Download Version 074 – Midterm 2

Version 074 – Midterm 2 – OConnor (05141)
This print-out should have 18 questions.
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the next column or page – find all choices
before answering. V1:1, V2:1, V3:3, V4:5,
V5:5.
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1
10. 5.08593 s
Explanation:
The initial x and y components of the velocity are
vx0 = v0 cos θ0 = 12.3824 m/s
vy0 = v0 sin θ0 = 9.81414 m/s .
To find t, we can use the relation
1 2
gt
2
y = vy0 t −
001 (part 1 of 2) 10 points
A stone is thrown from the top of a building
upward at an angle of 38.4◦ to the horizontal
and with an initial speed of 15.8 m/s, as in the
figure. The height of the building is 53.2 m.
The acceleration of gravity is 9.8 m/s2 .
y
v0
θ0
x
with y = −53.2 m and vy0 = 9.81414 m/s
(we have chosen the top of the building as the
origin):
1
(9.81414 m/s) t − (9.8 m/s2 ) t2
2
= −53.2 m .
2 2 2
(4.9 m /s )t − (9.81414 m/s) t
−53.2 m = 0
Applying the quadratic formula, since
(−9.81414 m/s)2 − 4(4.9 m/s2 )(−53.2 m)
= 1139.04 m2 /s2 ,
h
then
How long is the stone “in flight”?
1. 3.17932 s
2. 3.56725 s
3. 3.63387 s
p
9.81414 m/s ± 1139.04 m2 /s2
t=
9.8 m/s2
= 4.44528 s
002 (part 2 of 2) 10 points
What is the speed of the stone just before it
strikes the ground?
1. 26.0056 m/s
4. 3.69763 s
2. 27.5092 m/s
5. 3.98715 s
3. 29.3169 m/s
6. 4.15309 s
4. 30.052 m/s
7. 4.3606 s
5. 33.1771 m/s
8. 4.44528 s correct
6. 34.7925 m/s
9. 4.78976 s
7. 35.9494 m/s correct
Version 074 – Midterm 2 – OConnor (05141)
8. 36.8744 m/s
4. 40.1959 N
9. 38.0433 m/s
5. 41.1196 N
10. 40.1408 m/s
Explanation:
The y component of the velocity just before
the stone strikes the ground can be obtained
using the equation
8. 52.8437 N
vy = vy0 − g t
9. 54.3434 N
with t = 4.44528 s :
2
6. 42.8758 N
7. 49.7272 N
10. 56.8695 N
Explanation:
vy = (9.81414 m/s) − (9.8 m/s2 ) (4.44528 s)
= −33.7496 m/s .
Since vx = vx0 = 12.3824 m/s, the required
speed is
q
v = vx2 + vy2
q
= (12.3824 m/s)2 + (−33.7496 m/s)2
= 35.9494 m/s .
003 (part 1 of 3) 10 points
A block of mass 1.91332 kg lies on a frictionless table, pulled by another mass 3.40141 kg
under the influence of Earth’s gravity.
The acceleration of gravity is 9.8 m/s2 .
Given :
m1 = 1.91332 kg ,
m2 = 3.40141 kg ,
µ = 0.
a
T
T
m1
N
and
m1 g
m2
a
m2 g
The net force on the system is simply the
weight of m2 .
Fnet = m2 g
= (3.40141 kg) (9.8 m/s2 )
= 33.3338 N .
1.91332 kg
004 (part 2 of 3) 10 points
What is the magnitude of the acceleration a
of the two masses?
1. 5.19695 m/s2
µ=0
3.40141 kg
What is the magnitude of the net external
force F acting on the two masses?
1. 29.6699 N
2. 5.63451 m/s2
3. 5.80805 m/s2
4. 6.27197 m/s2 correct
2. 31.0125 N
5. 6.36673 m/s2
3. 33.3338 N correct
6. 6.49314 m/s2
Version 074 – Midterm 2 – OConnor (05141)
7. 6.60275 m/s2
8. 6.82578 m/s2
9. 7.87789 m/s2
10. 8.06403 m/s2
Explanation:
From Newton’s second law,
006 (part 1 of 3) 10 points
A block is at rest on the incline shown in the
figure. The coefficients of static and kinetic
friction are µs = 0.7 and µk = 0.59, respectively.
The acceleration of gravity is 9.8 m/s2 .
19
Fnet = m2 g = (m1 + m2 ) a .
Solving for a,
m2
g
a=
m1 + m 2
3.40141 kg
=
(9.8 m/s2 )
1.91332 kg + 3.40141 kg
= 6.27197 m/s2 .
