physics 111 spring 2006 exam # 1
1/8 PHYSICS 1307 FALL 2007 EXAM # 2 March 4 , 2008 SOLUTIONS
CONCEPTUAL QUESTIONS 1. Why can't we define a potential energy associated with friction? Potential energy is associated with conservative forces, where the amount of work is independent of the path followed by the particle. Since the work done by the friction force depends of the trajectory followed thus we cannot assign a potential to this force.
2. A pendulum bob swings back and forth on the end of a string, describing a circular arc. Does the tension force in the string do any work? Explain. From basic geometry we know that the radius of a circle is perpendicular to the tangent of the circle at the same point. Hence the force of tension of a string, which is in the direction of any radius, is perpendicular to the displacement, which is tangent to the circle. Thus no work is performed by the force of tension.
3. If the potential energy is zero at a point, must the force be also zero at that point? Give an example The answer is no. We can assign the zero of potential energy to any point. In particular we generally have chosen the ground to be the zero of potential energy. At this level the force of gravity still acts, but in this point it cannot perform any work. 4. The momentum of a system of pool balls is the same before and after they are hit by the cue ball. Is it still the same after one of the balls strikes the edge of the table? Explain.
2/8 When the ball strikes the edge, the edge strikes back the ball by Newton's third law. Therefore there is a change of momentum of magnitude F ext t= p . This is the reason that momentum is not conserved when when the ball hits the edge of the table.
5. Two identical satellites are going in opposite directions in the same circular orbit when they collide headon. Describe their subsequent motion if the collision is elastic. Since the two identical satellites have the same mass, we know (as it was discussed in class) the final velocities, for each one of the satellites, are going to have the same magnitude that the initial ones but in opposite directions, i.e. They collide and then they recede from each other with inverted relative velocity.
NUMERICAL QUESTIONS 6. A meteorite plunges to Earth, embedding itself 75 cm in the ground. If it does 140 MJ of work in the process. The magnitude of the average force exerted on the ground is ____MN A. 227
B. 187
C.57
D. 5.7
E. 1.87
W =Fdcos 140×10 6 J =F ave 0.75 m 8
1.87×10 N=F ave
7. A 60kg skateboarder comes over the top of a hill at 5.0 m/s and reaches 10.0 m/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
Using the workenergy theorem we have: 1 1 1 2 2 2 2 W ext = KE = m v f − m v i = 60 kg 10 m / s − 5 m / s =2250 J 2 2 2
3/8 8. 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
C. 180
B. 100
D. 270
E. 300
Using the conservation of energy we can write: KE=− PE 1 1 1 1 2 2 2 2 m v f − m vi = k x i − k x f 2 2 2 2 1 1 10,000 kg vi 2= 40,000 N / m25 m 2 2 2 v i =50 m/ s=180 km/ h
9. A popcorn kernel in a hot pan burst into two pieces, with masses of 91 mg and 64 mg. The more massive piece moves horizontally at 47 cm/s along the positive ydirection. The velocity of the second piece of popcorn is ___cm/s
A. 67 i
B. −67 i
Pi=0 y
D. −67 j
C. 0 v1 m1
Pf=0
m2 v2 Using conservation of linear momentum along the yaxis we obtain: 0 = m 1 v1 m2 v2 −m 1 v 1 m2
= v2
−91 mg 47 cm / s =−67 cm / s= v2 64 cm / s
E. 67 j
4/8 10. A glass ball is traveling directly at a steel ball. An observer at rest in the lab sees the glass ball moving east at 10 m/s and the steel ball moving west at 20.0 m/s. The velocity at which the two balls approach or recede from each other after undergoing an elastic collision is ____m/s. Assume east as positive direction. A. 10 m/s
B. 20 m/s
C. 0
E. 30 m/s
D. 10 m/s
Using the fact that after a elastic collision the relative velocity switches direction: v1 i −v2 i =− v1 f − v 2 f
we get,
10 m/s−−20 m/ s=−v1 f −v2 f 1
2
−30 m/ s=v f −v f
This means that the balls recede from each other with a speed of 30 m/s.
11. Two objects collide and after the collision they stick together. Each particle has a mass of 0.25 kg. The initial speed of A is 20 m/s and B is initially at rest. The collision lasts for 8.0 ms. The magnitude of the average force of each particle on the other during this time is ___ N.
