physics 111 spring 2006 exam # 1
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PHYSICS 1307 SUMMER 2008 EXAM # 1 Monday, June 9, 2008 SOLUTIONS
CONCEPTUAL QUESTIONS
1. Observing the graph to the right, determine in which interval the velocity was not constant. Explain. Since velocity is the slope of the curve in a x vs t graph we can look for the interval where the slope is not constant. That interval is BC.
x
O
A
B
C
t
2. Find the time during which the object switch v direction. Explain. The time is between 4 and 5 where the velocity changes sign. It goes from negative to positive, which means that the particle represented switched direction.
t 1 2 3 4 5 6 7 8 9 10 11
2/10 3.True or false. An object with constant nonzero acceleration can never stop and stay stopped. Justify your answer. True. By Newton's second law, an object that experiences a net force different from zero also experiences an acceleration different from zero. So it cannot stay stopped.
4.Where, in a parabolic motion, are the vectors velocity and acceleration perpendicular? Justify your answer.
At the highest point in the trajectory. Observe that at this point, the ycomponent of the velocity is zero. Thus the velocity is exactly horizontal and thus perpendicular to the gravitational acceleration. Any other point has vertical velocity different from zero and so the velocity is not perpendicular to the acceleration.
5. You press a book flat against a vertical wall with your hand. Draw the free body diagram of the situation. (Do not forget the force of friction) Ff
FN
Fp
Fg
3/10
NUMERICAL QUESTIONS 6. The magnitude and direction of the vector sum are :
y A. B. C. D. E.
(8,5)
2 , 4 50 2 , 5 50 2 , 1250 2 , 3150 2 , 3550
x (1,1) (7,6)
The vector sum A+B= ( 1,1). Using Pythagorean theorem we have ∣A B∣= 12−12= 2 −1 0 0 and the direction is equal to =arctan 360 =315 . Notice that we need to add 360° since 1
the answers provided are all positive.
7. At t = 0 a particle starts from rest at x=0 and y=0 and moves with an acceleration a = 4.0 i 3.0 j m / s2 . The position at t = 2.0 s is : _______m A. (0,0)
B. (4,3)
C. (8, 6)
1 2
D.(9,5)
E.(16,12)
1 2
2 2 2 The vector position is given by x = 4.0 t i 3.0 t jm / s where we have used the equation
for motion with constant acceleration in each component. We just need now to evaluate this vector at t = 2.0 s: 1 1 2 2 2 x = 4.0 2s i 3.02s jm/ s 2 2 x =8 i 6 j m
4/10 8. A Cessna aircraft has a liftoff speed of 120 km/h. The takeoff run is about 240 m. The minimum constant acceleration required for liftoff is ___m/s². A. 5.2
B. 2.3
C. 1.5
D. 0.3
E. 0.5
Since the Cessna starts from rest we have that the acceleration is given by final velocity is the takeoff velocity. Therefore a=
33 m/ s2 2×240 m
vf 2 2d
=a where the
=2.31 m / s
9. A force F=2.0 i5.0 j moves an object with net displacement equal to S =2.0 i 3.0 j The work done is _____J. A. 5
B. 13
C. 19
D. 25
E. 32
The work done is equal to the scalar product between the force and the acceleration, thus: S=2.0 i5.0 j⋅2.0 i 3.0 j =19J W =F⋅
10. You are driving home from school steadily at 95 km/h for 130 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 3 hours and 20 minutes. Your average speed was___km/h. A. 65
B. 78
C. 80
D. 88
E. 90
In order to solve this problem we need the total distance of the trip. This is equal to the sum of the partial distances. We have the first one : 130 km so we just have to find the second one. Since we have the velocity, 65 km/h, the time elapsed during the second leg is the last piece of information needed. We can obtained from the data given for the whole trip and the first part: 1 130 km 3 h− =2 h . Therefore the second distance is 2 h×65km /h=130km . The total 3 95 km /h 260 km s = =78 km/h 1 distance is 260 km and the average speed is 3 h 3
5/10 11. The position of a ball rolling in a straight line is given by x=2.0−3.6 t 1.1 t 2 where x is in meters and t in seconds. The instantaneous velocity at t =3.0 s is _____m/s A. 3.0