005 (part 3 of 3) 10 points
What is the magnitude of the tension T of the
rope between the two masses?
1. 9.62636 N
3
kg
µ
What is the frictional force acting on the
19 kg mass?
1. 51.5525 N
2. 60.5674 N
3. 63.8114 N
4. 78.6916 N
2. 9.75315 N
5. 89.3359 N
3. 10.0116 N
6. 91.7786 N
4. 10.8381 N
7. 95.9001 N correct
5. 12.0003 N correct
8. 164.617 N
6. 14.1684 N
9. 181.3 N
7. 14.5665 N
10. 261.536 N
Explanation:
8. 16.3735 N
9. 16.5868 N
10. 17.2402 N
Explanation:
Analyzing the horizontal forces on block
m1 , we have
X
Fx : T = m 1 a
= (1.91332 kg) (6.27197 m/s2 )
= 12.0003 N .
31◦
N
Ff
31◦
mg
Version 074 – Midterm 2 – OConnor (05141)
The forces acting on the block are shown
in the figure. Since the block is at rest, the
magnitude of the friction force should be equal
to the component of the weight on the plane
of the incline
Ff = M g sin θ
= (19 kg) (9.8 m/s2 ) sin 31◦
= 95.9001 N .
007 (part 2 of 3) 10 points
What is the largest angle which the incline
can have so that the mass does not slide down
the incline?
1. 19.7989 ◦
2. 20.8068
◦
3. 22.7824
◦
4. 24.2277
◦
5. 27.0216
◦
6. 30.1137
◦
7. 33.0239
◦
8. 34.992
9. 36.1294
4
the incline if the angle of the incline is 39◦ ?
1. 0.854362 m/s2
2. 1.12114 m/s2
3. 1.24553 m/s2
4. 1.47742 m/s2
5. 1.49405 m/s2
6. 1.59314 m/s2
7. 1.61954 m/s2
8. 1.67388 m/s2 correct
9. 1.8754 m/s2
10. 2.06274 m/s2
Explanation:
When θ exceeds the value found in part 2,
the block starts moving and the friction force
is the kinetic friction
Fk = µk N = µk M g cos θ.
Newton’s equation for the block then becomes
◦
correct
◦
10. 36.8699 ◦
Explanation:
The largest possible value the static friction
force can have is Ff,max = µs N , where the
normal force is N = M g cos θ. Thus, since
Ff = M g sin θ,
M g sin θm = µs M g cos θm
tan θm = µs
θm = tan−1 (µs )
= tan−1 (0.7)
= 34.992◦ .
008 (part 3 of 3) 10 points
What is the acceleration of the block down
M a = M g sin θ − Ff
= M g sin θ − µk M g cos θ
and
a = g [sin θ − µk cos θ]
= (9.8 m/s2 ) [sin 39◦ − (0.59) cos 39◦ ]
= 1.67388 m/s2 .
009 (part 1 of 1) 10 points
As viewed by a bystander, a rider in a “barrel
of fun” at a carnival finds herself stuck with
her back to the wall.
ω
Version 074 – Midterm 2 – OConnor (05141)
Which diagram correctly shows the forces
acting on her?
5
The acceleration of gravity is 980 cm/s2 .
What is the coefficient of static friction between the coin and the turntable?
1. 0.30012
2. 0.340136 correct
1.
3. 0.364431
4. 0.398288
2.
5. 0.445269
6. 0.507203
7. 0.522573
8. 0.565149
3.
9. 0.6387
4.
correct
5. None of the other choices
6.
Explanation:
The normal force of the wall on the rider
provides the centripetal acceleration necessary to keep her going around in a circle. The
downward force of gravity is equal and opposite to the upward frictional force on her.
Note: Since this problem states that it is
viewed by a bystander, we assume that the
free-body diagrams are in an inertial frame.
010 (part 1 of 1) 10 points
A coin is placed 30 cm from the center of a
horizontal turntable, initially at rest. The
turntable then begins to rotate. When the
speed of the coin is 100 cm/s (rotating at a
constant rate), the coin just begins to slip.
10. 0.663265
Explanation:
The normal force on the coin is N = m g .
The force provided by friction immediately
before slippage is given by
f = µN = µmg.
At this moment the centripetal acceleration
(there is no tangential acceleration) provides
a force
m v2
F =
.
r
Summation of forces yields
m v2
, therefore
r
v2
µ=
rg
(100 cm/s)2
=
(30 cm) (980 cm/s2 )
= 0.340136 .
