# PHYSICS 107 MIDTERM EXAMINATION

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PHYSICS 107
MIDTERM EXAMINATION
October 25, 2000
7:30–9:00 pm
When you are told to begin, check that this examination booklet contains all the
numbered pages from 2 through 10.
The exam contains three problems that count equally although some problems may
be harder than others.
Do not panic or be discouraged if you cannot do every problem; there are both easy
and hard parts in this exam. Keep moving and finish as much as you can!
on how well I can understand your solution even when you write down the correct answer.
DO ALL THE WORK YOU WANT GRADED IN THIS EXAMINATION BOOKLET!
Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the Honor
Code during this examination.
Signature
c 2000, Edward J. Groth
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 2
Problem 1. Gravity Launcher. The diagram shows a launcher that we might have
used in lab instead of the ones we actually used. It makes use of gravity rather than a
spring to launch a projectile towards the target. The projectile has mass m and slides on
the launcher which is a frictionless track. The projectile starts from rest at height h above
the launch point. The launch direction is an angle θ above the horizontal and the launch
point is a height H above the target and a horizontal distance D from the target. Just as
in lab, ignore air resistance.
(a) Assume the projectile leaves the launch point with speed v0 . How long does its flight
last before it hits the target? Your answer may include any or all of m, v0 , θ, g, D,
and H. (5 points)
Solution
We consider the vertical motion and take the initial height to be H and the final
height to be 0. Then the equation relating the initial and final heights and the time is
1
0 = H + v0 sin θt − gt2 ,
2
which gives
1
t=
g
q
v0 sin θ +
v02 sin2
θ + 2gH
.
Note that we had to make a choice for the proper sign. The other sign would give a time
before the projectile left the launcher.
End Solution
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 3
Problem 1. Continued.
(b) Find an expression for D in terms of any or all of m, v0 , θ, g, and H. (4 points)
Solution
The horizontal distance is just the horizontal velocity times the time (no horizontal
component of acceleration).
v0 cos θ
D=
g
q
2
2
v0 sin θ + v0 sin θ + 2gH .
End Solution
(c) Recall that in lab we calculated how far to compress the spring so that the projectile
would hit the target. In this case, the parameter we can adjust is the height h at
which the projectile is released on the launcher. Find h so that the projectile hits the
target. Your answer may include any or all of m, θ, g, D, and H. Note that v0 is not
allowed in the answer! (6 points)
Solution
By conservation of energy, we have mv02 /2 = mgh, so v0 =
our expression for D.
√
D=
2gh cos θ
g
p
√
2gh. We plug this into
q
2gh sin θ +
2
2gh sin θ + 2gH
p
√
D = 2h cos θ sin θ + 2 h cos θ h sin2 θ + H ,
p
D
− h sin θ = h2 sin2 θ + hH .
2 cos θ
Square both sides,
D2
Dh sin θ
−
+ h2 sin2 θ = hs sin2 θ + hH .
2
4 cos θ
2 cos θ
The h2 terms cancel and we can solve for h,
h=
D2
.
4 cos2 θ(H + D tan θ)
End Solution
c 2000, Edward J. Groth
SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 4
Problem 1. Continued.
(d) (You should be able to do this part even if you didn’t get the previous parts.) At
the lowest point of the launcher track, the radius of curvature is R = 0.2 m. The
projectile, with mass m = 0.1 kg, is released from a height y = 1.0 m above the
bottom (lowest point) of the track. Use g = 9.8 m s−2 . What are the magnitude,
direction, and units of the force exerted on the projectile by the track at this point?
(5 points)
Solution
The track exerts a normal force, N , pointed upwards. There is no horizontal force
since the track is frictionless. The weight, mg, of the projectile points downwards. At the
bottom, the projectile is following a circle with radius of curvature R, so it has an upward
acceleration v 2 /R, where v is its speed at the bottom. The speed, v, at the bottom may
be found from conservation of energy.
1
mgy = mv 2 ,
2
So,
v2
y
N − mg = ma = m = 2mg ,
R
R
y
1
N = mg 2 + 1 = 0.1 &middot; 9.8 &middot; 2 + 1 = 10.8 N upwards
R
.2
End Solution
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 5
hinged together at the top by a frictionless hinge. The ladder
rests on a smooth (frictionless) floor and the legs are kept from
spreading apart by a horizontal cable connecting the two legs.
You are standing on the top of the ladder (violating a whole
bunch of CPSC warning labels!). Some data: the left leg rests
on the floor a distance a = 0.5 m to the left of the point directly
below the hinge and the right leg rests b = 0.25 m to the right of
that point. The ladder is H = 2 m tall and the cable connecting
the legs is h = 1.5 m above the floor. The acceleration of gravity
is g = 9.8 m s−2 . The stepladder is a lightweight aluminum
ladder so its mass may be ignored. Finally, you have put on
some weight since quiz 5 and your mass is now M = 90 kg.
(Note that the figure is not to scale!)
