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Material Type: Quiz; Class: General Physics I: Mechanics; Subject: Physics; University: Illinois Institute of Technology; Term: Spring 2002;
Typology: Quizzes
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Quizes: Spring 2002
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
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.
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.]
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.
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.
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
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.
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 )
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
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.
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
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
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.
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
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.
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 ) =
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:
m + M + I / R^2
mg
m + M + I / R^2
mg
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.
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.
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 +
v^2
h =
v^2 2 g
mR^2
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.
Solution: The force of gravitational attraction between the bodies is F = Gm^2 /( 2 R )^2. Thus,
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
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.
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,
mv^2 R
Gm^2 ( 2 R )^2
v^2 =
Gm 4 R
v = 1. 3 × 10 −^6 m / s
The period is
2 π r v
2 π 103
= 4. 8 × 109 sec = 170 years
Spring 2002 Examinations:
PHYS 123 - 001/002 TEST 1 25 February 2002
v 0 = 200
km hr
1 m / s
= 55 m / s
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
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
T = macent =
mv^2 R
5 · v^2 1 v^2 = 200 v = 14. 1 m / s
The corresponding frequency of revolution is
f =
v 2 π R
2 π · 1
= 2. 25 Hz
~ 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^2 x + a^2 y + a^2 z =
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
mv^20 −
mv^2 =
The energy concerted into heat is
The average force on the bullet is given by
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
= 4. 5 m / s^2
The coefficient of kinetic friction is
μk =
f N
ma mg
a g
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
k (−/ 2 )^2
k =
= 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