A. 250
B. 312
C. 125
D. 500
E. 625
We need first to find the impulse or the change in momentum of any of the two colliding particles. m m =5 kg s s the final momentum is P f =m Am B v f =0.25 kg 0.25 kg v f . Equating these two expressions and solving for the final velocity we obtain: v f =10 m/ s . The initial momentum of the system is P i=mA v A=0.25 kg 20
Now we are ready. Initial momentum for the particle A P iA=m A v A=0.25 kg 20
m m =5 kg s s
m . Thus the impulse s m m m experienced by A is P fA−P iA=m A v fA −v fB =0.25 kg 10 −20 =−2.5 kg . The average s s s m −2.5 kg P s force is F= = =−312.5 N t 0.008 s final momentum for the particle A P fA= mA v fA =0.25 kg 10
5/8 12. A force F =67 i 23 j 55 k N is applied to a body as it moves in a straight line from r 1= 16 i 31 j to r 2 =21 i10 j14 k m . The work done by the force is _____J. A. 622
B. 1785
C. 2407
D. 4192
E. 5785
W = F⋅ d = F⋅r 2 −r 1 W = 67 i 23 j 55 k N ⋅ 5 i −21 j 14 k m =622 J
The ballistic pendulum is a device used to measure the speed of a projectile. A projectile with mass 100 g is fired into a large block with mass 10 kg suspended like a pendulum. As a result of the collision the pendulumprojectile system swings up to a maximum height of 25 cm. 13. The initial velocity of the projectile is ____m/s A. 2.2
B. 22
C. 125
D. 224
E. 350
Using the result derived in class we can write: M m m 10 kg0.100 kg
vi = 2 g h vi = 2× 9.8 m / s2 ×0.25 m
0.100 kg
=224 m / s
14. The loss of kinetic energy during the collision is _____J A. 0
B. 125
C.596
D. 1254
E. 2484
Using conservation of linear momentum before and after the collision we obtain: m vi =M m v f m vi
=v f M m 0.1 kg224 m/ s =2.22 m/ s 10.1 kg
6/8
thus the difference in kinetic energy is : 1 1 2 2 KE f − KE i = M m v f − m v i =−2484 J 2 2
15. How long (in seconds) will it take a 1750W motor to lift a 300kg piano to a sixthstory window 10.0 m above? A. 25
B. 10
C. 17
Work mg h P= = . Thus the time taken is t t
D.20
E.30 2
mg h 300 kg×9.8 m/ s ×10 m t= = =16.8 s P 1750 W
16. A 1000 kg racing car starts from rest in the pit area and accelerates at uniform rate to a speed of 35 m/s. The average power delivered by the engine is ____kW. Assume that the coefficient of kinetic friction is equal to 0.3. A. 51 P= F⋅v= F f
B. 40 v f vi
2
=
mg v f 2
C. 35
D. 12
E. 2.0
= 0.3×1000 kg× 9.8 m / s2 ×17.5 m / s =51450 W
17. 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
C. 241
D. 43
E. 43
W Ext = K PE=mg h=−5kg×9.8 m/ s2 ×5×sin 10 0 =−42.54 J . Note that since the box is
sliding down with constant velocity the change in kinetic energy is zero.
7/8 18. The angle between the two vectors A= 2 i 3 j , B=−3 i 6 j is ____¡ A. 0
B. 30
C. 45
D. 60
E. 90
Using the definition of the scalar product we can write A⋅B=∣A∣∣B∣cos then: A⋅ B =cos ∣A∣∣ B∣ 12
49 936
=0.50=cos
so the angle is 60¡.
A billiard ball moving with speed v=5.0 m/s along the positive xaxis strikes headon an equal mass ball initially at rest. After the collision the balls are observed to scatter according with the picture below.
v i =5.0 m/s
y 1
1
x
2
2
65¡ 30¡
19. The speed after the collision of the first ball is ____m/s A. 0
B. 2.5
C. 3.4
D. 4.0
E. 4.5
Using a result given in class (written in more elegant way) : v1 tan 2 5.00 m / s tan 30 0 f vi = = =2.51 m / s tan 2 cos 1 sin 1 tan 30 0 cos 65 sin 65
20. The speed of the second ball after the collision is ____m/s. A. 0
B. 2.5
C. 3.4
D. 4.0
E. 4.5
8/8 Using conservation of momentum along the yaxis (also given in class) we write: v 1 f sin 1 = v2 f sin 2 2.51
m s
sin 650
sin 30 0
= 4.55 m / s
Write your answers here: 6. B
7. A
8. C
9.D
10.E
11.B
12.A
13.D
14.E
15.C
16. A
17.D
18.D
19.B
20.E
CONCEPTUAL QUESTIONS 1. Why can't we define a potential energy associated with friction? Potential energy is associated with conservative forces, where the amount of work is independent of the path followed by the particle. Since the work done by the friction force depends of the trajectory followed thus we cannot assign a potential to this force.
