Grant Morby, Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A(n) 5.53 g bullet is red into a(n) 1.59 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8 m/s2 . If the pendulum rises a vertical distance of 5.78 cm, calculate the initial speed of the bullet. Correct answer: 307.094 m/s. Explanation: Basic Concepts: Applying conservation of mechanical energy from just after the collision until the end of the swing is reached, we have 1 (M + m) V 2 = (M + m) g h 2 where M is the mass of the pendulum, m the mass of the bullet, h the vertical height through which the pendulum swings, and V is the velocity of the (pendulum plus bullet) immediately after the collision. The equation above reduces to V= 2gh. (1)
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The following gure shows a Ferris wheel that rotates 3 times each minute and has a diameter of 17 m. The acceleration of gravity is 9.8 m/s2 .
What is the centripetal acceleration of a rider? Correct answer: 0.838916 m/s2 . Explanation: The period of the Ferris wheel is T= (60 s) = 20 s . (3)
The speed of the wheel is v= 2r T 2 (8.5 m) = 20 s = 2.67035 m/s ,
so the centripetal acceleration is a= v2 r (2.67035 m/s)2 = (8.5 m) = 0.838916 m/s2 . 003 (part 2 of 4) 10 points What force does the seat exert on a 51 kg rider at the lowest point of the ride? Correct answer: 542.585 N. Explanation: The force exerted by the seat balances the gravity and provides the centripetal force, so Fl = m [g + a] = (51 kg) (9.8 m/s2 + 0.838916 m/s2 ) = 542.585 N .
Now apply conservation of momentum from just before to just after the collision. We have m v0 = (M + m) V , (2)
where v0 is the velocity of the bullet just prior to collision. Thus, m v0 = (M + m) 2 g h , so (M + m) v0 = 2gh m (1.59 kg) + (0.00553 kg) = (0.00553 kg) 2 (9.8 m/s2 ) (0.0578 m) = 307.094 m/s . 002 (part 1 of 4) 10 points
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond 004 (part 3 of 4) 10 points What force does the seat exert on a 51 kg rider at the highest point of the ride? Correct answer: 457.015 N. Explanation: The gravity is partly balanced by the force exerted by the seat and this resultant provides the centripetal force, so Fl = m [g a] = (51 kg) [(9.8 m/s2 ) (0.838916 m/s2 )] = 457.015 N .
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point B, swing around the circular vertical loop B C B of radius 14 m, then go on towards further adventures (not shown). When a car goes through the top of the loop (point C), the passengers feel weightless (for just a moment). What is the height hA of the starting point A above the loops bottom B? Correct answer: 35 m. Explanation: Let : R = 14 m and g = 9.8 m/s2 . A passenger feels weightless when his acceleration a is precisely equal to the freefall acceleration g. At the top of the loop (point C), passengers have no tangential acceleration while their normal acceleration is directed straight down and has magnitude v2 . R Weightlessness happens when this acceleration equals g, therefore the cars should go through point C at speed aN = vC = Rg.
005 (part 4 of 4) 10 points What force (magnitude) does the seat exert on a rider when the rider is halfway between top and bottom? Correct answer: 501.628 N. Explanation: In this case, the force exerted by the seat has two components: the vertical one balancing the gravity and the horizontal one providing the centripetal force. Thus we have Fm = m g 2 + a2 (9.8 m/s2 )2 + (0.838916 m/s2 )2
= (51 kg)
= 501.628 N . 006 (part 1 of 1) 10 points Consider a frictionless roller coaster such as depicted below. The acceleration of gravity is 9.8 m/s2 . A C hA 14 m B Passenger cars start at point A with zero initial speed, accelerate as they go down to
In the absence of friction, the mechanical energy of a car is conserved as it follows the coasters route, thus KA + U A = K C + U C m2 m2 vA + m g h A = v + m g 2 R. 2 2C Therefore, in light of the initial condition vA = 0, we have m2 v m g hA = m g 2 R + 2C m Rg = mg2R + 2 5 = mgR 2 5 hA = R 2 5 = (14 m) 2 = 35 m . =
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond 007 (part 1 of 1) 10 points An object attached to the end of a string swings in a vertical circle of radius 1.7 m as shown. At an instant when = 39 , the speed of the object is 6 m/s and the tension in the string is 17 N. The acceleration of gravity is 9.8 m/s2 .
