Download Solutions to Physics 110 Sample Mid-Term Exam (hand

Solutions to Physics 110 Sample Mid-Term Exam (hand-out 2/18/00)
You will have 75 minutes to complete the mid-term examination. There are 10 problems
from which you should choose to answer 7. Each problem counts 10 points. If you
answer more than 7 problems, only the first 7 will be graded. You may use one sheet of
paper (front and back) with equations, so that this is not a test of your ability to memorize
equations. The average time it takes to complete a question is 5 – 10 minutes, as they
vary in difficulty. Please explain how you get your answers to be able to obtain
appropriate credit.
You may use as the acceleration due to gravity either 10 m/s2 or 32 ft/s2. The
gravitational constant is 6.67 x 10-11 Nm2/kg2.
The solution is written below each problem.
1) a) Can astronauts floating in orbit tell which objects within their space-ship would be
heavy or light on Earth even though everything in the ship is effectively weightless?
Explain.
b) Could an astronaut use a “weight-less” hammer in space to drive a nail through a piece
of wood? Explain.
A1)
a) Yes. In order to know what an object would weigh if it were on earth one must
determine the mass of the object. The object’s weight is then W = mg, where g = 9.8 m/s2
the acceleration due to earth’s gravity. Astronauts can determine in space by various
means the relative mass of an object, which will reflect whether the object is light or
heavy on earth. One way to do this would be to push the object and feel how much force
it takes you to move it from rest and then to compare this with the force it would take to
move the second object. The more force that is needed, the more massive the object.
[To be more quantitative, one could pull or push the object using a calibrated spring
(scale) and observe the reading (on the scale). The more massive the object, the larger
the reading for the force needed to accelerate the object. Another method is to swing the
mass in a circle at a constant angular velocity, with the spring scale connected between
your hand and the object. The scale reading will give you the centripetal force and you
can determine the object’s mass from the angular velocity and the distance from the
center of the circular motion to the mass.]
b) Yes. To drive a nail through a piece of wood simply requires pounding the nail with
the head of the hammer. Whether on earth or “weightless” the momentum that is
transferred in a given time (i.e. the force exerted) from the hammer to the nail drives it
into the wood.
2) A 5 kg bowling ball is rolling down the alley at a speed of 4 m/s.
a) What is its momentum?
b) What is it’s kinetic energy?
c) If you exert a horizontal force on the ball for 0.5 s, how much force must you exert on
the ball in the horizontal direction to get it up to this speed (4 m/s)?
A2)
a) momentum p = m v = (5 kg) (4 m/s) = 20 kg m/s
b) kinetic energy = KE = ½ m v2 = ½ (5 kg) (4 m/s) 2 = 10 kg m2 / s2 = 10 Joules
c) F = m a = m (vf – vi) / (0.5 s) = (5 kg) (4 m/s) / (0.5 s) = 40 kg m / s2 = 40 N
3) A marble is dropped from the top of a skyscraper. Is the work done by Earth’s
gravitational force on the marble equal to, smaller than, or larger than that done by the
marble’s gravitational force on Earth? Explain.
A3) The work done by the earth’s gravitational force on the marble is greater than the
work done by the marble’s gravitational force on the earth. The gravitational force
between two masses (m1 , m2) whose centers are a distance r apart is F = G m1 m2 / r2 .
At the surface of the earth the diagram below applies for the forces of the problem, where
m1
F1
FE
earth
the F1 = FE = F , where F1 and FE are equal and opposite forces (of the earth on the
marble and vice versa). Using g as the acceleration due to earth’s gravity the force of the
earth on the marble is F1 = m1 g . The force of the marble on earth is FE = mE a , where
a is the acceleration of the earth due to the gravitational force of the marble on the earth.
Since F1 and FE are equal in magnitude and opposite in direction, then m1 g = - mE a.
Finally since mE >> m1 , aE << g . Now to compare the work that is done, note that W =
F x d, where d is the distance over which the force is applied. Since the forces are equal
and the acceleration of the marble is much greater than the acceleration of the earth, the
distance travelled by the marble is significantly larger than the distance traveled by the
earth due to the force of the marble on it. Thus, the work done by the earth’s
gravitational force on the marble is greater than the work done by the marble’s
gravitational force on the earth.
4) Identify all the action-reaction pairs that exist for a horse pulling a cart. Include Earth,
but ignore air resistance. Make sure your notation makes clear which force acts on which
object.
A4) (Please excuse the drawing!)
F1
n
-n
W
-W
-F1
H
-T
T
-H
The force of the horse on the cart is F1. The reaction force of the cart on the horse is –F1.
The normal force of the earth on the cart is n and the reaction force is -n . The force of
gravity on the cart from the earth is W. The reaction force on the earth is –W. The force
of gravity on the horse from the earth is H. The reaction force on the earth is –H. On
each of the hooves of the horse that is in contact with the earth, there is a lateral force of
the hoof on the earth which is labeled T in the diagram. The reaction force of the earth
on the hoof is – T. This force is only drawn for one hoof, although it will exist and be a
different value for each hoof in contact with the earth (as well as its reaction force).
