Your Name: PHYSICS 101 MIDTERM. Please circle your section 1 9 am Nappi 2 10 am McDonald 3 10 am Galbiati 4 11 am McDonald 5 12:30 pm Pretorius

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1 1 Your Name: PHYSICS 101 MIDTERM October 25, hours Please circle your section 1 9 am Nappi 2 10 am McDonald 3 10 am Galbiati 4 11 am McDonald 5 12:30 pm Pretorius Problem Score 1 /13 2 /20 3 /20 4 /15 5 /18 6 /14 Total /100 Instructions: When you are told to begin, check that this examination booklet contains all the numbered pages from 2 through 15. The exam contains 6 problems. Read each problem carefully. You must show your work. The grade you get depends on your solution even when you write down the correct answer. BOX your final answer. Do not panic or be discouraged if you cannot do every problem; there are both easy and hard parts in this exam. If a part of a problem depends on a previous answer you have not obtained, assume it and proceed. Keep moving and finish as much as you can! Possibly useful constants and equations are on the last page, which you may want to tear off and keep handy Rewrite and sign the pledge: I pledge my honor that I have not violated the Honor Code during this examination. Signature

2 2 Problem 1: Grab Bag (a) [3 pts] Three balls, A, B, and C are thrown from the top of a cliff of height h = 100 m overlooking a flat land. The speed of the the three balls is identical: v A = v B = v C = 10 m/s. The angle in the figure is φ=π/4. V C φ φ V A V B h a.1 (2 pts) Which one of the three balls will have the largest speed at the moment it makes contact with the ground? Box your answer and explain your reasoning on the side. Ball A Ball B Ball C They all reach ground with the same speed The three balls have equal initial kinetic energy. They will also have equal final kinetic energy, as they drop by the same height. So they all reach ground with the same speed. a.2 (1 pts) Which of the three balls will reach the ground in the shortest time? Box your answer and explain your reasoning on the side. Ball A Ball B Ball C They all reach ground at the same time Ball B and C reach ground at the same time, before ball A. Ball B reaches the ground first. What matters here is only the motion in the y direction. Since ball B is the only one with an initial velocity directed towards ground, it will reach the ground first.

3 3 (b) [4 pts] A car on wheels rolls without friction on on horizontal surface. Bob and Billy are both playing on the car, and each of the two boys has a mass m of 50 kg. The mass M of the car is 100 kg. Bob Billy (b.1) [5 pts] Bob and Billy run to one end of the car in the unison. Their speed relative to the car is v = 10 m/s just before they jump off at the same time. Calculate the velocity u 1 of the car relative to the ground after the two boys have jumped off.

4 4 (b.2) [5 pts] This second time Bob anticipates Billy. Bob jumps off one end of the car with a velocity, relative to the car, v = 10 m/s. After Bob has jumped and is already off the car, it s the turn of Billy to jump. Billy jumps from the same end of the car, and is velocity relative to the car is also v = 10 m/s. Calculate the velocity u 2 of the car relative to the ground after both the boys have jumped off. (c) [5 pts] In which case, a or b, does the car attain the greater velocity relative to the ground?

5 5 Problem 2: Rock the Car! A car on wheels of mass m = 1000 kg is equipped with a pendulum of mass µ = 1 kg and length l = 1 m. A second car of mass M is on the same track. m l μ M a. [3 pts] With the two carts initially at rest, calculate the minimum speed that the pendulum bob of mass µ must possess at the bottom of its trajectory to perform one full revolution.

6 b. [3 pts] With the second car (mass M) initially still on the track, the first car (mass m) is traveling with constant velocity v = 5 m/s directed towards the second car, and the pendulum bob is vertical and at rest relative to the first car. When the first car hits the second car, the two stick together as a result of an inelastic collision. What is the minimum value of the mass M such that the pendulum, as a result of the collision, is able to perform a full revolution? 6

7 7 Problem 3. Roller Coaster. A cart of mass m is released with zero velocity at the top of the roller coaster and then runs in the loop of radius R shown in the picture. Picture to follow on Monday. a. [3 pts] If the cart has to stay on track in the loop of radius R, at what minimal height must it be released? I.e., find the minimum height h that will allow the cart to complete the loop of radius R. At the top of the loop: m v2 r = F N + mg. For zero normal force that gives v 2 = Rg. Using this equation and the conservation of energy: The solution of this equation is: mgh = 1 2 mv2 + mg2r = 1 mrg + mg2r 2 h = 5 2 R. b. [4 pts] Assume that the cart starts at an height H > h that allows it to complete the loop. Find the expression of the normal force as function of H and of the angle θ from the vertical. At each angle θ the equation of motion is: The conservation of energy gives: m v2 r = F N + mg cos θ. mgh = 1 2 mv2 + mgr(1 + cos θ). Eliminating v 2 one gets the result for the normal force we get: F N = mg ( 2H R 2 3 cos θ).

