PHY 1150

Second Hour Exam

March 11, 1998

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This is a multiple-choice exam. That means there is no partial credit. But it also means there are no multiple-part questions. Take your time. None of the questions are intended to be "trick questions". If you do a calculation and have something like 12.345 for an answer and I have 12.346, it is probably just a round-off error (mine or yours).

All I want is the "scantron" sheet, the "bubble sheet", or the answer sheet. You can keep the questions from the exam. The solutions to the questions will be posted on the world wide web tomorrow. The results of the exam will be posted, by social security number, outside my door (Room 126) shortly after they are back from Academic Testing Services.

For every question, also consider, as a possible answer,

E)none of the above.

Possibly useful information(if you think of other equations that might be useful, come see me and I'll put them on the board for everyone):

_{i }+ v _{i
} t + (^{1}/_{2}) a
t^{2}
v = v v sin = opp/hyp cos = adj/hyp tan = opp/adj
w = mg g = 10 m/s F v = r a F F G = 6.67 x 10 M R |
P = W / t KE = ( W = KE PE F = - k x PE = r F sin
= 0
s = r
v = I I =
m =
=
2
= X Y KE = ( F W = q |

For every question, also consider, as a possible answer,

E)none of the above.

**1. How much work is done by gravity on a
skier who weighs 600 N as the skier moves 500 m along a hill that
makes an angle of 30° with the horizontal?**

**Be careful with the angle theta. Theta is
the angle ****between F and D****
so that means it is 60 ^{o} in this case!**

A) 50,000 J

B) 100,000 J

C) 150,000 J

D) 200,000 J

** **

**2. An electric bill lists 1284 kW-h as
the energy used for a month. ****How
many joules of energy is this?****
[ If the charge is 11.5 cents per kW-h, what is the total bill?
]**

E = 1284 kW-h [ 10^{3}W / kW ] [ 3 600 s / h ]

E = 4.62 x 10^{9}W-s [ J / W-s ]

E = 4.62 x 10^{9}J

**This is excerpted from homework problem
6.19 but the answer in the homework solution to 6.19 is incorrect.
The incorrect answer in the homework solution was 4.62 x 10
^{6} J. Either the correct answer of
**

A) 14 766 J

B) 1.48 x 10^{6}J

C) 4.62 x 10^{6}J

D) 9.24 x 10^{6}J

E) none of the above

** **

**3. The output of a common student
laboratory laser is about 250 milliwatts (250 mW). How much energy is
delivered in one hour?**

**This is homework problem 6.22 but the
solution there has an error. The
****in****correct answer in the
homework solution would be B). Either B) or
****E)****
was counted as correct.**

A) 0.250 W

B) 0.90 W

C) 1.20 W

D) 15.0 W

E) none of the above

** **

**4. What is the kinetic energy of a 1,750
kg car traveling at 80 km/h?**

A) 1.95 x 10^{4}J

B) 1.40 x 10^{5}J

C) 4.32 x 10^{5}J

D) 1.12 x 10^{7}J

** **

**5. A ski slope is inclined at an angle of
20° and is 600 m long. Find the change in potential energy of a
70-kg skier who goes down this slope. **

A) 0.7 x 10^{5}J

B) 1.4 x 10^{5}J

C) 3.5 x 10^{5}J

D) 4.2 x 10^{5}J

** **

**6. In loading a certain dart gun, the
dart is pushed in 12 cm before the gun is cocked. A force of
20 N is required at the end. What is the spring constant of the
spring in the gun?**

A) 16.7 N/m

B) 167 N/m

C) 1 670 N/m

D) 1.67 N/m

** **

**7. An arrow with a mass of 0.2 kg is shot
straight up into the air so it goes to a height of 40 m. It then
comes down and penetrates 25 cm vertically into the ground. What
average upward force does the ground exert on the arrow while
stopping it?**

**At the top of the arrows path it has
total energy equal to its gravitational potential
energy,**

**All this gravitational potential energy
is converted into kinetic energy just before the arrow strikes the
ground. Then it is converted into heat energy as the ground does work
on the arrow in bringing it to rest,**

**F = E / d**

**F = 78.4 J / 0.25 m**

**F = 313.6 N**

**Here is a similar problem from the
homework:**

A) 78 N

B) 87 N

C) 314 N

D) 413 N

** **

**8. At what speed must a 3,600 kg truck
travel to have the same momentum as a 1,200 kg that is traveling at
90 km/h?**

**p _{truck} = m _{truck} v
_{truck} = m _{car} v _{car} = p
_{car}**

**(3600 kg) v _{truck} = (1200 kg)
(90 km/h)**

**v _{truck} = (1200 kg) (90 km/h)
/ 3600 kg**

**v _{truck} = 30 km /
h**

**Here is a similar homework
problem:**

A) 30 km/h

B) 60 km/h

C) 90 km/h

D) 120 km/h

** **

**9. A 150 g baseball initially traveling
at 30 m/s is struck by a bat and leaves in the opposite direction at
35 m/s. What is the impulse delivered by the bat?**

Impulse = change in momentum

Impulse = p = p_{f}- p_{i}

p = (0.150 kg) (- 35 m/s) - (0.150 kg) (30 m/s)

p = (0.150 kg) (- 65 m/s)

p = - 9.75 kg m / s

A) 3.45 kg m/s

B) 5.97 kg m/s

C) 7.95 kg m/s

D) 9.75 kg m/s

** **

**10. A 600 kg cannon fires a 5 kg
cannonball with a horizontal muzzle velocity of 120 m/s. What is the
recoil velocity of the cannon?**

