Download 1988

Name ____________________
Mr. Willie
Hastings AP Physics 1
Midterm Exam Review-2
3. Which of the following quantities is a scalar
that is always positive or zero?
(A) Power
(B) Work
(C) Kinetic energy
(D) Linear momentum
(E) Angular momentum
1. The displacement x of an object moving
along the x-axis is shown above as a function of
time t. The acceleration of this object must be
(A) zero
(B) constant but not zero
(C) increasing
(D) decreasing
(E) equal to g
View from Above
2. The horizontal turntable shown above rotates
at a constant rate. As viewed from above, a coin
on the turntable moves counterclockwise in a
circle as shown. Which of the following vectors
best represents the direction of the frictional
force exerted on the coin by the turntable when
the coin is in the position shown?
(A)
(B)
(C)
(D)
(E)
4. A block of weight W is pulled along a
horizontal surface at constant speed v by a force
F, which acts at an angle of  with the
horizontal, as shown above. The normal force
exerted on the block by the surface has
magnitude
(A) W - Fcos
(B) W - Fsin
(C) W
(D) W + Fsin
(E) W + Fcos
5. A 2-kilogram block rests at the edge of a
platform that is 10 meters above level ground.
The block is launched horizontally from the edge
of the platform with an initial speed of 3 meters
per second. Air resistance is negligible. The time
it will take for the block to reach the ground is
most nearly
(F) 0.3 s
(G) 1.0 s
(H) 1.4 s
(I) 2.0 s
(J) 3.0 s
6. A horizontal force F is used to pull a 5kilogram block across a floor at a constant speed
of 3 meters per second. The frictional force
between the block and the floor is 10 newtons.
The work done by the force F in 1 minute is
most nearly
(K) 0J
(L) 30J
(M) 600 J
(N) 1,350J
(O) 1,800J
7. A tennis ball of mass m rebounds from a
racquet with the same speed v as it had initially,
as shown above. The magnitude of the
momentum change of the ball is
(A) 0
(B) mv
(C) 2mv
(D) 2mv sin
(E) 2mv cos
8. A diver initially moving horizontally with
speed v dives off the edge of a vertical cliff and
lands in the water a distance d from the base of
the cliff. How far from the base of the cliff
would the diver have landed if the diver initially
had been moving horizontally with speed 2v?
(F) d
(G) 2 d
(H) 2d
(I) 4d
(J) It cannot be determined unless the height
of the cliff is known.
9. Two bodies of masses 5 and 7 kilograms are
initially at rest on a horizontal frictionless
surface. A light spring is compressed between
the bodies, which are held together by a thin
thread. After the spring is released by burning
through the thread, the 5-kilogram body has a
1
speed of meter per second. The speed of the
5
7-kilogram body is
1
m/s
12
1
(L) m/s
7
1
(M)
m/s
35
1
(N)
m/s
5
7
(O)
m/s
25
(K)
Questions 10-11
12. Mars has a mass
1
that of Earth and a
10
1
that of Earth. The acceleration of a
2
falling body near the surface of Mars is most
nearly
(B) 0.25 m/s2
(C) 0.5 m/s2
(D) 2 m/s2
(E) 4 m/s2
(F) 25 m/s2
diameter
A block oscillates without friction on the end of
a spring as shown above. The minimum and
maximum lengths of the spring as it oscillates
are, respectively, xmin and xmax. The graphs below
can represent quantities associated with the
oscillation as functions of the length x of the
spring.
13. A uniform rope of weight 50 newtons hangs
from a hook as shown above. A box of weight
100 newtons hangs from the rope. What is the
tension in the rope?
(A) 50 N throughout the rope
(B) 75 N throughout the rope
(C) 100 N throughout the rope
(D) 150 N throughout the rope
(E) It varies from 100 N at the bottom of the
rope to 150 N at the top.
10.Which graph can represent the total
mechanical energy of the block-spring system as
a function of x?
