Physics 123
Quizes and Examinations
Spring 2002
Porter Johnson
“Physics can only be learned by thinking, writing, and worrying.”
-David Atkinson and Porter Johnson (2002)
“There is no royal road to geometry.”
-Euclid
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Guidelines and tips
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community
Ask the community for help and clear up your study doubts
University Rankings
Discover the best universities in your country according to Docsity users
Free resources
Our save-the-student-ebooks!
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
General Physics I: Mechanics - Quiz 1 | PHYS 123, Quizzes of Physics
Material Type: Quiz; Class: General Physics I: Mechanics; Subject: Physics; University: Illinois Institute of Technology; Term: Spring 2002;
Typology: Quizzes
Pre 2010
1 / 27
This page cannot be seen from the preview
Don't miss anything!
Related documents
Partial preview of the text
Download General Physics I: Mechanics - Quiz 1 | PHYS 123 and more Quizzes Physics in PDF only on Docsity!
Physics 123
Quizes and Examinations
Spring 2002
Porter Johnson
“Physics can only be learned by thinking, writing, and worrying.”
-David Atkinson and Porter Johnson (2002)
“There is no royal road to geometry.”
-Euclid
Quizes: Spring 2002
- PHYS 103 - 001 QUIZ 1 05 February 2002 An engineer standing on a bridge throws a penny straight up with a speed of v 0 = 20 meters/second, from a height h = 100 meters above the water. [Neglect air resistance.] - What is the speed of the penny when it hits the water below? - How long does it take the penny to hit the water?
Solution: The final speed can be calculated from the relation
v^2 = v^20 + 2 gh = 202 + 2 ( 10 )) 100 = 2400 v = 49 m / s
The time of travel can be calculated from the relation
h = v 0 t +
gt^2
5 t^2 − 20 t = 100 t^2 − 4 t − 20 = 0 t = 2 ±
24 = 6. 9 sec
We have chosen the positive square root in the last relation, since the time of travel must be positive. Alternatively, we may calculate the travel time t from the relation
h =
v + v 0 2
t 100 = 14. 5 t t = 6. 9 sec
- Draw a diagram showing the trajectory of the rock from its launch point to the ground.
- What is the (vector) velocity of the rock when it hits the ground?
- How long (in seconds) does it take the rock to hit the ground?
Solution: The trajectory or the rock is parabolic, starting out horizontally at the win- dow and becoming concave downward. The coördinates of the rock at time t are
y = 0 −
gt^2 x = v 0 t
At the time of flight, we obtain − 100 = − 5 t^2 , or t =
20 = 4. 4 sec. The horizontal position at which the rock strikes the ground is x = v 0 t = ( 30 ) 4. 4 = 130 m. The velocity components of the rock when it strikes the ground are
vy = − gt = − 44 m / s vx = v + 0 = 30 m / s
The rock thus strikes the ground at a speed of 54 m / s , at an angle of below the horizontal.
- PHYS 123 - 002 QUIZ 2 21 February 2002
A rock, thrown out of the open window of a building that is high above the (level) ground, is given an initial speed of 30 meters per second, traveling horizontally. It hits the ground 3.0 seconds after the launch. [You make take g = 10 m / sec^2 , if you wish.]
- Draw a diagram showing the trajectory of the rock from its launch point to the ground.
- What is the (vector) velocity of the rock when it hits the ground?
- From what height (in meters) was the rock launched?
Solution: The trajectory or the rock is parabolic, starting out horizontally at the win- dow and becoming concave downward. In a time t = 3 sec , the rock travels a horizontal distance x = vxt = 30 · 3 = 90 m , and a vertical distance h = 1 / 2 gt^2 = 5 ( 3 )^2 = 45 m. The velocity components when it strikes the ground are vx = 30 m / s and vy = 10 · 3 = 30 m / s. It thus moves with a speed of 42 m / s , at an angle 45 o below the horizontal.
- PHYS 123 - 001 QUIZ 3 05 March 2002
A smaller box of mass m 1 is placed on top of a larger box of mass m 2 , which, in turn, is placed on a frictionless table, as shown. The coefficient of static friction of the smaller box on the rough surface between the two boxes is given by μs. A horizontal force F is applied to the top box.
- Draw a diagram showing all the forces acting on each box.
- Determine the acceleration of the bottom box, assuming that the top box does not slip on the surface of the bottom box.
- In terms of the parameters m 1 , m 2 , μs , and g , determine the maximum force F that be applied before the top box slips.