A) 12 m/s

B) 6 m/s

C) 2 m/s

D) 1 m/s

** **

**11. A 600 g glider sits at rest on an air
track when it is struck by a 250 g glider traveling 25 cm/s. The two
couple and stick together. What is their common, final
velocity?**

A) 3.35 cm/s

B) 5.35 cm/s

C) 7.35 cm/s

D) 9.80 cm/s

** **

**12. A 200-g glider on an air track moves
to the right at 50 cm/s. A 300-g glider on an air track moves to the
left at 90 cm/s. The two gliders collide
****in****elastically -- that is,
they collide and stick together. What is their common, final velocity
vf?**

**P _{f} = (M_{tot})
v_{f} = m_{1} v_{1} + m_{2}
v_{2} = P_{i}**

**M _{tot} = m_{1} +
m_{2} = 200 g + 300 g**

**M _{tot} = 500 g**

**( 500 g ) v _{f} = (200 g) ( 50
cm/s ) + (300 g) ( - 90 cm/s)**

**( 500 g ) v _{f} = 10 000 g cm/s -
27 000 g cm/s**

**( 500 g ) v _{f} = - 17 000 g
cm/s**

**v _{f} = ( - 17 000 g cm/s ) / 500
g**

**v _{f} = - 34
cm/s**

**this is to the ****left**

A) 14 cm/s to the left

B) 17 cm/s to the right

C) 28 cm/s to the right

D) 34 cm/s to the left

** **

**13. Exhaust gasses leave a certain rocket
with a mass flow rate of 1 250 kg/s. What must be the exhaust speed
of these gasses in order to lift a 50 000 kg rocket from its launch
pad? (use g = 10 m/s ^{2}) ?**

**F _{thrust} = M g**

**F _{thrust} = M g = (50 000 kg)
(10 m/s^{2}) **

**F _{thrust} = 500 000
N**

**F _{thrust} = [ m
/ t ]
u**

**[ m
/ t ]
u = 500 000 N**

**[ 1 250 kg / s ] u = 500 000
N**

**u = 500 000 N ] / [ 1 250 kg /
s]**

**u = 400 m/s**

A) 200 m/s

B) 300 m/s

C) 400 m/s

D) 500 m/s

** **

**14. A man who weighs 800 N climbs 4.0 m
up a 6.0 m ladder that leans against a smooth (i e,
friction****less****) wall at an
angle of 60 ^{o} with the horizontal as shown here. The ladder
is uniform so its center of mass is 3 m from the foot. What must be
the minimum coefficient of static friction between the ground and the
foot of the ladder if it is not to slip?**

**The first condition of equilibrium deals
with only the forces:**

**F _{wall} =
F_{f}**

**F _{net,y} = F_{N} - 800 N
- 400 N = 0**

**F _{N} = 1200
N**

**The secondition of equilibrium deals with
torques:**

**Calculate torques about the foot of the
ladder:**

F_{f}: = 0

F_{N}: = 0

400N:_{ccw}= (3.0 m) (400 N) sin 30^{o}= (3.0 m)(400N)(0.5) = 600 m-N

800N:_{ccw}= (4.0 m) (800 N) sin 30^{o}= (4.0 m)(800N)(0.5) = 1600 m-N

F_{wall}:_{cw}= (6.0 m) ( F_{wall}) sin 60^{o}= (6.0 m)( F_{wall})(0.866) = 5.2 m F_{wall}

_{cw}=_{ccw}

5.2 m F_{wall}= 600 m-N + 1600 m-N

5.2 m F_{wall}= 2200 m-N

F_{wall}= 2200 m-N / 5.2 m

F_{wall}= 423 N

**And we already know**

**F _{f} = 423 N**

**F _{f} = F_{N}**

**=
F _{f} / F_{N}**

**=
423 N / 1200 N**

**=
0.35**

A) 0.15

B) 0.35

C) 0.55

D) 0.75

** **

** **

**15. A solid disk, with moment
of inertia I = ( ^{1}/_{2}) M R^{2}, rolls
without slipping down a hill from a height of 1.0 m. The disk starts
at rest. What is its linear speed at the bottom of the
hill?**

**E _{f} =
E_{i}**

**KE _{f} = PE
_{i}**

**( ^{1}/_{2}) M
v^{2} + (^{1}/_{2}) I
^{2} = M g h_{i}**

**( ^{1}/_{2}) M
v^{2} + (^{1}/_{2}) [
(^{1}/_{2}) M R^{2} ] [ v / R ]
^{2} = M g h_{i}**

**( ^{1}/_{2}) v^{2}
+ (^{1}/_{2}) [ (^{1}/_{2})
R^{2} ] [ v / R ] ^{2} = g
h_{i}**

**( ^{3}/_{4}) v^{2}
= g h_{i}**

**v ^{2}
=(^{4}/_{3}) g h_{i}**

**v ^{2}
=(^{4}/_{3}) (9.8 m/s^{2}) ( 1.0
m)**

**v ^{2} = 13.07 m^{2}/s
^{2}**

**v = 3.62 m/s**

A) 1.81 m/s

B) 3.62 m/s

C) 7.52 m/s

D) 9.80 m/s| Return to PHY 1150's Home Page |(c) Doug Davis, 1998; all rights reserved