(A) A (B) B (C) C (D) D (E) E
10. Which graph can represent the kinetic energy
of the block as a function of x?
(A) A (B) B (C) C (D) D (E) E
16. The speed of the ball at point II is most
nearly
(A) 3.0 m/s
(B) 4.5 m/s
(C) 9.8 m/s
(D) 14 m/s
(E) 20 m/s
14. To weigh a fish, a person hangs a tackle box
of mass 3.5 kilograms and a cooler of mass 5
kilograms from the ends of a uniform rigid pole
that is suspended by a rope attached to its center.
The system balances when the fish hangs at a
1
point
of the rod’s length from the tackle box.
4
What is the mass of the fish?
(A) 1.5 kg
(B) 2kg
(C) 3 kg
(D) 6 kg
(E) 6.5 kg
Questions 15-16
17. When an object of weight W is suspended
from the center of a massless string as shown
above, the tension at any point in the string is
(A) 2Wcos
(B)
W cos
2
(C) 2Wcos
W
(D)
2 cos 
W
(E)
cos
A ball swings freely back and forth in an arc
from point I to point IV, as shown above. Point
II is the lowest point in the path, III is located 0.5
meter above II, and IV is 1 meter above II. Air
resistance is negligible.
15. If the potential energy is zero at point II,
where will the kinetic and potential energies of
the ball be equal?
(A) At point II
(B) At some point between II and III
(C) At point III
(D) At some point between III and IV
(E) At point IV
18. An object weighing 4 newtons swings on the
end of a string as a simple pendulum. At the
bottom of the swing, the tension in the string is 6
newtons. What is the magnitude of the
centripetal acceleration of the object at the
bottom of the swing?
(F) 0
1
(G) g
2
(H) g
3
(I) g
2
5
(J) g
2
19. A satellite of mass M moves in a circular
orbit of radius R at a constant speed v. Which of
the following must be true?
I.
The net force on the satellite is equal to
Mv 2
and is directed toward the center
R
of the orbit.
II. The net work done on the satellite by
gravity in one
revolution is zero.
III. The angular momentum of the satellite
is a constant.
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
20. A truck traveled 400 meters north in 80
seconds, and then it traveled 300 meters east in
70 seconds. The magnitude of the average
velocity of the truck was most nearly
(A) 1.2m/s
(B) 3.3 m/s
(C) 4.6m/s
(D) 6.6 m/s
(E) 9.3 m/s
21. A projectile is fired with initial velocity v0 at
an angle o with the horizontal and follows the
trajectory shown above. Which of the following
pairs of graphs best represents the vertical
components of the velocity and acceleration, v
and a, respectively, of the projectile as functions
of time t?
AP Physics 1
Midterm Review 2
Free Response
1. The cart shown above is made of a block of mass m and four solid rubber tires each of
mass m/4 and radius r. Each tire may be considered to be a disk. (A disk has rotational
inertia ½ ML2 , where M is the mass and L is the radius of the disk.) The cart is released
from rest and rolls without slipping from the top of an inclined plane of height h. Express
all algebraic answers in terms of the given quantities and fundamental constants.
(a) Determine the total rotational inertia of all four tires.
(b) Determine the speed of the cart when it reaches the bottom of the incline.
Free Response 1. (continued)
(c) After rolling down the incline and across the horizontal surface, the cart collides with
a bumper of negligible mass attached to an ideal spring, which has a spring constant k.
Determine the distance xm the spring is compressed before the cart and bumper come to
rest.
(d) Now assume that the bumper has a non-negligible mass. After the collision with the
bumper, the spring is compressed to a maximum distance of about 90% of the value of xm
in part (c). Give a reasonable explanation for this decrease.
Annotated Answers to APP1 MT Rev2 MC
1. A. The slope of the x vs t graph is the velocity,
which is constant. Then the acceleration is zero.
2. D. The net force, provided by the friction,
must be inward for uniform circular motion.