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radial direction is m g sin . The radial component of the resultant force is then Fr = m ar = T + m g sin , m Finally, m= T v2 R = g sin (17 N) (6 9.8 m/s2 sin(39 ) (1.7 m) m/s)2 v2 g sin R =T. or
1.7 m 17 N 39 9.8 m/s
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What is the mass m of the object? Correct answer: 1.13264 kg. Explanation: Let : v r g = 39 , = 6 m/s , = 1.7 m , and = 9.8 m/s2 .
R g
T v mg
6m /s
= 1.13264 kg . 008 (part 1 of 1) 10 points A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed shown in the gure below. The crate is pulled a distance of 7.42 m on the incline by a 150 N force. The acceleration of gravity is 9.8 m/s2 .
/s 6m 1. 13 kg = 0.3
15
0N
7
35
Basic Concepts: Fr = m a r = m v2 r
What is the change in kinetic energy of the crate? Correct answer: 284.284 J. Explanation: Let : F = 150 N , d = 7.42 m , = 35 , m = 13 kg , g = 9.8 m/s2 , = 0.37 , and v = 1.6 m/s .
Solution: The centripetal acceleration is v2 ar = . According to Newtons second law, r the total force in the radial direction should be equal to Fr = m ar . The forces acting on the mass are the tension T on the string, directed radially inward, and the weight force, whose component in the
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond
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N
v
F
N
mg
009 (part 1 of 1) 10 points A 2 kg steel ball strikes a wall with a speed of 14.9 m/s at an angle of 34.8 with the normal to the wall. It bounces o with the same speed and angle, as shown in the gure. 14 .9 y m/ s 34.8 34.8
2 kg 2 kg
x
The work-energy theorem with nonconservative forces reads Wf ric + Wappl + Wgravity = K To nd the work done by friction we need the normal force on the block from Newtons law Fy = N m g cos = 0 Thus N = m g cos . Wf ric = m g d cos
1
/s 9m 4.
If the ball is in contact with the wall for 0.11 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 444.913 N. Explanation:
= (0.37) (13 kg) (9.8 m/s2 ) (7.42 m) cos 35 = 286.51 J .
Let : M = 2 kg , v = 14.9 m/s , = 34.8 .
and
The work due to the applied force is Wappl = F d = (150 N) (7.42 m) = 1113 J , and the work due to gravity is Wgrav = m g d sin = (13 kg) (9.8 m/s2 ) (7.42 m) sin 35 = 542.207 J ,
The y component of the momentum is unchanged. The x component of the momentum is changed by Px = 2 M v cos . Therefore, using impulse formula, P t 2 M v cos = t 2 (2 kg) (14.9 m/s) cos 34.8 = 0.11 s F = 444.913 N . F= Note: The direction of the force is in negative x direction, as indicated by the minus sign. 010 (part 1 of 1) 10 points Note: Take East as the positive direction.
so that K = Wf ric + Wappl + Wgrav = (286.51 J) + (1113 J) + (542.207 J) = 284.284 J .