Solution: The forces on the top block are its weight m 1 g (downward), the normal force from the second block N 1 = mg (upward), the applied force F (to the right), and static friction fs (to the left). The vertical forces on the bottom block are its weight m 2 g (downward), the normal force from the top block N 1 = m 1 g (downward), and the normal force from the table N 2 = ( m 1 + m 2 ) g (upward). In addition, there is static friction from the upper block fs to the right. We apply Newton’s second law to the horizontal motions to obtain
F − fs = m 1 a fs = m 2 a
- PHYS 123 - 001 QUIZ 4 28 March 2002
An m = 10 kg block is released from rest on a θ = 30 o^ frictionless incline. Below the block on the incline there is a spring, which can be compressed by 2. 0 cm by a force of 200 Newtons. The block stops momentarily after it compresses the sprng by d = 6. 0 cm.
- How far does the block move up the incline from its initial rest position to this stopping point?
- What is the speed of the block just as it touches the spring?
Solution: The spring constant is k = 200 /( 0. 05 ) = 104 N / m. If the ball slides a dis- tance ℓ before coming to the spring, then it follows from conservation of mechanical energy that
kd^2 − mg (ℓ + d ) sinθ
10 ( 10 )(ℓ + 0. 06 )
( 104 )( 0. 06 )^2
ℓ = 0. 30 m
According to energy conservation, the speed of the block when it touches the spring is
1 2
mv^2 = mg ℓ sin θ
v^2 = 2 ( 10 )( 0. 3 )
v = 1. 7 m / s
- PHYS 123 - 002 QUIZ 4 28 March 2002
An m = 10 kg block is launched up a θ = 30 o^ frictionless incline, using a spring that is compressed a distance of x = 10 cm. The spring can be com- pressed by 1. 0 cm by a force of 200 Newtons. The block stops momentarily after it travels some distance up the incline.
- How far does the block move up the incline from its initial rest position to this stopping point?
- What is the speed of the block just after it leaves the spring?
Solution: The spring constant is k = 200 /( 0. 01 ) = 2 × 104 N / m. We use conservation of mechanical energy to determine the distance ℓ tht the block goes beyond the string:
1 2
kx^2 + 0 + 0 = 0 + 0 + mg ℓ sin θ 1 2
( 2 × 104 )( 0. 1 )^2 = ( 10 )( 10 )ℓ
ℓ = 2 m
The speed of the block when it leaves the spring is given by
1 2
mv^2 =
kx^2
v^2 = ( 2 × 104 )( 0. 1 )^2 / 10 = 20 v =^4. 5 m / s
- PHYS 123 - 001 QUIZ 5 09 April 2002
A particle of mass m = 1. 0 kg , which is initially moving freely in the hori- zontal direction with a speed of v 0 = 10 meters/second, collides elastically with a target particle of equal mass, and then travels at an angle of θ = 37 o to the initial direction.
- Determine the speed of the incident particle, after the collision.
- Determine the speed and direction of motion of the target particle.
Solution: Let v be the final speed of the projectile, V the final target speed, and φ the angle of travel of the target relative to the + x -axis. The laws of conservation
Let v be the final speed of the projectile, V the final target speed, and φ the angle of travel of the target relative to the + x -axis. The laws of conservation of momentum and energy are
mv 0 = mv cos θ + mV cos φ 0 = mv sin θ − mV sin φ 1 2
mv^20 =
mv^2 +
mV^2
From the energy consevation relation we obtain V =
v^20 − v^2 =
8 m / s. From the momentum consevation we obtain
v^2 x + v^2 y = ( v 0 − V cos φ)^2 + ( V sin φ)^2 36 = 100 + 64 − 160 cosφ cos φ = 128 / 160 = 0. 8 φ = 37 o
- Phys 123-001 QUIZ 6 25 April 2002
A particle of mass m is hung by a massless string, which is wrapped over a frictionless massive pulley. The string is attached to a mass M that is initially at rest on a frictionless table. The pulley has radius R and moment of inertia I about its central axis. It rotates freely about its central axis. Assume that the string does not slip on the pulley.
- Show all forces acting on the pulley and the two masses.
- Determine the downward acceleration of the mass m.
- Determine the tensions in the string, both above and below the pulley.
