Download PHY 231 Midterm Exam II Form 1 Name

PHY 231
Midterm Exam II
Name: ________________________
Form 1
PID: ___________________
1. A billiard ball collides in an elastic head-on collision with a second stationary identical ball.
After the collision which of the following conditions applies to the first ball?
a.
b.
c.
d.
e.
maintains the same velocity as before
has one half its initial velocity
comes to rest
moves in the opposite direction
none of the above
Cf. extensive discussion of this
problem in my lectures – the only
way to satisfy both energy and
momentum conservation in this
particular case is to have the two balls
exchange roles: the 1st one comes to
a stop, and the 2nd one assumes the
initial velocity of the 1st.
(c)
2. If the momentum of an object is tripled, its kinetic energy will change by what factor?
a.
b.
c.
d.
e.
one third
one half
two
three
nine
momentum p = mv
K.E. = mv2/2
tripling p Þ tripling v Þ K.E. goes
up by nine-fold
(e)
3. During a football game, a 90-kg halfback running north with a speed of 10 m/s is tackled by a
120-kg defensive lineman running south at 4 m/s. The collision is inelastic. Compute the velocity
of the two players just after the tackle.
a.
b.
c.
d.
e.
3 m/s south
2 m/s south
0 (they come to a standstill)
2 m/s north
3 m/s north
Momentum is conserved in the collision.
initial momentum = m1v1 +m2v2
final momentum = (m1+m2) Vf
equating the two, one can solve for Vf.
Cf. Example 6.5 in the book.
(d)
PHY 231
Midterm Exam II
Form 1
4. A Ferris wheel starts at rest and builds up to a final angular velocity of 0.7 rad/s while rotating
through an angular displacement of 4.9 rad. What is its average angular acceleration?
a.
b.
c.
d.
e.
0.10 rad/s2
0.05 rad/s2
1.80 rad/s2
0.60 rad/s2
0.07 rad/s2
Equation 7.7 of the book applies to this
situation
Therefore: α = ω2 / 2 θ
(initial angular speed is zero)
with ω = 0.7 rad/s and θ = 4.9 rad, one
can plug in and get the answer.
(b)
5. A 1500 kg car rounds an unbanked curve with a radius of 52 m at a speed of 12 m/s. What
minimum coefficient of friction must exist between the road and tires to prevent the car from
slipping? (g = 9.8 m/s2)
a.
b.
c.
d.
e.
0.18
0.30
0.28
0.37
none of the above
This problem was worked out in class. It is
also given as Example 7.6 in the book.
centripetal force needed = mv2 / R.
This must be supplied by the acceleration due
to frictional force = µs N = µs mg
Equating the two, one gets:
(c)
µs = v2 / Rg
6. Consider a child playing on a swing. As she reaches the lowest point in her swing:
a.
b.
c.
d.
e.
the tension in the rope is equal to her weight.
the tension in the rope is equal to her mass times her acceleration.
her acceleration is downward and equal to g (9.8 m/s2).
her acceleration is equal to g plus v2/L where L is the length of the swing.
none of the above.
Sol. : (e)
Explanation:
acceleration at this point = v2 / L
tension at this point = mg + m v2 / L
, or
tot. force = tension – weight = mass x acceleration
7. An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius
(4000 miles). At this location, the acceleration due to gravity is what factor times the value of g
at the Earth’s surface?
a.
b.
c.
d.
e.
There is no acceleration since the satellite is in orbit.
2
1
At this point distance to the center of the earth r = 2RE
1/2
Since gravity acceleration is proportional to 1/r2 , (e) is the
1/4
right answer.
PHY 231
Midterm Exam II
Form 1
8. A ventilation fan with a moment of inertia of 0.034 kg-m2 has a net torque of 0.11 N-m applied
to it. If it starts from rest, what angular momentum will it have 8.0 s later?
a.
b.
c.
d.
e.