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond A(n) 82 kg sherman jumps from a dock into a 124 kg rowboat at rest on the West side of the dock. If the velocity of the sherman is 4.2 m/s to the West as he leaves the dock, what is the nal velocity of the sherman and the boat? Correct answer: 1.67184 m/s. Explanation: Let West be negative: Let : m1 = 82 kg kg , m2 = 124 kg kg , and vi,1 = 4.2 m/s m/s . v 7 kg 6 m/s 7 kg d v = 0 7 kg 153 N/m
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vf 7 kg
The boat and sherman have the same nal speed, and vi,2 = 0 m/s, so m1 vi,1 + m2 vi,2 = (m1 + m2 ) vf m1 vi,1 = (m1 + m2 ) vf
v =0 7 kg D
153 N/m
vf =
m1 vi m1 + m 2 (82 kg) (4.2 m/s) = 82 kg + 124 kg
= 0.47 Find the compressed distance d. Correct answer: 1.08983 m. Explanation: The principle we are going to use to solve this problem is that the change in the total energy of the system is equal to the work done by the nonconservative forces Wnc = K + U . The frictional force is f = mg = 0.47 (7 kg) (9.8 m/s2 ) = 32.242 N . In our case, the nonconservative force is the frictional force, therefore Wnc = f d = (32.242 N) d . The change in kinetic energy is K = 0 1 2 m vi 2 (1)
= 1.67184 m/s , which is 1.67184 m/s to the West. 011 (part 1 of 1) 10 points A 7 kg mass slides to the right on a surface having a coecient of friction 0.47 as shown in the gure. The mass has a speed of 6 m/s when contact is made with a spring that has a spring constant 153 N/m. The mass comes to rest after the spring has been compressed a distance d. The mass is then forced toward the left by the spring and continues to move in that direction beyond the unstretched position. Finally the mass comes to rest a distance D to the left of the unstretched spring. The acceleration of gravity is 9.8 m/s2 .
1 = (7 kg) (6 m/s)2 2 = 126 J ,
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond and the change in potential energy is 1 k d2 2 1 = (153 N/m) d2 2 = (76.5 N/m) d2 , so Wnc = K + U (32.242 N) d = (126 J) + (76.5 N/m) d2 (76.5 N/m) d2 + (32.242 d N) + (126 J) = 0 A d2 + B d + C = 0 . U = This is a quadratic equation, and since B2 4 A C = (32.242 N)2
1/2
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in term of their magnitudes and the angle between them. Given the data, we immediately calculate A= B= A2 + A2 + A2 = 11.8651, (3) z y x
2 2 2 Bx + By + Bz = 10.5854, (4)
and using eq. (1), A B = 51.32. Hence, according to eq. (2), cos AB = and therefore AB = arccos(0.408611) = 114.118 . (7) AB = 0.408611 (6) (5)
4 (76.5 N/m) (126 J) = 198.986 N , the positive solution is B + B 2 4 A C d= 2A (32.242 N) + (198.986 N) = 2 (76.5 N/m) = 1.08983 m . 012 (part 1 of 1) 10 points Vector A has components
A
B
Two vectors always lie in a plane. When these two vectors are plotted in this plane, we have
while vector B has components Bx = 8.6, By = 4, Bz = 4.7. A
What is the angle AB between these vectors? (Answer between 0 and 180 .) Correct answer: 114.118 . Explanation: Note: The magnitude of vector X is X . Consider two formul for the scalar product A B of two vectors: A B = A x Bx + A y By + A z Bz (1) in terms of the two vectors components, and also A B = A B cos AB (2)
013 (part 1 of 1) 10 points A(n) 1.6 kg block is pushed by an external force against a spring with spring constant 1321 N/m until the spring is compressed by 0.2 m from its uncompressed length (x = 0). The block rests on a horizontal plane that has a coecient of kinetic friction of 0.4. The external force is then rapidly removed so that the compressed spring can push the mass. The acceleration of gravity is 9.8 m/s2 .
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Ax = 5.9,
Ay = 7.9,
Az = 6.6,
4.1 1
B
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Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond Remember: The block is not attached to the spring. 1321 N/m 1.6 kg = 0.4 x=0 0.2 m x
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vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the oor drops away (see gure). The coecient of static friction between the person and the wall is and the radius of the cylinder is R.
