Announcements
• Homework 15 and 16 are due on Today at 6
PM.
Exam # 2 Review
8 March 2012
Pre-class quiz
•
•
Suppose an astronaut working on the outside of the International
Space Station found himself unattached from the station. This was
discovered when he accidently threw away a 2 kg hammer at 2
m/s. Does the astronaut remain at rest? If not, estimate how fast he
would be moving. How could he stop himself from drifting away?
The astronaut could not remain at rest. Because the astronaut and
the hammer are an isolated system. Momentum must be conserved
so if they start with zero momentum and then the astronaut throws
the hammer, he is going to move in the opposite direction as the
hammer with the same momentum. Momentum = mv. So the
astronaut's momentum must equal (2*2 or 4 kg m/s). If the
astronaut weights 70 kg. He is moving at 4/70 or .057 m/s. To stop
himself from drifting away he could throw another object in the
opposite direction at the same momentum (this would stop him).
• Exam # 2 has started. It runs through
Saturday, with a late day on Monday until
Noon.
• We will review the exam results next class
period, with some material from Chapter 10
(1-3).
Practice Exam
1A. A block slides on a rough horizontal surface and is attached to a
wall by a spring. The block is pulled to one side stretching the spring
as shown in (a) of the figure. It is then released from rest and moves
until it stops with the spring compressed as in (b) of the figure. There
IS friction between the block and the surface.
a
b
The total work done by the spring as the block moves from (a) to (b) is
1) Positive
2) Negative
3) Zero
4) Not possible to determine from the given information
Answer
• The net work done on the block is zero
(since it has no change in KE.) The
frictional force does negative work, gravity
and the normal force do zero work, so the
spring must have done positive work.
Practice Exam
1B. An object is dropped from the top of a tall
building. It reaches terminal speed about halfway to
the ground. For the entire drop, from the top of the
building to the ground, which of these are true?
1) gravity does more work than air resistance
2) air resistance does more work than gravity
3) gravity and air resistance do the same
amount of work
4) None of the above choices are correct.
Answer
Practice Exam
• There are two forces doing work on it,
negative work done by air resistance and
positive work done by gravity. The net
change in the KE of the object is positive
(started from zero and has some KE when
it hits.) Since the net change is positive,
gravity must have done more work on the
object than air resistance.
1C A compact car and a large truck collide head on and stick
together. Which undergoes the larger momentum change? (You
are to compare the magnitudes of the momentum change)
1) The Car
2) The Truck
3) The momentum changes are the same for both
vehicles
4) We cannot know without knowing the initial
velocities of the two vehicles.
5) We cannot know without knowing the final
velocities of the two vehicles.
6) None of these answers are correct
Answer
The net change in momentum for the system,
,
But
௖
௧ , so we have
௖
௧
Practice Exam
1D. Think fast! You’ve just driven around a curve in a
narrow, one-way street at 25 mph when you notice a car,
identical to yours, coming straight toward you at 25 mph.
You have two options: hitting the other car head on or
swerving into a massive concrete wall, also head on. If
your decision is based only on how it affects you, in the
split second before impact you decide to:
so
௖
௧
and both have the same magnitude of
momentum change.
Answer
The forces on you (and therefore the
damage) are determined by the change in
your momentum:
1. If you hit the wall your momentum will go
from mv to zero.
2. If you hit the other car, your momentum
will go from mv to zero.
It makes no difference which you hit.
1) Hit the other car
2) Hit the wall
3) Hit either one, it makes no difference
4) Consult your lecture notes
Practice Exam
1E. A certain weight lifter is described as “Powerful”. If that
description is correct with the physical meaning of “power”,
which of the below is the most correct statement about the
persons “powerful” muscles?
1) They can apply a large force for a short time
2) They can apply a large force for a long time
3) They can apply a small force for a short time
4) They can apply a small force for a long time
5) They can apply a large force over a large distance
6) They can apply a large force over a small distance
7) They can apply a small force over a large distance
8) They can apply a small force over a small distance
9) They can apply large forces
Answer
• This is a little ambiguous, but since
A large force over a small time would
probably be the best choice here.
Practice Exam
2A. A pumpkin of mass 25 kg is dropped
from a height of 20 m. When it arrives at the
ground it is brought to rest in 0.20 seconds.
What is average power required to stop the
pumpkin in that time?
