Physics 53
Summer 2008
Exam I
Solutions
In questions or problems not requiring numerical answers, express the answers in terms of the
symbols for the quantities given, and standard constants such as g. In numerical questions or
problems, use g = 10 m/s2.
Part A: Multiple choice questions. Check the best answer. Each question carries a value of 4 points.
1.
Concerning the general principles of mechanics, which of the following is NOT a
complete and correct statement?
In an inertial frame, a body subject to no net force moves with constant
velocity.
The total force on an object is equal to the rate of change of its momentum.
If body A exerts force F on body B, then body B also exerts force –F on
body A.
√
2.
The change in total mechanical energy is equal to the work done by
external forces. [Non-conservative forces, not external forces.]
A box is being pulled across a rough floor by application of a force F as shown.
As the angle θ is increased, starting from zero, which of the following does NOT
happen?
The normal force exerted by the floor decreases.
√
The kinetic friction force increases.
[Decreases.]
The horizontal component of F decreases.
The acceleration of the box increases.
1
θ
F
Physics 53
3.
Summer 2008
A block slides all the way around the inside of a vertical circular track as shown.
The track is frictionless. Points A, B, and C are at the top, side, and bottom of the
track respectively.
A
The minimum possible speed at A is zero.
The speed is the same at all three points.
The direction of the acceleration is toward the
center of the circle at all three points.
√
4.
The total mechanical energy of the system is the
same at all three points. [Only conservative
forces do work.]
B
C
A mass m is attached to a vertical spring of stiffness k hanging from a ceiling as
shown. Initially the mass is held a the height where the spring is at its
equilibrium length. Which of the following is NOT true?
If the mass is lowered slowly, it will hang at rest when the
spring has been stretched through distance mg / k .
If the mass is dropped from rest, it will stop momentarily
after falling distance 2mg / k .
√
If the mass is dropped from rest, it will have its maximum
acceleration when it has fallen distance mg / k . [Maximum speed at that
point, zero acceleration.]
If the mass is dropped from rest, it will return to its original height with
zero instantaneous velocity.
Part B: True-false questions. Mark each sentence with T of F. Each question has a value of 3 points.
1.
In a head-on collision between a small car and a large SUV the acceleration of the
two vehicles during the collision has the same magnitude.
F
2.
[Forces have same magnitude (3rd law) but masses are not equal.]
Curves on highways are banked to use a component of the normal force as part
of the radial force, thus reducing the need for friction.
T
3.
Pushing a box up a frictionless incline to height h at constant speed v requires the
same energy input as lifting it straight up to that height at that speed, but the
power input required is less if you use the incline.
T
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Physics 53
Summer 2008
Part C: Problems. Work problems in the space provided, indicating your method clearly. A correct final
answer supported by no argument, or a fallacious one, will receive little or no credit. Problems carry the
point values shown.
1.
A baseball right fielder needs to throw the ball as quickly as possible to third
base, at distance R. He can throw the ball with initial speed v0 sufficient to get it
all the way to the base in the air if he throws it at an elevation angle !1 .
a.
From the equations for projectile motion, derive the formula for the time
of flight of the ball T1 in terms of v0 , !1 and g.
b.
Use this result to prove that R = (2v0 2 / g)sin !1 cos!1 .
c.
An alternative is to throw at the same speed but a different elevation angle
! 2 so that it bounces from the ground at distance R/2, without changing
its horizontal velocity but reversing its vertical velocity, continuing
another distance R / 2 with the same trajectory. [See the drawings below.]
Show that total time T2 it takes to get the ball to the base by this method is
T2 = T1
cos!1
. (Since ! 2 < !1 , T2 < T1 .)
cos! 2
!1
!2
[15 points]
a.
Vertical motion: y = v0 sin !1 " t # 12 gt 2 . Set y = 0 and solve for t > 0 :
t = (2v0 / g)sin !1 . This is the time of flight T1 . [Or use v y = v0 sin !1 " gt to
find the time when v y = 0 at the maximum height. This is half of T1 .]
b.
Use x = v0 cos!1 " t : R = v0 cos!1 " (2v0 / g)sin !1 = (2v0 2 / g)sin !1 cos!1 .
c.
Since the range for the first half of the flight is R / 2 , we have for the total
range R = (4v0 2 / g)sin ! 2 cos! 2 . Comparing to (b) we see that
sin !1 cos!1 = 2sin ! 2 cos! 2 . From (a) we see that T2 /T1 = 2sin ! 2 /sin !1 .
Combining these two we find the claimed result. Or (easier) use the fact
that the total distance R is the same for both methods, so we have both
R = v0 cos!1 " T1 and R = v0 cos! 2 " T2 . The claimed results follows
immediately.
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2.
Summer 2008
You are inside a closed moving railroad car and cannot see the outside world.
You notice that a lamp suspended from the ceiling hangs at the angle shown, and
that you must lean at the same angle to remain standing. There are two
possibilities: (A) The car is on level ground but accelerating; (B) The car is
moving at constant speed on an incline at angle θ above the horizontal.
a.
Assuming that (A) is correct, what is the magnitude of the acceleration,
and what is its direction (to the left or the right)? Justify your answer.
b.
Describe a method by which
you can determine by
measurements (without
looking out of the car) which
of the two possibilities is the
actual case.