0.75 kg-m2/s
0.97 kg-m2/s
2.0 kg-m2/s
3.25 kg-m2/s
none of the above
torque τ = I α Þ α =τ / I
angular velocity ω = α t
angular momentum L = I ω
Therefore this problem involves a series of
plugging in numbers.
(e)
9. A figure skater on ice with arms extended, spins at a rate of 2.5 rev/s. After he draws his arms in,
he spins at 6 rev/s. By what factor does the skater’s kinetic energy change when he draws his
arms in ?
a.
b.
c.
d.
e.
2.4
1.0
0.80
0.42
0.12
(i) angular momentum conservation Þ I1ω1=
I2ω2 Þ I1 / I2 = ω2 /ω1
(ii) kinetic energy change by the factor
KE2 / KE1 = (I2ω2 2) / (I1ω12) = (I2 / I1) (ω2 /ω1)2
(a)
= ω2 /ω1 = 6/2.5
10. A cylinder (I = MR2/2) is rolling along the ground at 7 m/s. It comes to a hill and starts going up.
Assuming no losses to friction, how high does it get before it stops?
a.
b.
c.
d.
e.
3.75 m
0.30 m
2.50 m
0.50 m
none of the above
v
h
Use conservation of energy:
The reasoning is identical to that of
Example 8.12 (except that one reverses the
initial and final states).
Equating the initial K.E. due to linear and
rotational motion to the final state
potential energy, one gets ¾ mv2 = mgh,
which allows one to solve for h.
(a)
11. A 40 kg boy is standing on the edge of a stationary 30 kg platform that is free to rotate. The boy
tries to walk around the platform in a counterclockwise direction. As he does:
a. the platform doesn’t rotate.
b. the platform rotates in a clockwise direction just fast enough so that the boy remains
stationary relative to the ground.
c. the platform rotates in a clockwise direction while the boy goes around in a
counterclockwise direction relative to the ground.
d. both move counterclockwise with equal angular speed.
e. both go around with equal angular speed but in opposite directions.
Total angular momentum of (boy + platform) must be conserved. It is initially zero.
Only (c) is true. Can you find the flaws with all the other choices?
PHY 231
Midterm Exam II
Form 1
12. By what factor is the total pressure greater at a depth of 850 m of sea water than at the surface
where pressure is one atmosphere? (water density = 1000 kg/m3, 1 atmosphere pressure = 1.01 ×
105 N/m2 and g = 9.8 m/s2)
a.
b.
c.
d.
e.
100
83
74
19
none of the above
At the surface, P0 = 1.01 x 105 Pa.
At h = 850 m, Ph = P0 + ρgh.
The factor is Ph / P0 = 1.0 + ρgh / P0 .
Just plug in the numbers, you get the answer.
(b)
13. If the column of mercury in a barometer stands at 72.6 cm just before a storm, what is the
atmospheric pressure at that time? (The density of mercury is 13.6 × 103 kg/m3 and g = 9.8 m/s2)
a.
b.
c.
d.
e.
0.967 × 105 N/m2
1.03 × 105 N/m2
0.925 × 105 N/m2
1.073 × 105 N/m2
1.203 × 105 N/m2
P = ρgh, with ρ = 13.6 × 103 kg/m3, g = 9.8
m/s2, and h = 0.726 m .
(a)
14. A solid rock, suspended in air by a spring scale, has a measured mass of 9.0 kg. When the rock is
submerged in water, the scale reads 3.3 kg. What is the density of the rock? (water density =
1000 kg/m3)
a.
b.
c.
d.
e.
4.55 × 103 kg/m3
3.50 × 103 kg/m3
1.20 × 103 kg/m3
1.58 × 103 kg/m3
none of the above
See Example 9.6 in the book.
More elegantly, I showed in class that one
can arrive at the simple relation:
T2 / T1 = 1 – ρf / ρs
Where T1,2 are the tension or scale readings
in/out of the fluid; ρf is the density of the
fluid and ρs is the density of the solid block.
Therefore, you can directly solve for ρs,
knowing T2 / T1 = 3.3/9, and ρf =103 kg/m3.
(d)