8 8 c. [4 pts] Compute the apparent weight of the cart at the top of the loop and at the bottom of the loop. Show that the difference is always 6mg, where m is the mass and g is the acceleration of gravity.(to answer this question you do not necessarily need to use your answer to part b.) From the expression derived in part b, one finds: F N (θ = π) F N (θ = 0) = 6mg. Alternatively, one can compute the normal forces at the top and at the bottom of the loop and compare them: F N top = m v2 top r mg, F N bottom = m v2 bottom r From energy conservation one derives: + mg. v 2 top = 2g(H 2R) v 2 bottom = 2gH. Substituting these velocities in the expressions for F N top and F N bottom : F N top = 2mg H R 5mg, F N bottom = 2mg H R + mg. Hence: F N bottom F N top = 6mg.

9 9 Problem 4. Target Practice. An archer is shooting arrows, each with a mass m of 165 g, at a target that is a distance l=100 m away. The bullseye of the target is a distance d=1.52 m above the ground. When the bullseye is in the archer s line of sight, i.e. looking directly down the shaft of the arrow with the bow drawn as shown in the figure, the arrow makes an angle of 0 relative to the horizontal. l = 100 m d = 1.52 m a. [4 pts] The archer shoots an arrow and it leaves the bow with a speed v 0 of 48.5 m/s. Ignoring air resistance, at what angle relative to the horizontal must the archer loose the arrow so that it hits the bullseye? Hint: you might find the following identity useful: 2 sin θ cos θ = sin(2θ) Equations of constant acceleration applicable here: y = y 0 + v 0 sin θ t 1 2 gt2 x = x 0 + v 0 cos θ t To hit the bullseye, y y 0 = 0 and x x 0 = l = 100 m; v 0 = 48.5 m/s. Solving for θ gives: θ = 1 [ ] g(x x0 ) 2 sin 1 = 1 [ ] 9.8 m s m v0 2 2 sin 1 = 12.3 (48.5 m/s) 2

10 10 b. [6 pts] A steady wind starts blowing. The archer now shoots a second arrow with exactly the same angle and speed as calculated in part a. Assume that the effect of the wind on the arrow during its flight is as if a constant force of F = 0.118N ˆx was applied to the arrow, where the ˆx direction is along the horizontal as illustrated in the figure. Where, relative to the bullseye, does the second arrow hit? Again, equations of constant acceleration, but now we have a net acceleration in the ˆx direction: a x = F x m = N = m s kg Therefore: y = y 0 + v y0 t 1 2 gt2 x = x 0 + v x0 t a xt 2 Now we want to solve for y y 0, given x x 0 = l = 100 m, v 0 = 48.5 m/s, θ = 12.3, v x0 = v 0 cos θ, and v y0 = v 0 sin θ. From the second equation we find the time of flight: t = v x0 ± v x02 + 2a x (x x 0 ). a x We get two positive solutions, the smaller one represents the initial impact: t = 2.15 s. Plugging this solution into the first equation, we obtain (y y 0 ) = 0.46 m. I.e., the arrow hits 0.46 m below the bullseye.

11 11 Problem 5. The Velodrome. A velodrome has the form of a circular bowl whose flat bottom has a radius of 25 m. The side of the velodrome has a radius of curvature of 25 m, as shown in the figure below. Ignore air resistance in this problem. a. [2 pts] The smallest circle that the bicycle riders can travel in has a radius of 25 m. If the coefficient of static friction is µ s = 0.5, what is the largest speed v A for uniform circular motion (rolling without slipping) around this circle? For rolling without slipping, the friction is static friction, and the frictional force is perpendcular to the velocity vector v. For motion in a circle, the frictional force F fr provides the needed centripetal acceleration: F fr = mv 2 /r, where m is the combined mass of the bicycle and the rider. For motion in a horizontal plane, the normal force has magnitude N = mg. Hence, the maximum speed v of the circular motion is related by mv 2 /r = F fr = µ k N = µ k mg, and: v = µ k gr = = 11.1 m/s = 39.8 km/hr.