Solution: The net force on the mass M on the table is the string tension T 1 , so that T 1 = Ma. The net (clockwise) torque on the pulley is ( T 2 − T 1 ) R , so that ( T 2 − T 1 ) R = I α = Ia / rR. The net downward force on the mass m is mg − T 2 , so that mg − T 2 = ma. In summary:
T 1 = Ma ( T 2 − T 1 ) =
I
r^2
a mg − T 2 = ma
We add these equations to obtain
a =
mg m + M + I / R^2
Substituting this result, we obtain the tensions:
T 1 =
M
m + M + I / R^2
mg
T 2 =
M + I / R^2
m + M + I / R^2
mg
- PHYS 123 - 002 QUIZ 6 25 April 2002
A particle of mass m is hung by a massless string, which is wrapped over a frictionless massive pulley. The string is attached to an equal mass m that is initially at rest on a frictionless table, as shown. The pulley has radius R , and it rotates freely about its central axis. Assume that the string does not slip on the pulley. The downward acceleration of the particle is measured to be g /4.
- Show all forces acting on the pulley and the two masses.
- Determine the moment of inertia I of the pulley, in terms of m , g , and R.
- Determine the tensions in the string, both above and below the pulley.
Solution: The force on the mass on the table is T 1 = ma = mg /4. The force on the suspended mass is mg − T 2 = mg /4, so that T 2 = 3 g /4. Finally, the net torque on the pulley is ( T 2 − T 1 ) R = mgR / 2 = I α = Ia / R = Ig /( 4 R ). Con- sequently, the moment of inertia of the pulley is I = 2 mr^2.
- PHYS 123 - 001 QUIZ 7 30 April 2002
A round ball of mass m = 0. 1 kg and radius R = 0 .05 meters rolls without slipping down a track after being released from rest at a height h = 0. 5 meters above the level portion of the track. Hint: The ball is a uniform sphere, with moment of inertia about its center of 2/ 5 mR^2.
mgh = Kf =
mv^2 +
I ω^2
m +
I
R^2
v^2
h =
v^2 2 g
I
mR^2
- PHYS 123 - 001 QUIZ 8 07 May 2002
Two small objects of equal mass m , which lie far away from everything else in the universe, travel along the same circular orbit with the same speed v , and always lie at precisely opposite locations, as shown.
- Draw a diagram showing all the forces on this system.
- What is the radius R of the circle, expressed in terms of v , m , and G?
- Determine the rotational period T of this system, expressed in terms of G , m , and R.
Solution: The force of gravitational attraction between the bodies is F = Gm^2 /( 2 R )^2. Thus,
F =
mv^2 R
Gm^2 ( 2 R )^2
v^2 =
Gm 4 R R =
Gm 4 v^2
T =
2 π R v
4 π R^3 /^2 √ Gm
- PHYS 123 - 002 QUIZ 8 09 May 2002
Two small objects, each of mass 100 kilograms, are located far from every- thing else in the universe. Under mutual gravitational attraction, they travel along the same stationary circular orbit of radius 1000 meters, always lying at precisely opposite locations, as shown.
- Draw a diagram showing all the forces on this system.
- What are their (tangential) speeds in this orbit, in meters/second?
- Determine the rotational period T , in seconds.
Hint: G = 6. 67 × 10 −^11 [ Ntm^2 ]/ kg^2. Solution: The force of gravitational attraction between the bodies is F = Gm^2 /( 2 R )^2. Thus,
F =
mv^2 R
Gm^2 ( 2 R )^2
v^2 =
Gm 4 R
6. 67 × 10 −^11 · 100
4 × 103
= 1. 67 × 10 −^12
v = 1. 3 × 10 −^6 m / s
The period is
T =
2 π r v
2 π 103
- 3 × 10 −^6
= 4. 8 × 109 sec = 170 years
Spring 2002 Examinations:
PHYS 123 - 001/002 TEST 1 25 February 2002
- [25 points] An automobile of mass 1000 kg on the Autobahn is traveling at 200 kilometers/hour on level ground. Seeing a stopped car in his/her lane a distance of 500 meters ahead, the driver immediately locks the brakes. Assume that the car slides to rest on the road just before striking the stopped car. Determine the coefficient of kinetic friction for the car, μk , under these circumstances. Solution:
v 0 = 200
km hr
1 m / s
- 6 km / hr
= 55 m / s
- How long does the ball stay in the air above the launch point [in sec- onds]?
