MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01
Fall 2002
50-MINUTE EXAM 1 SOLUTIONS Wednesday, September 25, 2002
FAMILY (LAST) NAME
GIVEN (FIRST) NAME
STUDENT ID NUMBER
Please check (
) your class
INSTRUCTIONS: 1. FORMULA SHEETS are in the back of this exam. You may tear them off. There are also BLANK PAGES in case you need them. 2. This is a closed book exam. CALCULATORS, BOOKS, and NOTES are NOT ALLOWED. 3. Unless otherwise stated, to earn full credit you must show a valid DERIVATION and/or EXPLANATION of your answer, and you must express it in terms of the GIVEN VARIABLES.
Problem
Maximum
1
20
2
25
3
25
4
30
TOTAL
100
Score
Grader
R01 R02 R03 R04 R05 R06 R07 R08 R09 R10
MW MW MW MW MW MW MW MW MW MW
1:00 2:00 3:00 1:00 2:00 2:00 3:00 4:00 1:00 2:00
Brian Ross Wit Busza Ian Ellwood Ian Ellwood Tanim Islam Paul Joss Paul Joss Paul Joss Maria Chan Pavlos Kazakopoulos
R11 R12 R13 R14 R15 R16
MW 3:00 TR 1:00 TR 2:00 TR 3:00 TR 10:00 TR 11:00
Maria Chan Brian Ross Brad Plaster Brad Plaster Tanim Islam Qudsia Ejaz
R17
TR 12:00
Qudsia Ejaz
R18 R19 R20 R21 R22
TR 9:00 TR 10:00 TR 11:00 TR 2:00 TR 3:00
Walter Lewin Walter Lewin Walter Lewin Nathan Collins Pavlos Kazakopoulos
R23 R24 R25 R26
TR 11:00 TR 12:00 TR 1:00 TR 3:00
Vishesh Khemani Vishesh Khemani Nathan Collins James McBride
8.01 Exam 1, Fall 2002 Name
Problem 1: Romeo tossing pebbles* (20 points ) Romeo is tossing pebbles gently up to Juliet’s window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden at a distance h below her window and a distance from the base of the wall. How fast are the pebbles going when they hit her window? Take the acceleration of gravity to be g. (That is, g is a positive number describing the magnitude of the gravitational acceleration vector. In answering this question, you should assume that the point at which the pebbles are released is a distance h below the window and a distance from the wall.)
* The text and diagram of this problem were adapted from Physics for Scientists and Engineers, 3rd Edition, by Douglas C. Giancoli, Chapter 3, Problem 75.
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Problem 2: Description of motion (25 points ) For the following questions you can mark your answer by circling it. For these multiple choice questions you need not show your work, and there will be no partial credit. (a) (6 points ) For t < t0 a spaceship travels along a trajectory r(t) = [ 0, 0, v0 t ] . At time t = t0 the crew suddenly discovers that there is an obstacle directly ahead. A deflector rocket is fired at t = t0 , causing the ship to accelerate with acceleration a = [ a, 0, 0 ] , where a is a constant. What trajectory r(t) will the spaceship follow for t > t0 ? (i)
1
(iv)
1
2 2 at , 2 a(t
0, 0
− t0 )2 , 0, v0 t
(ii)
1
(v)
1
2 a(t
− t0 )2 , 0, 0
2 2 at ,
0, v0 (t − t0 )
(iii)
1
(vi)
1
2 2 at , 2 a(t
0, v0 t
− t0 )2 , 0, v0 (t − t0 )2
(b) (7 points ) At time t = 0, a bee is located at the origin of a coordinate system and is flying at velocity v = [ 3, 0, 0 ] m/s. She spots a flower at location r = [ 1, 0, 2 ] m, and wishes to reach the flower in 2 seconds. Assuming that she intends to maintain a uniform acceleration for the 2 seconds, what should the acceleration be? (i) [ 1, 0, 2 ] m/s2 (iv) [ −5, 0, 2 ] m/s2 (vii) [ 7, 0, 2 ] m/s2
, 0, 1 m/s2 2 (v) − 52 , 0, 1 m/s2 (viii) 72 , 0, 1 m/s2 (ii)
1
— Problem 2 continues on page 8 —
6
1
1 2
m/s2 (vi) − 54 , 0, 12 m/s2 (ix) 74 , 0, 12 m/s2
(iii)
4 , 0,
8.01 Exam 1, Fall 2002 Name
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Problem 2 Continued: (c) (6 points ) Asleep at the helm, Sulu awakens in embarrassment just as his starship Excelsior is careening past his destination, Star Base Delta 11, at a high speed v0 . His warp drive out of commission, Sulu is forced to rely on Newtonian mechanics to return his starship to its destination. Firing its rockets with maximum thrust in reverse, the Excelsior achieves a deceleration of magnitude amax . What will be the distance between the Excelsior and Delta 11 when the Excelsior comes to rest? (i)
v0
(ii)
amax
2v0 amax
(iii)
v0 2amax
(iv)
v02 amax
(v)
2v02 amax
(vi)
v02 2amax