2. A pendulum bob swings back and forth on the end of a string, describing a circular arc. Does the tension force in the string do any work? Explain. From basic geometry we know that the radius of a circle is perpendicular to the tangent of the circle at the same point. Hence the force of tension of a string, which is in the direction of any radius, is perpendicular to the displacement, which is tangent to the circle. Thus no work is performed by the force of tension.
3. If the potential energy is zero at a point, must the force be also zero at that point? Give an example The answer is no. We can assign the zero of potential energy to any point. In particular we generally have chosen the ground to be the zero of potential energy. At this level the force of gravity still acts, but in this point it cannot perform any work. 4. The momentum of a system of pool balls is the same before and after they are hit by the cue ball. Is it still the same after one of the balls strikes the edge of the table? Explain.
2/8 When the ball strikes the edge, the edge strikes back the ball by Newton's third law. Therefore there is a change of momentum of magnitude F ext t= p . This is the reason that momentum is not conserved when when the ball hits the edge of the table.
5. Two identical satellites are going in opposite directions in the same circular orbit when they collide headon. Describe their subsequent motion if the collision is elastic. Since the two identical satellites have the same mass, we know (as it was discussed in class) the final velocities, for each one of the satellites, are going to have the same magnitude that the initial ones but in opposite directions, i.e. They collide and then they recede from each other with inverted relative velocity.
NUMERICAL QUESTIONS 6. A meteorite plunges to Earth, embedding itself 75 cm in the ground. If it does 140 MJ of work in the process. The magnitude of the average force exerted on the ground is ____MN A. 227
B. 187
C.57
D. 5.7
E. 1.87
W =Fdcos 140×10 6 J =F ave 0.75 m 8
1.87×10 N=F ave
7. A 60kg skateboarder comes over the top of a hill at 5.0 m/s and reaches 10.0 m/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
Using the workenergy theorem we have: 1 1 1 2 2 2 2 W ext = KE = m v f − m v i = 60 kg 10 m / s − 5 m / s =2250 J 2 2 2
3/8 8. 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
C. 180
B. 100
D. 270
E. 300
Using the conservation of energy we can write: KE=− PE 1 1 1 1 2 2 2 2 m v f − m vi = k x i − k x f 2 2 2 2 1 1 10,000 kg vi 2= 40,000 N / m25 m 2 2 2 v i =50 m/ s=180 km/ h
9. A popcorn kernel in a hot pan burst into two pieces, with masses of 91 mg and 64 mg. The more massive piece moves horizontally at 47 cm/s along the positive ydirection. The velocity of the second piece of popcorn is ___cm/s
A. 67 i
B. −67 i
Pi=0 y
D. −67 j
C. 0 v1 m1
Pf=0
m2 v2 Using conservation of linear momentum along the yaxis we obtain: 0 = m 1 v1 m2 v2 −m 1 v 1 m2
= v2
−91 mg 47 cm / s =−67 cm / s= v2 64 cm / s
E. 67 j
4/8 10. A glass ball is traveling directly at a steel ball. An observer at rest in the lab sees the glass ball moving east at 10 m/s and the steel ball moving west at 20.0 m/s. The velocity at which the two balls approach or recede from each other after undergoing an elastic collision is ____m/s. Assume east as positive direction. A. 10 m/s
B. 20 m/s
C. 0
E. 30 m/s
D. 10 m/s
Using the fact that after a elastic collision the relative velocity switches direction: v1 i −v2 i =− v1 f − v 2 f
we get,
10 m/s−−20 m/ s=−v1 f −v2 f 1
2
−30 m/ s=v f −v f
This means that the balls recede from each other with a speed of 30 m/s.
11. Two objects collide and after the collision they stick together. Each particle has a mass of 0.25 kg. The initial speed of A is 20 m/s and B is initially at rest. The collision lasts for 8.0 ms. The magnitude of the average force of each particle on the other during this time is ___ N.