3. C. Power, work and kinetic energy are all
scalars, but only kinetic energy must be 0.
4. B. The normal force is lessened by the force,
in particular by the y–component of the force,
which is F sin .
5. C. Since the block is launched horizontally,
we don’t care at all about the 3 m/sec velocity;
y = 1/2 gt2. That is, 10 = 5t2, or t = 1.4 sec.
6. E. Since the block is moving at a constant
velocity, the force applied must be equal to the
force of friction. The work done is Fx, and
x = vt = 3 m/sec*60 sec = 180 m. Then Work
= (10 N)(180 m) = 1800 J.
7. E. The initial momentum p = (mv cos , mv
sin ), and the final momentum p/ = (–mv cos ,
mv sin ). The change in momentum p = p/ – p
= (–2mv cos , 0). The magnitude of p is just
2mv cos . (We know that the vertical
momentum cannot be changed, because when the
ball hits the racquet, the force exerted is, surprise
surprise, the normal force. That means that there
is no force in the y direction, and as Faverage =
p/t, we must have no change in the y
component of the momentum.)
8. C. Since the diver is moving horizontally in
both cases, the time t to hit the water will be the
same in both cases. The first distance travelled is
d = vt, and the second distance travelled is 2vt =
2d.
9. B. Conservation of momentum! Initially the
total momentum is zero; so finally the
momentum must be zero. That
is, p/ = (5 kg)(1/5 m/sec) + (7 kg)*v7/ = 0;
or v7/ = – 1/7 m/sec.
10. E. Conservation of energy. There shouldn’t
be any change in the total energy as the mass
moves back and forth.
11. D. E = K + U = K + 1/2 kx2 = constant, so K
= (constant) – 1/2 kx2, which is an upside-down
parabola. Alternatively, use the fact that the
speed at the endpoints is zero, so the K graph
must hit the x axis at the endpoints. The answer
cannot be C because the force varies smoothly,
and so the acceleration varies smoothly. In
C, we have the acceleration jumping suddenly at
the equilibrium point (x = 0); this is impossible.
12. D. Recall F = mg = GMm/R2. Then g =
GM/R2, so gMars : gEarth = MMars/R2Mars :
MEarth/R2Earth = (MMars /MEarth)*(REarth / RMars )2
=(1/10)*(2/1)2 = 4 : 10.
13. E. The rope is not itself massless. The
bottommost fibers of the rope have to support
only the box; the tension in the bottom of the
rope is just the box’s weight, 100 N. The
topmost fibers of the rope have to support the
box as well as the rope, or 150 N. The tension
varies from a maximum of 150 N at the top to
100 N at the bottom.
14. C. If the pole is not to rotate, we must have
the clockwise torques equal the counterclockwise torques. The definition of torque is
distance times force times sine of the angle
between the force and the radius vectors. (The
radius is drawn from the pivot point to the point
where the force is applied.) In the present case,
clock = (5 kg)(g)(l/2)(sin 90) = 5g l/2
counter-clock = (3.5 kg)(g)(l/2)(sin 90) + mg(l/4)(sin
90) = 3.5g l/2 + mgl/4
so we have to have 5gl/2 = 3.5gl/2 + mgl/4 or
1.5l/2 = ml/4 ; m = 3 kg.
15. C. At point C, the ball is halfway up to its
maximum value, so U = mgy is half of its
maximum value, attained at 1
meter. If U = half of Umax, then the other half is
K, and U = K at that point.
16. B. A “green light” problem; it’s easy, and
you should do it quickly. At the top, E = U =
mgh; at the bottom, E =K = 1/2 mv2. Set them
equal as usual, and find v2 = 2gh = 20, or v = 4.5
m/sec.
17. D. We need Tleft + Tright + mg = 0. But Tleft =
(–T sin , T cos ) and similarly Tright = (T sin ,
T cos ) That means
2T cos – mg = 0, or T = mg/(2 cos ).