B. 4.0
C. 5.0
D. 6.0
E. 7.0
The instantaneous velocity is given by the derivative of the position with respect to time, thus: dx =−3.62.2 t dt v 3.0s=−3.62.2 3.0=3.0 m/ s
12. A stone is dropped from the top of a cliff and it is seen to hit the ground below after 3.75 s. The high of the cliff is ______m A. 18
B. 23
C. 45
D. 57
E. 69
1 h= g t 2 2 1 h= ×9.8 m/ s2×3.75 s2=69 m 2
13. An athlete executing a long jump leaves the ground at a speed of 10.0 m/s and an angle of 35° (relative to the horizontal direction). At the highest point, the magnitude and direction of her velocity are ___ m/s and ___º A. 10.0, 35
B. 0.0, 0
C. 5.7, 35
D. 8.2, 0
E. 5.7, 0
y Observing the diagram, we see that at the highest point that the only component of the velocity is which means that the velocity is v0x equal to 10.0 m/s Cos(35°)= 8.2 m/s
10.0 m/s 35° v0y v0x
v0x x
6/10 14. The mass of 747 jetliner is 4.5×105 kg . The aircraft is accelerating at 3.8 m/s² as it climbs at an angle of 22º above the horizontal. The vertical lifting force is _______N A. 5.0×106
B. 4.9×106
C. 4.4×106 D. 3.8×106
E. 3.2×106
Since we are only interested in the vertical direction (lifting force) we only draw the forces along this direction:
FL
F L −m g=m a
+
=m gm×3.8m /s 2 sin220 F F =4.5×10 5 kg×9.8 m /s 21.2 m/s 2 =5.05×106 N F
Fg
15. The coefficient of static friction between hard rubber and normal street pavement is about 0.90. The maximum angle that a hill can have so you can leave your car parked safely is ______º
A. 8
C. 42
B. 13
D. 50
E. 59
From an example of the textbook we know that the tangent of the critical angle is equal to the static coefficient of friction. So c =arctan0.9 0 c =41.98
16. Two blocks of mass 65 kg and 125 kg are in contact and at rest on a horizontal surface. A force F= 650 N is applied on the 65 kg block. If the coefficient of kinetic friction is 0.18, the acceleration of the system is______m/s². A. 3.4 B. 2.8
C. 2.2
D. 1.8
E. 1.7
7/10
F=650 N
65 kg
125 kg
The above system behaves like a single block of mass 190 kg (why?, convince yourself of this !). The free body diagram for the entire system is:
∑ F x =ma
FN
F −Ff =ma F−m1m2 g=m1m2 a F −m1m2 g =a m1 m2 650 N−190kg×0.18×9.8m/ s2 2 =1.66m/ s 190 kg
F
Ff 190 kg
Fg
17. Two blocks on horizontal surface are tied to each other as it is shown in the figure below. A force F=45 N is applied to block 1. If the coefficient of kinetic friction is 0.25, the force of tension in the rope between the blocks is ____N A. 45
B. 37
C. 30
D. 22
E. 17 F
10 kg
5 kg
2
1
Let us focus our attention in the second block. The free body diagram is
8/10
∑ F x =m2 a
FN
FT −Ff =m2 a F T −m2 g=m2 a FT =m2 ga
10 kg
Ff
FT
Thus we need the accelerating of the whole system which is equal to : F−m1m2 g =a m1 m2 45 N−15kg×0.25×9.8m /s 2 =0.55m /s 2 15 kg
Fg
So the force of tension is : F T =m2 g a=10kg×0.25∗9.8m /s 20.55m/ s2 =30N
18. A 60kg skateboarder comes over the top of a hill at 5.0 m/s and reaches 10m/s at the bottom of the hill. The work done on the skater is___J A. 2250
B. 1250
C.570
D. 250
E. 167
By the workenergy theorem we can write: 1 1 1 W ext = KE = m v f 2− m vi 2 = 60 kg10 m/s2−5 m/ s 2 =2250 J 2 2 2
19. A Navy jet of mass 10,000 kg lands on an aircraft carrier and snags a cable to slow it down. The cable is attached to a spring with spring constant 40,000 N/m. If the spring stretches 25 m to stop the plane, the landing speed of the plane is ____km/h A. 50
B. 100
C. 180
D. 270
E. 300
9/10 Using the conservation of energy we can write: KE=− PE 1 1 1 1 2 2 2 2 m v f − m vi = k xi − k x f 2 2 2 2 1 1 10,000 kg v i 2= 40,000 N / m25 m2 2 2 vi =50 m/ s=180 km /h
20. A 5 kg box is sliding down a 10° incline at constant velocity. After 5 m down the incline the work done by the force of friction acting on the box is ____ J . A. 0
B. 241
D. 43
C. 241
E. 43
W Ext = K PE=mg h=−5kg×9.8 m / s2 ×5×sin 100 =−42.54 J . Note that since the box is
sliding down with constant velocity the change in kinetic energy is zero.