R
After the block is released, how far along the plane will the block move before coming to a stop? Correct answer: 4.21237 m. Explanation: Basic Concepts: Spring Potential Energy. Frictional Forces. Work-Energy Theorem.
What is the minimum tangential velocity needed to keep the person from slipping downward? 1. v = 2gR gR gR 2gR gR gR 2gR gR gR correct 1 2 gR
Let :
L = 0.2 m , m = 1.6 kg , k = 1321 N/m , = 0.4 .
and
2. v = 2 3. v = 1
Solution: Examining the vertical forces, we observe that the normal force is N = m g and so the friction force is f = m g . Call the distance the block travels d . Then the work W done by the non-conservative friction force is W = K + U 1 f d = (0 0) + 0 k L2 2 1 m g d = k L2 . 2 Solving for d gives k L2 d= 2mg (1321 N/m) (0.2 m)2 = 2 (0.4) (1.6 kg) (9.8 m/s2 ) = 4.21237 m . 014 (part 1 of 1) 10 points An amusement park ride consists of a large
4. v = 5. v = 6. v = 2 7. v = 8. v = 9. v = 10. v =
Explanation: Basic Concepts: m v2 r max Frictional force: fs N = fs Solution: The maximum frictional force due to friction is fmax = N , where N is the inward directed normal force of the wall of Centripetal force: F =
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond the cylinder on the person. To support the person vertically, this maximal friction force max fs must be larger than the force of gravity m g so that the actual force, which is less than N , can take on the value m g in the positive vertical direction. Now, the normal v2 force supplies the centripetal acceleration R on the person, so from Newtons second law, N= Since
max fs = N =
8
Explanation: Let : m1 m2 r g = 3 kg , = 37 kg , = 5 m , and = 9.8 m/s2 .
m v2 . R m v2 mg, R
The linear velocity v can be expressed in terms of the distance it travels each revolution: 2r v= . T Consider the forces acting on each mass: T N m1 g T m2 g
the minimum speed required to keep the person supported is at the limit of this inequality, which is
2 m vmin = m g, R
or
1 2
vmin =
gR
.
The tension T in the string provides the centripetal force required to keep m1 moving in a circular path. Applying Fr = m ar to m1 , we obtain T = m 1 ac = m 1 v2 r and
015 (part 1 of 1) 10 points A 3 kg mass moves in a circular path of 5 m radius on a frictionless horizontal table. It is attached to a string that passes through a frictionless hole in the center of the table. A second 37 kg mass is attached to the other end of the string. The acceleration of gravity is 9.8 m/s2 . v 3 kg 5m
Fy = m ay = 0 to m2 , so m2 g T = 0 v2 r 4 2 r m2 g = m 1 T2 2m r 4 1 T2= m2 g m1 r T = 2 m2 g = 2 (3 kg) (5 m) (37 kg) (9.8 m/s2 )
m2 g = m 1
37 kg
= 1.27794 s . 016 (part 1 of 1) 10 points The vectors A and B are given by A = 1.6 + 6.62 i j B = 2.68 + 1.51 i j
Determine the period (the time for one revolution). Correct answer: 1.27794 s.
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond Find the angle between A and B. Correct answer: 74.189 . Explanation: The magnitudes of A and B are given by A= B= A2 + A2 = 6.81061 x y
2 2 Bx + By = 3.07612 .
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where h is the distance the pile driver falls, i.e., h = (3.17 m) + (0.116 m). Solving for the average force, F , we obtain F= mgh s (1450 kg)(9.8 m/s2 )(3.17 m + 0.116 m) = 0.116 m = 402535 N .