12 12 b. [1 pt] What is the angular velocity ω for the motion in part a? ω = v r = = 0.44 rad/s. c. [2 pts] A bicycle rider can pedal with a top speed of 15 m/s. Suppose he rides in a circle that passes through point B on the figure, and that he rides such that no friction is required to maintain uniform circular motion. What is the angle θ for this motion? The only forces on the bicycle are its weight mg and the normal force N. There is no vertical acceleration, so mg = N cos θ. The equation for uniform circular motion in the horizontal circle of radius R = r(1+sin θ) is mv 2 /R = N sin θ = mg tan θ. Hence, tan θ(1+sin θ) = v 2 /rg = (15) 2 /(25 9.8) = Using the trig functions on a calculator, we find that θ 33. d. [1 pt] What is the angular velocity ω for the motion in part c? ω = v R = v r(1 + sin θ) = 15 25(1 + sin 33 ) = 0.39 rad/s.

13 13 Problem 6. The Rube Goldberg Orange Juice Squeezing Machine. Milkman takes empty milk bottle (A), pulling string (B) which causes sword (C) to sever cord (D) and allow guillotine blade (E) to drop and cut rope (F) which releases battering ram (G). Ram bumps against open door (H), causing it to close. Grass sickle (I) cuts a slice off end of orange (J), at the same time spike (K) stabs prune hawk (L) who opens his mouth to yell in agony, thereby releasing prune and allowing diver s boot (M) to drop and step on sleeping octopus (N). Octopus awakens in a rage and, seeing diver s face which is painted on orange, attacks it and crushes it with tentacles, thereby causing all the juice in the orange to run into glass (O)... a. [1 pt] If the guillotine blade (E) falls 65 cm before cutting rope (F), what is the speed of the blade just before it cuts the rope? v = 2gh = m s m = 3.6 m/s.

14 14 b. [2 pts] The battering ram (G) has a mass of 52 kg. Each of the massless chains that support it have a length of 1.73 m and make an initial angle of 23 to the vertical. What is the momentum of the battering ram just before it bumps the door (H), if the chains are vertical at this moment? The battering ram falls through a vertical height of: h = l(1 cos θ) = 1.73 m (1 cos 23 ) = m. The speed v of the ram when the chains are vertical is given by: v = 2gh = ms m = 1.64 m/s. The momentum of the ram just before bumping the door is: P = mv = 52 kg 1.64 m/s = 85.4 kg-m/s. The ram is moving horizontally at this time, so the momentum vector is horizontal. c. [1 pt] The battering ram travels parallel to the wall and bumps into the door at a point 67 cm out from the wall. What is the angular momentum of the battering ram about the axis of rotation of the door? The door is initially at rest and is open by 90 with respect to the wall. L ram = mvr sin 90 = 52 kg 1.64 m/s 0.67 m 1 = 57.1 kg-m 2 /s d. [1 pt] The door has a mass of 33 kg and a width of 75 cm. What is the moment of inertia of the door about its axis of rotation? I = Mw 2 /3 = 33 kg (0.75 m) 2 /3 = 6.2 kg-m 2.

15 15 e. [5 pts] What is the angular velocity of the door about its axis of rotation just after it has been bumped by the battering ram, assuming that the collision is elastic? During the collision there can be an impulsive force on the hinges of the door. So, momentum is NOT conserved during the collision. But, since the force on the hinges exerts no torque on the door (about the axis of rotation of the door), angular momentum about the axis of rotation of the door is conserved: mvr sin 90 = mv r sin 90 + Iω, where v = speed of ram after the collision, I = moment of inertia of the door (found in part d), and ω = desired angular velocity of the door after the collision. Conservation of energy during the elastic collision implies that: From the first equation, we find: 1 2 mv2 = 1 2 mv Iω2. v = v Iω mr, and hence v 2 = v 2 2 Iv mr ω + I2 m 2 r 2 ω2, and from the second equation we have: v 2 = v 2 Iω2 m. Thus, v 2 2 Iv mr ω + I2 m 2 r 2 ω2 = v 2 Iω2 m. The quadratic equation for ω is: ( I 2 m 2 r 2 + I m ) ω 2 2 Iv mr ω = 0. The solution ω = 0 is not relevant. The nontrivial solution is: ω = v r m/s = 1 + I/mr m 2 = 3.7 rad/s. 6.2 kg m kg (0.67 m) 2

16 16 POSSIBLY USEFUL CONSTANTS AND EQUATIONS You may want to tear this out to keep at your side L = Iω I = Σm i ri 2 x = x 0 + v 0 t + at 2 /2 PE = mgh KE = 1 2 Iω2 KE = 1 2 mv2 ω = ω 0 + αt ω 2 = ω α θ θ = ω 0 t αt2 v = v 0 + at F t = p F = GMm/r 2 F = µn s = Rθ τ = F l sin θ Στ = Iα v = Rω p = mv a c = v 2 /r W = F s cos θ v 2 = v a x W nc = KE + PE a = Rα I = 1 2 mr2 [disk] I = 2 5 mr2 [sphere] R Earth = 6400 km M Earth = kg G = Nm 2 /kg 2