21. A baseball (m=140 g) traveling 35 m/s moves the fielder's glove backward 25 cm when the ball is caught. The magnitude of the average force exerted by the ball on the glove is ____N A. 308
B. 250
C. 154
D. 512
The workenergy theorem gives us: 1 1 1 W ext = KE = m v f 2− m vi 2 = 0.140 kg0−35m/ s2 =−85.75 J 2 2 2 −Fs=−F×0.25 m=−85.75 J F=343N
E.343
10/10 22. A freight train has a mass of 1.5×10 7 kg . If the locomotive can exert a constant pull of 7.5×105 N . The time that the locomotive takes to reach 80 km/h from rest is ____min. A. 3
B. 4.5
C. 5.6
D. 7.4
E. 8.2
vf −v0 =t a vf −v0 =t F m 80/ 3.6m/s =444.44 s 5 7.5×10 N 7 1.5×10 kg
Write your answers here: 6. D
7. C
8. B
9. C
10. B
11. A
12. E
13. D
14. A
15. C
16. E
17. C
18. A
19. C
20. D
21. E
22. D
PHYSICS 1307 SUMMER 2008 EXAM # 1 Monday, June 9, 2008 SOLUTIONS
CONCEPTUAL QUESTIONS
1. Observing the graph to the right, determine in which interval the velocity was not constant. Explain. Since velocity is the slope of the curve in a x vs t graph we can look for the interval where the slope is not constant. That interval is BC.
x
O
A
B
C
t
2. Find the time during which the object switch v direction. Explain. The time is between 4 and 5 where the velocity changes sign. It goes from negative to positive, which means that the particle represented switched direction.
t 1 2 3 4 5 6 7 8 9 10 11
2/10 3.True or false. An object with constant nonzero acceleration can never stop and stay stopped. Justify your answer. True. By Newton's second law, an object that experiences a net force different from zero also experiences an acceleration different from zero. So it cannot stay stopped.
4.Where, in a parabolic motion, are the vectors velocity and acceleration perpendicular? Justify your answer.
At the highest point in the trajectory. Observe that at this point, the ycomponent of the velocity is zero. Thus the velocity is exactly horizontal and thus perpendicular to the gravitational acceleration. Any other point has vertical velocity different from zero and so the velocity is not perpendicular to the acceleration.
5. You press a book flat against a vertical wall with your hand. Draw the free body diagram of the situation. (Do not forget the force of friction) Ff
FN
Fp
Fg
3/10
NUMERICAL QUESTIONS 6. The magnitude and direction of the vector sum are :
y A. B. C. D. E.
(8,5)
2 , 4 50 2 , 5 50 2 , 1250 2 , 3150 2 , 3550
x (1,1) (7,6)
The vector sum A+B= ( 1,1). Using Pythagorean theorem we have ∣A B∣= 12−12= 2 −1 0 0 and the direction is equal to =arctan 360 =315 . Notice that we need to add 360° since 1
the answers provided are all positive.
