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MASSACHUSETTS INSTITUTE OF TECHNOLOGYDepartment of Physics

Physics 8.01 Fall 2001

50-MINUTE EXAM 2 SOLUTIONSMonday, October 22, 2001

FAMILY (LAST) NAME

GIVEN (FIRST) NAME

STUDENT ID NUMBER

Your class (check one) =⇒

INSTRUCTIONS:

1. SHOW ALL WORK. All work is to be done in thisbooklet. Extra blank pages are provided.

2. FORMULA SHEETS are in the back of this exam.You may tear them off.

3. This is a closed book exam. CALCULATORS,BOOKS, and NOTES are NOT ALLOWED.

4. For full credit, all answers must be expressed interms of the GIVEN VARIABLES, unless otherwisestated.

5. Do all FOUR (4) problems.6. PRINT your NAME on each page of this booklet

which you use for your solutions.7. Exams will be COLLECTED 5 minutes before the

hour.

Problem Maximum Score Grader

1 30

2 30

3 25

4 15

TOTAL 100

R01 MW 1:00 Jason Seely

R02 MW 2:00 Wit BuszaR03 MW 3:00 Brian PattR04 MW 1:00 Bruno Coppi

R05 MW 2:00 Bruno Coppi

R06 MW 2:00 Brian PattR07 MW 3:00 Vishesh KhemaniR08 MW 4:00 Vishesh KhemaniR09 MW 1:00 Michael FeldR10 MW 2:00 Michael FeldR11 MW 3:00 Michael FeldR12 TR 1:00 Norbert SchulzR13 TR 2:00 Norbert SchulzR14 TR 3:00 Sekazi Mtingwa

R15 TR 10:00 Jeffrey Bowers

R16 TR 11:00 Jeffrey Bowers

R17 TR 12:00 Yoav Bergner

R18 TR 9:00 Ali Nayeri

R19 TR 10:00 Ali Nayeri

R20 TR 11:00 Yoav Bergner

R21 TR 2:00 Ronak BhattR22 TR 3:00 Ronak BhattR23 TR 11:00 Hugh Manini

R24 TR 12:00 Hugh Manini

R25 TR 1:00 Sekazi Mtingwa

R26 TR 3:00 James McBride

8.01 Exam 2 Solutions, Fall 2001Name

Problem 1: Peter’s Elevator Ride (30 points)Each part is worth 5 points. Please mark your answers by circling them. For these

multiple choice questions you need not show your work, and there will be no partialcredit.

Peter, who has a mass of 80 kg, steps into an elevator and travels from the 4th tothe 3rd floor, a total vertical distance of 4 m. Peter’s trip in the elevator begins with theelevator at rest on the 4th floor, and ends after the elevator has come to rest on the 3rdfloor. In the following questions, we seek answers that apply in the frame of reference ofthe building. Take g = 10 m/s2.a) During the trip described above, how much work does the force of gravity do on

Peter?

(i) 3200 J (ii) −3200 J (iii) 3.2× 106 J (iv) −3.2× 106 J (v) zero

b) During this trip, Peter’s gravitational potential energy changes by how much? (Apositive number indicates an increase; a negative number indicates a decrease.)


c) How much work does the elevator floor do on Peter during the trip?


d) When the elevator is midway between the floors, its downward speed is 1 m/s. Fromthe time that the elevator begins at rest to this half-way point, how much work doesgravity do on Peter?

(i) 1600 J (ii) −1600 J (iii) 1640 J (iv) −1640 J(v) 1560 J (vi) −1560 J (vii) 1.60× 106 J (viii) −1.60× 106 J(ix) 1.64× 106 J (x) −1.64× 106 J (xi) 1.56× 106 J (xii) −1.56× 106 J(xiii) zero

e) Continue thinking about the situation described in part (d). From the time that theelevator begins at rest to this half-way point, how much work does the elevator floordo on Peter?

(i) 1600 J (ii) −1600 J (iii) 1640 J (iv) −1640 J(v) 1560 J (vi) −1560 J (vii) 1.60× 106 J (viii) −1.60× 106 J(ix) 1.64× 106 J (x) −1.64× 106 J (xi) 1.56× 106 J (xii) −1.56× 106 J(xiii) zero

f) Suppose now that Peter is carrying a briefcase, with a mass of 5 kg. From the timethat the elevator begins at rest on the 4th floor to the time it comes to rest againon the 3rd floor, how much work does Peter do on the briefcase?

(i) 200 J (ii) −200 J (iii) 2× 105 J (iv) −2× 105 J (v) zero

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Problem 2: Collisions on a frictionless track (30 points)

An initially stationary block of mass MA slides downward on a curved track. Thenet change in its height is denoted by h, with h > 0. The track levels off, as shown inthe diagram, so that the block is sliding on a horizontal section when it collides with astationary cart of mass MB. Take the acceleration of gravity as g (g > 0), and assumethat friction is negligible in all stages of this problem.