Answer: 2.45 x 104 W
Practice Exam
Practice Exam
2B. How much potential energy (in joules) is
stored in a spring if it is stretched 30 cm
from equilibrium and its force constant is
160 N/m?
2C. A board is initially balanced on a cross
beam. A 20 kg mass is placed on the board
1.5 m from the cross beam. If the board is to
remain balanced, the center of mass should
remain above the cross beam. How far (in
meters) from the cross beam must the 80 kg
person stand to keep the board balanced?
Answer: 7.2 J
Answer: 0.375 m
Practice Exam
Practice Exam
2D. A 900 kg car traveling at 65 km/hr has a
head-on collision with a 1250 kg car. Both
cars come to a stop at the point of collision.
What was the original speed of the second
car (in km/hr)?
2E. A 50 kg box is dragged 40 m along a flat
horizontal floor by a force of 20 Newtons
that is directed 30 degrees above the
vertical and in the direction of motion. What
is the work done (in Joules) by the applied
force?
Answer: 46.8 km/hr
Answer: 693 J
Practice Exam
3. Thomas the Tank Engine (treat as a point
particle of mass Mt) is approaching
“Gordon’s hill” (height H and forming a near
perfect incline at θ degrees) with a constant
velocity. Wet leaves have covered the track
making them frictionless.
Practice Exam
3A. What is the minimum initial velocity vo
that Thomas must have to get over the top
of the hill? A symbolic answer in terms of
given variables and fundamental constants
is required.
Answer:
଴
Practice Exam
3B. On a different day, there are fewer leaves on the
tracks and thus Thomas can supply some push up
the slope. If, as a result of the “push”, there is a net
constant force of F acting on Thomas parallel to and
up the slope, what minimum velocity vm must
Thomas have at the bottom of the hill to just reach
the top with the help of that extra push? A symbolic
answer in terms of given variables and fundamental
constants is required.
Answer:
௠
ி
ெ೟ ୱ୧୬ ఏ
Practice Exam
3C. Under the conditions of part B and an
initial velocity of 20 m/s at the bottom of the
hill, calculate Thomas’s final velocity at the
top of the hill. The following numbers may
be helpful: Mt = 4000 kg, H = 45 m, θ = 20
degrees, and F = 8000 N.
Answer: 6.65 m/s
Practice Exam
Practice Exam
4. Two “point particle” masses are attached to identical
strings (length 3 m) and attached to the same point as
shown in the figure. One mass is pulled to the side with
the string kept straight and the second is left stationary
with the string vertical. The smaller mass (m = 200 g) is
pulled back to a height of 20 cm (h1) and is released from
rest. It strikes the second larger mass (M= 500 g) at the
bottom of its trajectory.
4A. There is Velcro on the surfaces of the
masses so that they stick together on contact.
What is the height h2 (their mutual height when
the come momentarily to a stop)? A symbolic
answer in terms of given variables and
fundamental constants is required.
Answer:
ଶ
ଶ
௠
௠ାெ
ଵ
Practice Exam
Practice Exam
4B. After many demonstrations of this
apparatus, the Velcro begins to wear out and is
no longer able to keep the two masses together
and the masses bounce off each other. The
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smaller mass rebounds to a height equal to భ .
ସ
What is the maximum height of the second
mass in this case? Please give a numerical
answer here.
4C. What is the change of energy of the two
mass system in this second case? A
numeric answer is appropriate here.
Answer: 0.0588 J
Answer: 7.20 cm
Practice Exam
Practice Exam
5. A large spring loaded gun (mass mg and
spring constant k) is made to fire bowling balls
(mass mb). It is loaded and mounted to a canoe
(mass mc and length L). The gun is placed at
one end of the canoe and the gun is fired
horizontally in the direction of the far end of the
canoe. The canoe is initially at rest and moves
in the water without friction. In the calculations,
treat each object as a point particle at its
respective center of mass.
5A When the fired bowling ball passes the
far end of the canoe, how far has the canoe
moved? A symbolic answer in terms of the
variables given and fundamental constants
is required.
Answer:
௠್
௠್ ା௠೎ ା௠೒
Practice Exam
5B. The fired bowling ball is seen to have a
horizontal velocity of vb as it passes over the far
end of the canoe (the observer measuring the
velocity is on the shore). At that instant, how
fast is the canoe moving (vc)? A symbolic
answer in terms of the variables given and
fundamental constants is required.
Answer:
௖
௠್
௠೎ ା௠೒
Practice Exam
5C. Given that mb = 10 kg, mg = 40 kg,
mc= 50 kg, L = 2.0 m, k = 16000 N/m, and
the ball’s horizontal velocity is measured at
vb= 25 m/s, how far was the spring in the
gun compressed?
Answer: 0.659 m
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