= = 1 and = 0, so ii jj ij A B = 1.6 + 6.62 2.68 + 1.51 i j i j jj ii = (1.6) (2.68) + (6.62) (1.51) + (1.6) (1.51) + (6.62) (2.68) = 5.7082 . The dot product is also A B = A B cos AB cos = AB 5.7082 = (6.81061) (3.07612) = 0.272465 , which implies = arccos(0.272465) = 74.189 . 017 (part 1 of 1) 10 points A 1450 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.17 m before contacting the beam, and it drives the beam 11.6 cm into the ground before coming to rest. The acceleration of gravity is 9.8 m/s2 . Using the work-energy theorem, calculate the magnitude of the average force the beam exerts on the pile driver while the pile driver is brought to rest. Correct answer: 402535 N. Explanation: The work energy theorem tells us that the change in potential energy of the falling pile driver equals the work done on the I beam. We can express this relation as mgh = F s, ij
018 (part 1 of 1) 10 points A(n) 22 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s2 . With what force does he press against the seat when the car moving at 7 m/s goes over a crest whose radius of curvature is 11 m? Correct answer: 117.6 N. Explanation: The boy is executing circular motion at the top of the crest, so the net force is centripetal. Two forces act on him: (1) his weight acting down and (2) the supporting normal force from the seat which acts up Fnet = M a M v2 Fg N = k , so r M v2 k N = Fg r v2 =M g k r = (22 kg) (9.8 m/s2 ) = 117.6 N . The force with which he presses against the seat is equal but opposite to the normal force. 019 (part 1 of 1) 10 points A 2.11 kg block is pushed 1.67 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 64.2 with the horizontal. (7 m/s)2 (11 m) k
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond The acceleration of gravity is 9.8 m/s2 . Thus WF = (F sin ) (y) = (30.4047 N) (sin 64.2 )(1.67 m) 2.11 kg = 45.7145 J , since 1 J = 1 kg m2 /s2 .
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.2 64
If the coecient of kinetic friction between the block and wall is 0.506, nd the work done by F . Correct answer: 45.7145 J. Explanation: Given : m = 2.11 kg , = 0.506 , = 64.2 , and y = 1.67 m . v n fk
020 (part 1 of 1) 10 points Suppose the incline is frictionless for the system shown. The angle of inclination is 25 , the spring constant is 145 N/m and the mass of the block is 5.7 kg. The block is released from rest with the spring initially unstretched. The acceleration of gravity is 9.8 m/s2 .
F
x kg =0
25
145 N/m
5.7
.2 64
How far x does it move down the incline before coming to rest? Correct answer: 0.32562 m. Explanation: Let : = 25 , k = 145 N/m , m = 5.7 kg , and = 0. Energy is conserved, so Ui = m g x sin = Uf = Hence, solving for x x= 2 m g sin k 2 (5.7 kg) (9.8 m/s2 ) sin 25 = 145 N/m k x2 2
mg
The block is in equilibrium horizontally, so Fx = F cos n = 0 , so that n = F cos Since the block moves with constant velocity, Fy = F sin m g fk = 0
F sin m g F cos = 0 F (sin cos ) = m g , so mg F= sin cos (2.11 kg) (9.8 m/s2 ) = sin 64.2 0.506 cos 64.2 = 30.4047 N .
F
= 0.32562 m .
Morby, Grant Quiz 2 Due: Mar 8 2006, 10:00 pm Inst: Drummond 021 (part 1 of 1) 10 points A car of weight 1730 N operating at a rate of 121 kW develops a maximum speed of 29 m/s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant, what is the cars maximum speed on an incline of 1 in 20; i.e., if is the angle of the incline with the horizontal, sin = 1/20 ? Correct answer: 28.411 m/s. Explanation: If f is the resisting force on a horizontal road, then the power P is P = f vhorizontal , P f= vh (121000 W) = (29 m/s) = 4172.41 N . so that
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On the incline, the resisting force is F = f + m g sin = f + And, Fv=P, P v= F = P so W W P + = . 20 vh 20
P W + vh 20 (121000 W) = (121000 W) (1730 N) + (29 m/s) 20 = 28.411 m/s .