7. At t = 0 a particle starts from rest at x=0 and y=0 and moves with an acceleration a = 4.0 i 3.0 j m / s2 . The position at t = 2.0 s is : _______m A. (0,0)
B. (4,3)
C. (8, 6)
1 2
D.(9,5)
E.(16,12)
1 2
2 2 2 The vector position is given by x = 4.0 t i 3.0 t jm / s where we have used the equation
for motion with constant acceleration in each component. We just need now to evaluate this vector at t = 2.0 s: 1 1 2 2 2 x = 4.0 2s i 3.02s jm/ s 2 2 x =8 i 6 j m
4/10 8. A Cessna aircraft has a liftoff speed of 120 km/h. The takeoff run is about 240 m. The minimum constant acceleration required for liftoff is ___m/s². A. 5.2
B. 2.3
C. 1.5
D. 0.3
E. 0.5
Since the Cessna starts from rest we have that the acceleration is given by final velocity is the takeoff velocity. Therefore a=
33 m/ s2 2×240 m
vf 2 2d
=a where the
=2.31 m / s
9. A force F=2.0 i5.0 j moves an object with net displacement equal to S =2.0 i 3.0 j The work done is _____J. A. 5
B. 13
C. 19
D. 25
E. 32
The work done is equal to the scalar product between the force and the acceleration, thus: S=2.0 i5.0 j⋅2.0 i 3.0 j =19J W =F⋅
10. You are driving home from school steadily at 95 km/h for 130 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 3 hours and 20 minutes. Your average speed was___km/h. A. 65
B. 78
C. 80
D. 88
E. 90
In order to solve this problem we need the total distance of the trip. This is equal to the sum of the partial distances. We have the first one : 130 km so we just have to find the second one. Since we have the velocity, 65 km/h, the time elapsed during the second leg is the last piece of information needed. We can obtained from the data given for the whole trip and the first part: 1 130 km 3 h− =2 h . Therefore the second distance is 2 h×65km /h=130km . The total 3 95 km /h 260 km s = =78 km/h 1 distance is 260 km and the average speed is 3 h 3
5/10 11. The position of a ball rolling in a straight line is given by x=2.0−3.6 t 1.1 t 2 where x is in meters and t in seconds. The instantaneous velocity at t =3.0 s is _____m/s A. 3.0
B. 4.0
C. 5.0
D. 6.0
E. 7.0
The instantaneous velocity is given by the derivative of the position with respect to time, thus: dx =−3.62.2 t dt v 3.0s=−3.62.2 3.0=3.0 m/ s
12. A stone is dropped from the top of a cliff and it is seen to hit the ground below after 3.75 s. The high of the cliff is ______m A. 18
B. 23
C. 45
D. 57
E. 69
1 h= g t 2 2 1 h= ×9.8 m/ s2×3.75 s2=69 m 2
13. An athlete executing a long jump leaves the ground at a speed of 10.0 m/s and an angle of 35° (relative to the horizontal direction). At the highest point, the magnitude and direction of her velocity are ___ m/s and ___º A. 10.0, 35
B. 0.0, 0
C. 5.7, 35
D. 8.2, 0
E. 5.7, 0
y Observing the diagram, we see that at the highest point that the only component of the velocity is which means that the velocity is v0x equal to 10.0 m/s Cos(35°)= 8.2 m/s
10.0 m/s 35° v0y v0x
v0x x
6/10 14. The mass of 747 jetliner is 4.5×105 kg . The aircraft is accelerating at 3.8 m/s² as it climbs at an angle of 22º above the horizontal. The vertical lifting force is _______N A. 5.0×106
B. 4.9×106
C. 4.4×106 D. 3.8×106
E. 3.2×106
Since we are only interested in the vertical direction (lifting force) we only draw the forces along this direction:
FL
F L −m g=m a
+
=m gm×3.8m /s 2 sin220 F F =4.5×10 5 kg×9.8 m /s 21.2 m/s 2 =5.05×106 N F
Fg