Solution: The basic equation for trajectory motion are:
vx = v 0 cos θ x = v 0 cos θ t vy = v 0 sin θ − gt
y = v 0 sin θ t −
gt^2
The maximum altitude occurs when vy = 0, at the time t = v 0 sin θ/ g = 15 sin 30 o / 9. 8 = 0. 76 sec. The corresponding altitude is
ymax = v 0 sin θ(
v 0 sin θ g
g (
v 0 sin θ g
)^2 =
v^20 sin^2 θ 2 g
= 2. 87 m
The landing y = 0 occurs at a time t = 2 v 0 sin θ/ g = 1. 53 sec , so that the range is
R = v 0 cos θ(
2 v 0 sin θ g
) = 15 cos 30 o ( 1. 53 ) = 19. 9 m
- [25 points] An m = 5 kilogram mass is attached to a massless cord. It slides on a frictionless horizontal plane along a circular path of radius R = 1 meter, the cord being tied to a point at the middle of the plane. The cord can safely withstand a tension of F = 1000 Newtons without breaking. What is the maximum number of revolutions per second permitted for safe operation? Solution: The tension in the cord is
T = macent =
mv^2 R
5 · v^2 1 v^2 = 200 v = 14. 1 m / s
The corresponding frequency of revolution is
f =
T
v 2 π R
2 π · 1
= 2. 25 Hz
- [Extra Credit; 10 points] Calculate the angle between the [three-dimensional] vectors ~ A and ~ B , where
~ A = 2ˆ i + 3 ˆ j + 4ˆ k ~ B = 3ˆ i + 4 ˆ j + 2ˆ k
Note that ˆ i , ˆ j , and ˆ k are unit vectors in the x , y , and z directions, respectively. Solution: Let us compute the scalar product:
~ A · ~ B = axbx + ayby + azbz = 2 × 3 + 3 × 4 + 4 × 2 = 26
The respective magnitudes of these vectors are
| A | =
a^2 x + a^2 y + a^2 z =
| B | =
b^2 x + b^2 y + b^2 z =
Since ~ A · ~ B = | A || B | cosθ, if follows that 26 = (
29 )^2 cos θ, or cos θ = 26 /29, so that θ = 26 o. Alternatively, one may use the relation |~ A × ~ B | = | A || B | sinθ
∆ Eb =
mv^20 −
mv^2 =
( 0. 01 )( 1000 )^2 −
( 0. 01 )( 300 )^2
= 5000 − 450 = 4550 J
The energy concerted into heat is
∆ Elost = ∆ Eb − ∆ Erecoil = ∆ Eb −
MV^2 = 4550 −
1 ( 7 )^2 = 4525 J
The average force on the bullet is given by
Fb = ∆ Eb /ℓ = 4550 / 0. 3 = 1. 5 × 104 N
- [25 points] A car (mass m ) is traveling around a horizontal circular track at a speed of v = 30 meters per second. The radius of curvature of the track is R = 200 meters. - Draw a diagram showing all forces acting on the car. - Calculate the acceleration of the car in the curved track. - What minimum coefficient of kinetic friction μk is required to keep the car from sliding off the track?
Solution: The forces acting on the car are its weight W = mg (downward), the normal force N = mg from the track (upward), and kinetic friction f (toward center of track). The net force acts toward the center of the track:
f =
mv^2 R
= mac
Thus
ac =
v^2 R
( 30 )^2
= 4. 5 m / s^2
The coefficient of kinetic friction is
μk =
f N
ma mg
a g
- [25 points] A block of mass m = 2 kg is released from rest on a frictionless inclined plane tilted at an angle of θ = 30 o^ to the horizontal. It slides down the track for ℓ = 1 meter before beginning to compress a spring, and comes to rest after compressing the spring by x = 20 cm. - What is the spring constant k of the spring, in Newtons per meter? - How fast was the block going when it first touched the spring?
Solution: The block slides down the inclined plane a total distance ℓ + x = 1. 2 m be- fore stopping. The initial gravitational potential energy is converted entirely into energy stored in the spring. Thus
mg (ℓ + x ) sin θ =
kx^2
2 × 9. 0 × 1. 2 ×
k (−/ 2 )^2
k =
2 × 9. 8 × 1. 2
= 588 N / m
The speed of the block when it just touches the spring is also determined from energy conservation:
mv^2 = mg ℓ sinθ
v^2 = 2 g ℓ sinθ = 2 × 9. 8 ×
v = 3. 13 m / s
- [25 points] A rocket of initial mass m = 500 kilograms, momentarily at rest in free space, is expelling spent fuel at the rate of −∆ m /∆ t = 10 kg / sec out of its back, at a speed of ve = 500 meters/second. Calculate the magnitude of the initial forward acceleration a of the rocket, in meters/second-squared. Solution: The total momentum of the rocket plus exhauseted gas is conserved. Thus, over a short time ∆ t , the mass of expelled gas is −∆ m , and we obtain