = .3(q) --> q = 67 radians.
    b) wo = vo/r = 10/.3 = 33 rad/s. Then use w2 =wo2 + 2a q --> a = -8.3 rad/s2