(d) (6 points ) Continuing from the previous part, Sulu must now plan a course back to Delta 11, at a distance . This time, however, his starship must slow to zero velocity just as it reaches the Star Base, so that it will be prepared to dock. He therefore accelerates with an acceleration of magnitude amax halfway to Delta 11, and then decelerates with a deceleration of magnitude amax for the second half of the trip. How long will it be from the time he is at rest at a distance to the time he reaches Delta 11? (i)
2 amax
(v) 2
(ii)
amax
(iii)
amax (vi) amax
2amax 1 (vii) 2 amax
8
(iv)
4amax 1 (viii) 4 amax
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Problem 3: Blocks and springs (25 points) A block of mass M0 slides on a frictionless horizontal plane. It is attached to a spring of spring constant k0 , which is attached at its other end to a fixed support. Assume that the spring obeys Hooke’s law. The block is observed to slide back and forth along a line with a period T0 . (a) (6 points ) Suppose the block of mass M0 is replaced by another block of mass M = 2M0 . What is the new period? (i) 4T0 (iv) (vii)
√
2T0
T0 2
√ (ii) 2 2T0 (v) T0 T0 (viii) √ 2 2
(iii) 2T0 T0 (vi) √ 2 (ix)
T0 4
(b) (6 points ) Suppose that the original block of mass M0 is restored, but the spring is replaced by one of spring constant 2k0 . What is the new period in this case? (i) 4T0 (iv) (vii)
√
2T0
T0 2
√ (ii) 2 2T0 (v) T0 T0 (viii) √ 2 2
(iii) 2T0 T0 (vi) √ 2 (ix)
T0 4
— Problem 3 continues on page 13 —
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Problem 3 Continued: (c) (6 points ) Suppose now that the original spring of spring constant k0 is replaced, but an additional spring of identical spring constant is attached between the block and the fixed support. What is the new period in this case? (i) 4T0 (iv) (vii)
√
2T0
T0 2
√ (ii) 2 2T0
(iii) 2T0 T0 (vi) √ 2
(v) T0 T0 (viii) √ 2 2
(ix)
T0 4
(d) (7 points ) Suppose now that the additional spring is disconnected, but is then placed between the first spring and the block, as shown. What is the new period in this case?
(i) 4T0 (iv) (vii)
√
2T0
T0 2
√ (ii) 2 2T0 (v) T0 T0 (viii) √ 2 2
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(iii) 2T0 T0 (vi) √ 2 (ix)
T0 4
8.01 Exam 1, Fall 2002 Name
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Problem 4: Hanging from a rotating wheel (30 points ) A wheel of radius R is mounted so that it can rotate freely about a vertical axis, as shown. A rope (massless and inextensible) of length is tied to the rim, and at the other end of the rope is a small ball of mass M. The ball is observed to hang at an angle θ relative to the vertical as it swings around with the wheel, making a circle in a horizontal plane. Take the acceleration of gravity as g (where g > 0).
(a) (10 points) Find the tension T in the rope. Be sure that your answer includes no variables other than those given: R, , M, θ, and g. (b) (20 points) Find the period P of rotation that is necessary to cause the ball to hang at a given angle θ. (Again, be sure to use only the given variables.)
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Solutions by Prof. Wit Busza 17
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01
Fall 2002
50-MINUTE EXAM 1 FORMULA SHEET Wednesday, September 25, 2002 Equations introduced in Chapter 1:
dr v = ; dt
dv d2r a = = 2; dt dt
r(t1 ) = r0 +
t1
v dt ;
0
v(t1 ) = v0 +
t1
a dt .
0
For constant acceleration a, if r = r0 and v = v0 at time t = 0, then v(t) = v0 + at 1 r(t) = r0 + v0 t + at2 . 2 For one-dimensional motion with constant acceleration a: v 2 = v02 + 2a(x − x0 ) . For circular motion at constant speed v: a=
v2 , r
where r is the radius of the circle, and the acceleration is directed towards the center of the circle. If an object has position r and velocity v, its position and velocity relative to an observer with position r0 and velocity v0 are given respectively by r = r − r0 ,
v = v − v0 .
Average velocity and acceleration are given by vaverage =
∆r , ∆t
aaverage =
1
∆v . ∆t
Equations introduced in Chapter 2: F = ma F = − GMm ˆ r r2 F = 1 Qq ˆ r 4π0 r2
(Newton’s second law); (the gravitational force between two particles); (the electrostatic force between two particles);
Fx = −kx
(Hooke’s law);
d2 x = −ω 2 x 2 dt
(for a particle near a point of stable equilibrium; equation leads to simple harmonic motion);
x = A sin ωt
(a solution to the above equation; any solution can be written this way if we choose t = 0 when x = 0);
ω = 2πf
(relation between angular frequency and frequency);
T =
2π 1 = f ω
(period of an oscillator).
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