5) Your friend is sitting in a tree at the edge of a river. Your friend grabs a rope (Tarzanstyle) which is hanging down from a branch and swings out over the river. Assume that
the vertical height of the swing when your friend starts off from the tree is 12 meters
above the lowest point of the swing’s arc and that your friends weight is 600 N. Neglect
air resistance and friction.
a) What is your friend’s potential energy (due to the swing) in Joules when she just gets
on the swing in the tree?
b) What is her kinetic energy in Joules at the lowest point of the arc?
c) What is her velocity at a height of 4 meters above the lowest point of the arc?
A5)
a) PE = m g h = (600N) (12 m) = 7200 N m = 7200 Joules
b) KE at lowest point = PE at highest point of take-off = 7200 Joules
c) At a height of 4 meters above the lowest point of the arc, by conservation of energy
KE (h = 4 m) + PE (h = 4 m) = PE (h = 12 m) + KE (h = 12 m)
KE (h = 4 m) + PE (h = 4 m) = PE (h = 12 m) ,
since KE (h = 12 m) = 0
KE (h = 4 m) = PE (h = 12 m) - PE (h = 4 m) = m g (12 m – 4 m) = (600 N) (8 m)
= 4800 N m = 4800 Joules
at
h=4m
To calculate the velocity use KE = ½ m v2 = 4800 Joules at h = 4 m.
This gives for the velocity v = √ [(2) (KE) / m] = √ [(2 (4800 J) / 60 kg] = 12.8 m/s
6) You are aboard a Ferris wheel of radius 30 m that is turning with a constant speed.
a) Draw a diagram of the forces on your body as you reach the bottom of the ferris wheel.
b) Draw a diagram of the forces on your body when you are on the side (half way up in
height) of the ferris wheel.
c) You become weightless at the top of the trajectory. What is the speed of the ferris
wheel?
Total force F = m ac = W + N (vectorially).
A6)
a)
b)
ac
N
ac
N
F = mac = N - W
W
F = mac = N
W
c)
N
W
ac
Weightless if F = m ac = W = mg .
ac = v2 / r = g → v = √ (g r) = √ [ (10 m /s 2) (10 m)]
= 10 m / s
F = mac = W - N
this corresponds to w = v / r = (10 m / s) / (10 m)
= 1 radian / s
7) What is the smallest force with which you can raise an object of mass m on Earth into
the air?
A7) The smallest force that can raise an object of mass m on earth into the air is F > m
g, where g is the acceleration due to gravity. This means any force greater than mg.
8) A 10,000 kg truck is traveling at 24 miles per hour northward. At an intersection it
collides with a 8,000 kg truck which is traveling at 30 miles per hour eastward. The two
trucks stick together. What is their velocity (direction and speed) after the collision
neglecting friction?
A8)
p1 = m1 v1 = (10000 kg) (24 mi/hr) = 24,000 kg mi / hr
NORTHWARD
before
p2 = m2 v2 = (8000 kg) (30 mi/hr) = 24,000 kg mi / hr
EASTWARD
p2
p1
after
p = p1 + p2 = √ [p12 + p22]
= √ [(24,000)2 + (24,000)2 ]
= 336,000 kg mi / hr
direction is northeast as seen on left
v = p / (m1 +m2 ) = (336,000 kg mi /hr) / 18,000 kg
= 18.7 mi / hr
9) A merry-go-round is rotating freely with a constant angular velocity before you get on.
Once you have stepped on, does the angular velocity increase, decrease or remain the
same? Explain.
A9) The angular velocity decreases. If the merry-go-round is rotating freely with a
constant angular velocity, then it has an angular momentum which is L = I ω, where I is
the rotational moment of inertia and ω is the angular velocity. Once you step onto the
merry-go-round at a distance r from the center, you add your mass m at r and therefore
increase the rotational moment of inertia from I to I + mr. Angular momentum must be
conserved (meaning that it is unchanged between the time you stepped on to after you
stepped on). Since I has increased since you stepped on, then the angular velocity must
decrease to conserve angular momentum (i.e. to leave it unchanged).
10) When I first moved into an apartment after graduate school, the owners were worried
about the weight and pressure of my waterbed on the floor. This was a second floor
apartment. The waterbed is 2 meters on one side and 0.25 meters in depth. Neglecting the
frame and given that the density of water is 1000 kg/m3:
a) Calculate the weight and pressure of the waterbed on the floor.
b) If a grand piano weighs the same as the waterbed and all of its weight is supported
equally by three legs which are each 100 cm2 thickness, i.e. the area in contact with the
floor, what is the pressure of the piano on the floor through one leg?
A10)
a) Volume = V = (length) (width) (depth) = (2m) (2m) (0.25 m) = 1 m3 water.
Density = d = 1000 kg / m3 for water.
Mass of water in waterbed = M = V d = (1 m3) (1000 kg / m3) = 1000 kg
Neglect the frame etc.
Weight = W = M g = (1000 kg) (9.8 m/s2) = 9800 N
Pressure = P = F / A = (9800 N )/ (4 m2) = 2450 Pascals
b) pressure = P = F / A
Area of 3 legs = 3 x 100 cm2 = 3.00 x 10-2 m2
P = (9800 N) / (3.00 x 10-2 m2) = 3.27 x 105 Pascals.
This is the pressure on the floor through one leg.