18. B. First off, the mass of the object is m =
mg/g = 4 N/10 m/sec2 = 0.4 kg. At the bottom of
the swing, we have T – mg = mv2/R = macentrip
= 6 – 4 = 2 N; so a = 2 N/0.4 kg = 5 m/sec2 = g/2.
would peel away from the mother body, and
would no longer be in orbit. III is true, and is just
a restatement of Kepler’s Second Law; that equal
areas are swept out in equal times.
19. E. I. is, of course, true. II is also true, because
the displacement (along the circumference) is
always at right angles to the force (gravity), so
Fx cos = 0. Another way to see that the work
done by the net force in uniform circular motion
must be zero is the following argument. Say that
the work was not zero. Then by the WorkEnergy Theorem, the kinetic energy would
increase. But if the speed increased, the mass
20. B. Average velocity v = r/t = (400,
300)/150 = (500 @ 37º)/150 = (3.3 m/sec @
37º).
21. D. The acceleration in the y direction is just
–g, a constant; and the y velocity starts positive,
falls through zero, and winds up negative.
Free Response 1 Solution
15 points total
Distribution of
points
(a) 2 points
For determining the rotational inertia of each tire
1
1𝑚 2 1
𝐼 = 𝑀𝐿2 =
𝑟 = 𝑚𝑟 2
2
24
8
For the correct total rotational inertia for all 4 tires
1
𝐼𝑡𝑜𝑡 = 4𝐼 = 𝑚𝑟 2
2
1 point
1 point
(b) 7 points
For an indication of the conservation of mechanical energy
Etop = Ebottom ; ∆U=-∆K ; or equivalent
1 point
For correct expressions for energies at the top
Ktop = 0; Utop _ mgh + 4(mgh/4)= 2mgh
1 point
For a correct expression for potential energy at the bottom and
that kinetic energy at the bottom is the sum of
translational and rotational kinetic energies
Ubottom = 0; Kbottom = Ktrans + Krot
1 point
For a correct expression for translational kinetic energy at the bottom
Ktrans = 1/2 (2m)υ2 = mυ2
1 point
For a correct expression for rotational kinetic energy at the bottom
Krot = Iω2/2
1 point
(b) continued
Distribution of
points
For recognition of the relationship between translational and rotational velocity
ω=v/r
Substituting these expressions to determine total kinetic energy at the bottom
1 1
𝑣2 5
𝐾𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑚𝑣 2 + ( 𝑚𝑟 2 ) 2 = 𝑚𝑣 2
2 2
𝑟
4
Setting potential energy at the top equal to the kinetic energy at the bottom
5mυ2/4 = mgh
For the correct solution for v
1 point
1 point
8
𝑣 = √ 𝑔ℎ
5
(c) 4 points
For recognition that energy is conserved (Although it is an inelastic collision,
the mass of the bumper is negligibly small, thus its kinetic energy is negligible
and there is no loss of energy.)
1 point
For a correct expression for potential energy of the spring at maximum
compression
UK = kxm2/2
1 point
For applying conservation of energy by equating the potential energy of
1 point
the spring at maximum compression EITHER to the gravitational potential
energy of the cart and wheels at the top of the inclined plane OR to the
kinetic energy of the cart and wheels at the bottom of the inclined plane
1
1
5
2
2
𝑘𝑥𝑚
= 2𝑚𝑔ℎ
OR
𝑘𝑥𝑚
= 4 𝑀𝑣 2
2
2
For a correct solution of either of these equations for xm (including a correct
substitution for v from part (b) for the second equation) or an answer consistent
with work done in (b)
𝑥𝑚 = 2√
1 point
𝑚𝑔ℎ
𝑘
(d) 2 points
For an explanation that discusses the inelastic collision with a loss of
mechanical energy or a reduced velocity resulting in a smaller compression.
The discussion should have been correctly stated. If there were any incorrect
statements, then 1 point was subtracted. Points were neither added or
subtracted for irrelevant statements, such as references to friction.
2 points