θ
[10 points]
a.
The cord hangs in the direction of g eff = g ! a , for which the vertical and
horizontal components are geff cos! = g and geff sin ! = a . Dividing the
equations we find a/ g = tan ! , or a = g tan ! . The direction of a is to the
right (since –a is to the left). See the vector diagram below.
b.
If the car is moving at constant speed on the incline, the magnitude of
gravitational forces would be unaffected. One of these is the tension in the
cord, which would be equal to mg. If the car is accelerating, this tension
would be larger: mgeff = m g 2 + a2 . Another thing to measure is the time
it takes an object to fall from the ceiling to the floor, which would be
smaller in the accelerating car.
g eff ! g
–a
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Physics 53
3.
Summer 2008
You are driving along at speed v when a large tree falls across the road in front of
you. There is a side street just before the tree that you can turn into, if you make
a right turn of radius R. Or you can apply the brakes and hope to stop before you
hit the tree. The figure shows the view from above.
a.
What minimum coefficient of static friction do you
need between your tires and the road so as to make
the right turn without skidding? [Give the answer in
terms of v, R and g.]
b.
With that coefficient of friction, what is the shortest
distance d within which you can stop your car by
applying the brakes and driving straight ahead
without slipping? [Give the answer in terms of R.]
[15 points]
a.
The radial acceleration must be ar = v 2 /R , so the friction force must be m
times this. But the maximum static friction force is fs = µs N = µs mg , so we
have µs mg = mv 2 /R , or µs = v 2 /Rg .
b.
The (backward) acceleration comes from the same maximum friction force
as in (a), so µs mg = ma , giving a = µs g . Use v f 2 = vi 2 + 2ax x (or energy
v2
. Using
2 µs g
(a) this gives x = R / 2 . [You can stop in a distance equal to half the radius
of the smallest curve you can make.]
considerations) with v f = 0 , vi = v and ax = ! µs g to find x =
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Physics 53
4.
Summer 2008
A small block of mass m starts from rest at height h on the left incline, sliding
down, then across a frictionless horizontal section, and onto the right incline. The
inclines have kinetic friction coefficient µ k = 1/3 , and the angle of the incline is
such that sin ! = 0.8 and cos! = 0.6 .
a.
Express the work done by friction as the block slides down the left incline
in terms of m, g and h. [Through what distance on the incline has it slid?]
b.
When the block comes momentarily to rest on the right incline, what is the
height h’ to which it has risen? Give the answer in terms of h.
c.
If the static friction coefficient of the inclines is µs = 1/ 2 , will the block
slide back down the right incline? Justify your answer.
!
!
[15 points]
a.
Call the distance it slides d. Then h = d sin ! . The friction force is
f k = µ k N = µ k mg cos! , and the work it does is W f = ! f k d = ! µ k mgh
cos"
.
sin "
Using the numbers we find W f = ! 14 mgh .
b.
When the block reaches the horizontal section its total energy is
E = mgh + W f = 34 mgh . In sliding up the other incline the work by friction
is W f! = " 14 mgh! . The final total energy is E! = mgh! , so we have
E! " E = W f! , or mgh! " 34 mgh = " 14 mgh! . Solving we find h! = 35 h .
c.
The maximum static friction force is fs = µs mg cos! = 0.3mg . The
component of weight down the plane is mg sin ! = 0.8mg , so the block will
slide back down.
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Physics 53
5.
Summer 2008
A ball of mass m is attached to a massless string L, which is attached to a ceiling
as shown. The ball is pulled back until it is against the ceiling, and then released
from rest. At the bottom of the swing, the string encounters a peg at distance d
below the ceiling. The upper part of the string stops moving, while the ball now
swings about the peg in a circle of radius L ! d .
a.
If the ball describes a complete circle about the peg, what is its speed at
the top of that circle, in terms of L, d and g?
b.
Suppose d is such that the speed at the top is as small as it can be for the
ball to make a complete circle. What is this value of d? [What is the string
tension in this case?]
c.
What is the speed at the bottom of the small circle, for this value of d?
d.
What is the tension in the string at that point in the motion?
[Take gravitational potential energy to be zero at the bottom of the swing.]
m
L
d
[20 points]
a.
Conservation of energy. Take the bottom of the swing to be y = 0 . Then
we have E(1) = mgL and E(2) = 12 mv 2 + mg ! 2(L " d) . Setting E(2) = E(1)
gives
b.
1
mv 2
2
+ mg ! 2(L " d) = mgL , or v 2 = 2g(2d ! L) .
At the top the forces are the tension (down) and the weight (down), so we
have T + mg = mar = mv 2 /(L ! d) . The minimum speed occurs when T = 0 ,
or v 2 = g(L ! d) . Setting this equal to the speed in (a) we find d = 35 L .
c.
Conservation of energy. Since E(3) = 12 mvB 2 , we find by setting E(3) = E(1)
that vB 2 = 2gL .
d.
The radial acceleration is ar = vB 2 /(L ! d) = 5vB 2 / 2L . The forces are the
tension (up) and the weight (down), so T ! mg = m " 5vB 2 / 2L = 5mg and
T = 6mg.
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Physics 53
Summer 2008
Median = 56
SD = 15.7
8