(a) Draw a free body diagram showing all the external forces on the ladder. (4 points)
Solution
The forces are the weight acting down and vertical forces from the floor acting straight
up on the left and right legs, FL and FR .
End Solution
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 6
Problem 2. Continued.
(b) Determine the magnitude and direction of the force the floor exerts on each leg of the
Solution
Since there is no friction, the forces point up . Of course, their sum must equal your
weight. To find the individual forces, we need to consider torques. If we take the origin
(pivot) where the left leg contacts the floor, then the weight makes a clockwise (negative)
torque with lever arm a, the force on the left leg has zero lever arm, and the force on
the right leg makes a counterclockwise (positive) torque with lever arm (a + b). Since the
torques must sum to zero, we have
−Mga + FR (a + b) = 0 .
which gives
FR = Mg
a
0.5
= 90 &middot; 9.8 &middot;
= 588 N ,
a+b
0.5 + 0.25
and
FL = 90 &middot; 9.8 − 588 = 294 N .
End Solution
(c) Determine the tension in the cable. Hint: isolate one leg and take the origin for
torques at the top hinge. (6 points)
Solution
Following the hint, consider the left leg. The forces that act at the top hinge, which
acts to make a clockwise torque with lever arm a and the tension in the cable makes a
counterclockwise torque with lever arm H − h. So,
−FL a + T (H − h) = 0
or
T = FL
a
0.5
= 294
= 294 N .
H −h
2 − 1.5
End Solution
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 7
Problem 2. Continued.
(d) Suppose the cable breaks, the legs fly out, and you come crashing down. Where do
you hit the floor and why? (State your answer relative the spot directly below you
when you are standing on the ladder.) (5 points)
Solution
You hit the floor directly below where you were standing. Since there is no friction,
the floor exerts no horizontal force on the ladder and of course gravity exerts no horizontal
force, so there is no horizontal acceleration of the center of mass of the system. This means
the horizontal velocity of the center of mass remains 0 and the horizontal position of the
center of mass doesn’t change. You are the location of the center of mass of the system,
so you hit the floor directly below where you started.
End Solution
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 8
Problem 3. World’s Fastest Man in 1996. At the Atlanta Olympics in 1996, Donovan
Bailey won the 100 meters in a World Record time of 9.84 s. Michael Johnson won the 200
meters in a World Record time of 19.32 s. A certain NBC announcer (can you guess which
one?) started to say things like Johnson should have the title of the World’s fastest man
since his time was less than twice Bailey’s time. Of course, the announcer forgot about
the need to accelerate from rest! Johnson’s “splits” (times to run each half of the 200 m
race) were 10.12 s for the first 100 m and 9.20 s for the second 100 m.
Let’s consider a simple model in which a sprinter accelerates with constant acceleration
until he reaches his top speed and then runs with constant speed to the end of the race.
(This ignores things like running on the curve and fatigue of the sprinter, but it might still
give us a handle on what happens in a sprint.) We’ll use the data for Michael Johnson.
The top speed is reached well within the first 100 meters of the race. You might want to
consider the plots you are asked to draw in part (f) as you do parts (a)–(e).
(a) Using data for the second split, determine Johnson’s top speed, vm . (2 points)
Solution
Distance divided by time: vm = 100/9.20 = 10.87 m/s .
End Solution
(b) In the first half of the race, suppose Johnson accelerated with constant acceleration
from rest to speed vm in time ta and then ran at constant speed vm for time tv . What
was Johnson’s average velocity in the constant acceleration portion of the first half of
the race? (3 points) (Hint: a formula is desired.)
Solution
Constant a, vinitial = 0, vfinal = vm , so vave = vm /2 .
End Solution
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 9
Problem 3. Continued.
(c) In terms of vm , ta , and tv , give expressions (not numbers!) for the distance covered
and the time elapsed in the first half of the race. (3 points)
Solution
Sum constant a and constant v parts: t = ta + tv , d = vm ta /2 + vm tv .
End Solution
(d) Using the fact that the distance and time for the first half of the race are 100 m and
10.12 s, determine numerical values for ta and tv . (3 points)
Solution
We have two equations in two unknowns. (We know t, d, and vm .) In fact, if we divide
the distance equation by vm , the left hand side becomes d/vm = t2 = 9.20 s, where t2 is
the time taken to run the second 100 m. So our two equations become 10.12 = ta + tv ,
and 9.20 = ta /2 + tv which are easily solved to give ta = 1.84 s and tv = 8.28 s .
End Solution
(e) What is the magnitude of the constant acceleration at the start of the race? (3 points)
Solution
vm = ata , so a = vm /ta = 5.91 m/s2 .
End Solution
c 2000, Edward J. Groth
Copyright SOLUTIONS, Physics 107 Midterm Examination, October 25, 2000
Page 10
Problem 3. Continued.
(f) On the diagram below make reasonably accurate sketches of Michael Johnson’s position, velocity and acceleration as functions of time according to the model we’ve
worked out. Be sure to indicate the scale for the velocity and acceleration graphs.
(6 points)
Solution
End Solution
c 2000, Edward J. Groth