15. The coefficient of static friction between hard rubber and normal street pavement is about 0.90. The maximum angle that a hill can have so you can leave your car parked safely is ______º
A. 8
C. 42
B. 13
D. 50
E. 59
From an example of the textbook we know that the tangent of the critical angle is equal to the static coefficient of friction. So c =arctan0.9 0 c =41.98
16. Two blocks of mass 65 kg and 125 kg are in contact and at rest on a horizontal surface. A force F= 650 N is applied on the 65 kg block. If the coefficient of kinetic friction is 0.18, the acceleration of the system is______m/s². A. 3.4 B. 2.8
C. 2.2
D. 1.8
E. 1.7
7/10
F=650 N
65 kg
125 kg
The above system behaves like a single block of mass 190 kg (why?, convince yourself of this !). The free body diagram for the entire system is:
∑ F x =ma
FN
F −Ff =ma F−m1m2 g=m1m2 a F −m1m2 g =a m1 m2 650 N−190kg×0.18×9.8m/ s2 2 =1.66m/ s 190 kg
F
Ff 190 kg
Fg
17. Two blocks on horizontal surface are tied to each other as it is shown in the figure below. A force F=45 N is applied to block 1. If the coefficient of kinetic friction is 0.25, the force of tension in the rope between the blocks is ____N A. 45
B. 37
C. 30
D. 22
E. 17 F
10 kg
5 kg
2
1
Let us focus our attention in the second block. The free body diagram is
8/10
∑ F x =m2 a
FN
FT −Ff =m2 a F T −m2 g=m2 a FT =m2 ga
10 kg
Ff
FT
Thus we need the accelerating of the whole system which is equal to : F−m1m2 g =a m1 m2 45 N−15kg×0.25×9.8m /s 2 =0.55m /s 2 15 kg
Fg
So the force of tension is : F T =m2 g a=10kg×0.25∗9.8m /s 20.55m/ s2 =30N
18. A 60kg skateboarder comes over the top of a hill at 5.0 m/s and reaches 10m/s at the bottom of the hill. The work done on the skater is___J A. 2250
B. 1250
C.570
D. 250
E. 167
By the workenergy theorem we can write: 1 1 1 W ext = KE = m v f 2− m vi 2 = 60 kg10 m/s2−5 m/ s 2 =2250 J 2 2 2
19. A Navy jet of mass 10,000 kg lands on an aircraft carrier and snags a cable to slow it down. The cable is attached to a spring with spring constant 40,000 N/m. If the spring stretches 25 m to stop the plane, the landing speed of the plane is ____km/h A. 50
B. 100
C. 180
D. 270
E. 300
9/10 Using the conservation of energy we can write: KE=− PE 1 1 1 1 2 2 2 2 m v f − m vi = k xi − k x f 2 2 2 2 1 1 10,000 kg v i 2= 40,000 N / m25 m2 2 2 vi =50 m/ s=180 km /h
20. A 5 kg box is sliding down a 10° incline at constant velocity. After 5 m down the incline the work done by the force of friction acting on the box is ____ J . A. 0
B. 241
D. 43
C. 241
E. 43
W Ext = K PE=mg h=−5kg×9.8 m / s2 ×5×sin 100 =−42.54 J . Note that since the box is
sliding down with constant velocity the change in kinetic energy is zero.
21. A baseball (m=140 g) traveling 35 m/s moves the fielder's glove backward 25 cm when the ball is caught. The magnitude of the average force exerted by the ball on the glove is ____N A. 308
B. 250
C. 154
D. 512
The workenergy theorem gives us: 1 1 1 W ext = KE = m v f 2− m vi 2 = 0.140 kg0−35m/ s2 =−85.75 J 2 2 2 −Fs=−F×0.25 m=−85.75 J F=343N
E.343
10/10 22. A freight train has a mass of 1.5×10 7 kg . If the locomotive can exert a constant pull of 7.5×105 N . The time that the locomotive takes to reach 80 km/h from rest is ____min. A. 3
B. 4.5
C. 5.6
D. 7.4
E. 8.2
vf −v0 =t a vf −v0 =t F m 80/ 3.6m/s =444.44 s 5 7.5×10 N 7 1.5×10 kg
Write your answers here: 6. D
7. C
8. B
9. C
10. B
11. A
12. E
13. D
14. A
15. C
16. E
17. C
18. A
19. C
20. D
21. E
22. D