(a) (6 points) What is the speed v1 of the block just before the collision?

(b) (6 points) If the block sticks to the cart, what is the speed v2 of the block/cartsystem immediately after the collision?

(c) (9 points) The moving block/cart system thencollides with a stationary cart of mass MC . Sup-pose that this collision is perfectly elastic (i.e.,kinetic energy is conserved). Let v3 denote thex-component of the velocity of the block/cartsystem immediately after the collision, and letv4 denote the x-component of the velocity ofthe single cart after the collision. Write downone equation or a set of equations that could be solved to determine v3 in terms ofMA, MB, MC , and v2. You need not solve these equations.

(d) (9 points) Another way to analyze the collision discussed in part (c) is to work in thecenter-of-mass or zero-momentum reference frame — the reference frame in whichthe center of mass is at rest, and the total momentum is zero. What is the velocityof this frame relative to the laboratory frame (the frame in which the track is atrest), and what are the velocities of each object, before the collision, when measuredin this frame? You may express your answer in terms of MA, MB, MC , and v2.

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Problem 3: A suitcase placed on a conveyor belt (25 points)

A suitcase of mass Mis placed on a level conveyorbelt at an airport. The co-efficient of static friction be-tween the suitcase and theconveyor belt is µs, and thecoefficient of kinetic frictionis µk, with µk < µs. Theconveyor belt moves withconstant speed u, and attime t = 0 the suitcase isplaced on the conveyor withspeed v = 0.

(a) (6 points) At t = 0, what is the total force �F acting on the suitcase?

(b) (6 points) How long does the suitcase take to reach the speed of the conveyor belt(i.e. at what time t does v(t) = u)?

(c) (7 points) What is the work done by friction on the suitcase during this time? Besure that you give the sign as well as the magnitude.

(d) (6 points) After the suitcase reaches the speed of the conveyor belt, what is the forceof friction that acts on it?

Answer:

(a) The normal force upward cancels the force of gravity downward, so the total forceacting on the suitcase at t = 0 is equal to the force of friction. Since the suitcasehas zero velocity and the conveyor belt is moving, there is a nonzero relative velocitybetween the suitcase and the conveyor belt, and hence the friction is kinetic. So

|�F| = µkN = µkMg .

The direction opposes the relative motion. Since the velocity of the suitcase relativeto the conveyor belt is in the negative x-direction, the force on the suitcase is in thepositive x-direction .

(b) Since the total force is in the x-direction and is constant,

vx(t) = axt =Fx

Mt .

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Thus, the time t at which the speed reaches u is given by

Fx

Mt = u =⇒ t =

M

Fxu .

If one substitutes Fx = µkMg, one finds that

t =u

µkg.

(c) To find the work done by friction, start by noting that the suitcase moves in thepositive x-direction, which is the same direction as the force, so the work is positive.Since Fx is constant, the work done on the suitcase is

W = Fx�,

where � is the distance traveled. Since the acceleration is uniform,

� =12axt2 =

12

Fx

Mt2 .

Combining the equations,

W =1

2MF 2

x t2 =1

2M(µkMg)2

(u

µkg

)2

=12Mu2 .

Alternatively, one could have seen that W = 12Mu2 from the work-energy theorem,

since 12Mu2 is the increase in kinetic energy of the suitcase.

Friction always acts to oppose the relative motion of two surfaces. In this case thatmeans that friction causes the suitcase to accelerate in the direction of its motion,so its kinetic energy increases and the work done by friction is positive.

(d) Once the suitcase reaches the speed of the conveyor belt, there is no relative motionbetween the suitcase and the conveyor belt, so the relevant frictional force is that ofstatic friction. Static friction will exert whatever force is necessary to prevent thesurfaces from slipping, up to a maximum magnitude of µs|�N|, where �N is the normalforce. In this case no horizontal force is necessary to keep the suitcase moving withthe conveyor belt, so the force of friction will be zero .

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Problem 4: Raining on a brick wall (15 points)

A brick wall has the shape of a rectangular slab, with height h, widthw, and length �, as shown in the diagram. Assume that rain falls straightdownward on the wall at a constant speed v, and that the amount of rainin the air per unit volume is ρ. (In SI units, ρ would be measured inkg/m3.) The wall has mass M , and the acceleration of gravity is denotedby g, with g > 0.

a) (5 points) If the rain falls straight downward, what is the rate R (massper unit time) at which water strikes the top surface?

b) (6 points) If the rain collects on the top surface and then runs off the side withnegligible velocity, what force �F does the rainwater exert on the top surface of thebrick wall? (When asked for a vector, you should always make sure that you havespecified the direction as well as the magnitude.)

c) (4 points) Under the circumstances described in parts (a) and (b), what is the normalforce �N that the ground exerts on the wall?

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Solution to Problem 3 by Prof. Alan Guth; other solutions by Prof. Wit Busza.11

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