A. 250
B. 312
C. 125
D. 500
E. 625
We need first to find the impulse or the change in momentum of any of the two colliding particles. m m =5 kg s s the final momentum is P f =m Am B v f =0.25 kg 0.25 kg v f . Equating these two expressions and solving for the final velocity we obtain: v f =10 m/ s . The initial momentum of the system is P i=mA v A=0.25 kg 20
Now we are ready. Initial momentum for the particle A P iA=m A v A=0.25 kg 20
m m =5 kg s s
m . Thus the impulse s m m m experienced by A is P fA−P iA=m A v fA −v fB =0.25 kg 10 −20 =−2.5 kg . The average s s s m −2.5 kg P s force is F= = =−312.5 N t 0.008 s final momentum for the particle A P fA= mA v fA =0.25 kg 10
5/8 12. A force F =67 i 23 j 55 k N is applied to a body as it moves in a straight line from r 1= 16 i 31 j to r 2 =21 i10 j14 k m . The work done by the force is _____J. A. 622
B. 1785
C. 2407
D. 4192
E. 5785
W = F⋅ d = F⋅r 2 −r 1 W = 67 i 23 j 55 k N ⋅ 5 i −21 j 14 k m =622 J
The ballistic pendulum is a device used to measure the speed of a projectile. A projectile with mass 100 g is fired into a large block with mass 10 kg suspended like a pendulum. As a result of the collision the pendulumprojectile system swings up to a maximum height of 25 cm. 13. The initial velocity of the projectile is ____m/s A. 2.2
B. 22
C. 125
D. 224
E. 350
Using the result derived in class we can write: M m m 10 kg0.100 kg
vi = 2 g h vi = 2× 9.8 m / s2 ×0.25 m
0.100 kg
=224 m / s
14. The loss of kinetic energy during the collision is _____J A. 0
B. 125
C.596
D. 1254
E. 2484
Using conservation of linear momentum before and after the collision we obtain: m vi =M m v f m vi
=v f M m 0.1 kg224 m/ s =2.22 m/ s 10.1 kg
6/8
thus the difference in kinetic energy is : 1 1 2 2 KE f − KE i = M m v f − m v i =−2484 J 2 2
15. How long (in seconds) will it take a 1750W motor to lift a 300kg piano to a sixthstory window 10.0 m above? A. 25
B. 10
C. 17
Work mg h P= = . Thus the time taken is t t
D.20
E.30 2
mg h 300 kg×9.8 m/ s ×10 m t= = =16.8 s P 1750 W
16. A 1000 kg racing car starts from rest in the pit area and accelerates at uniform rate to a speed of 35 m/s. The average power delivered by the engine is ____kW. Assume that the coefficient of kinetic friction is equal to 0.3. A. 51 P= F⋅v= F f
B. 40 v f vi
2
=
mg v f 2
C. 35
D. 12
E. 2.0
= 0.3×1000 kg× 9.8 m / s2 ×17.5 m / s =51450 W
17. 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
C. 241
D. 43
E. 43
W Ext = K PE=mg h=−5kg×9.8 m/ s2 ×5×sin 10 0 =−42.54 J . Note that since the box is
sliding down with constant velocity the change in kinetic energy is zero.
7/8 18. The angle between the two vectors A= 2 i 3 j , B=−3 i 6 j is ____¡ A. 0
B. 30
C. 45
D. 60
E. 90
Using the definition of the scalar product we can write A⋅B=∣A∣∣B∣cos then: A⋅ B =cos ∣A∣∣ B∣ 12
49 936
=0.50=cos
so the angle is 60¡.
A billiard ball moving with speed v=5.0 m/s along the positive xaxis strikes headon an equal mass ball initially at rest. After the collision the balls are observed to scatter according with the picture below.
v i =5.0 m/s
y 1
1
x
2
2
65¡ 30¡
19. The speed after the collision of the first ball is ____m/s A. 0
B. 2.5
C. 3.4
D. 4.0
E. 4.5
Using a result given in class (written in more elegant way) : v1 tan 2 5.00 m / s tan 30 0 f vi = = =2.51 m / s tan 2 cos 1 sin 1 tan 30 0 cos 65 sin 65
20. The speed of the second ball after the collision is ____m/s. A. 0
B. 2.5
C. 3.4
D. 4.0
E. 4.5
8/8 Using conservation of momentum along the yaxis (also given in class) we write: v 1 f sin 1 = v2 f sin 2 2.51
m s
sin 650
sin 30 0
= 4.55 m / s
Write your answers here: 6. B
7. A
8. C
9.D
10.E
11.B
12.A
13.D
14.E
15.C
16. A
17.D
18.D
19.B
20.E