F = f = µmg =
011 (part 1 of 2) 10 points
Consider the “loop-a-loop” setup, where a
mass m is sliding along a frictionless track
and the radius of the loop is R.
Version 074 – Midterm 2 – OConnor (05141)
C
m
R
A
D
Given h = 5 R, determine the speed of the
mass at A.
p
1. |vA | = 2 2 g R correct
p
2. |vA | = 2 g R
p
3. |vA | = 4 g R
p
4. |vA | = 3 g R
p
5. |vA | = 3 2 g R
p
6. |vA | = 4 2 g R
Explanation:
Applying the work energy theorem from C
to A we get
mgh − mgR =
1
2
m vA
2
mg(5 R − R) = 4 R m g =
m v2
= mg.
R
Conservation of energy between C and B implies
m g (h − 2 R) =
012 (part 2 of 2) 10 points
Find the critical initial height of the mass,
such that it would just barely pass the point
B.
1
1
m v2 = m g R
2
2
m v2
R
5
R
h = 2R + = R
2
2
mg =
013 (part 1 of 1) 10 points
A(n) 88.3 g ball is dropped from a height of
66.1 cm above a spring of negligible mass.
The ball compresses the spring to a maximum
displacement of 4.6768 cm.
The acceleration of gravity is 9.8 m/s2 .
h
1
2
m vA
2
2
vA
= 8gR
p
vA = 2 2 g R
1
1. h = R
2
2
2. h = R
3
5
3. h = R correct
2
3
R
2
1
5. h = R
3
Explanation:
The critical condition at B implies that
4. h =
B
h
6
x
Calculate the spring force constant k.
1. 324.238 N/m
2. 364.509 N/m
3. 511.447 N/m
4. 537.872 N/m
5. 560.028 N/m correct
6. 621.622 N/m
7. 702.946 N/m
Version 074 – Midterm 2 – OConnor (05141)
8. 758.352 N/m
9. 851.065 N/m
7
9. 1.48144 m/s
10. 1.5 m/s
Explanation:
10. 1028.17 N/m
Explanation:
Let :
Let : m = 88.3 g ,
h = 66.1 cm , and
x = 4.6768 cm .
Using conservation of energy, we have
m = 18.8 kg ,
M = 2170 kg , and
v = 140 m/s .
The cannon’s velocity immediately after it
was fired is found by using conservation of
momentum along the horizontal direction:
1 2
k x = m g (h + x)
2
M V + mv = 0
A revolutionary war cannon, with a mass of
2170 kg, fires a 18.8 kg ball horizontally. The
cannonball has a speed of 140 m/s after it has
left the barrel. The cannon carriage is on a
flat platform and is free to roll horizontally.
What is the speed of the cannon immediately after it was fired?
1. 0.797431 m/s
= 1.2129 m/s .
m
v
M
from which
where M is the mass of the cannon, V is the
2 m g (h + x)
velocity of the cannon, m is the mass of the
k=
x2
cannon ball and v is the velocity of the cannon
2 (0.0883 kg)(9.8 m/s2 )(0.661 m + 0.046768 m) ball. Thus, the cannon’s speed is
=
(0.046768 m)2
m
|v|
|V | =
= 560.028 N/m .
M
18.8 kg
=
(140 m/s)
2170 kg
014 (part 1 of 2) 10 points
2. 0.872308 m/s
3. 1.12371 m/s
⇒ −V =
015 (part 2 of 2) 10 points
The same explosive charge is used, so the total
energy of the cannon plus cannonball system
remains the same.
Disregarding friction, how much faster
would the ball travel if the cannon were
mounted rigidly and all other parameters remained the same?
1. 0.397969 m/s
4. 1.14343 m/s
2. 0.435308 m/s
5. 1.16174 m/s
3. 0.560479 m/s
6. 1.2129 m/s correct
4. 0.570196 m/s
7. 1.37624 m/s
5. 0.579618 m/s
8. 1.42581 m/s
6. 0.605144 m/s correct
Version 074 – Midterm 2 – OConnor (05141)
8
r
m →
v
.
M man
³m´ →
=−
v man . correct
M
7. 0.686424 m/s
3. VEarth = −
8. 0.71115 m/s
4. VEarth
9. 0.738726 m/s
5. VEarth = + v man .
µ ¶
M →
v man .
6. VEarth = +
m
→
10. 0.747671 m/s
Explanation:
By knowing the speeds of the cannon and
the cannon ball, we can find out the total
kinetic energy available to the system
Knet =
1
1
m v2 + M V 2 .
2
2
This is the same amount of energy available
as when the cannon is fixed. Let v 0 be the
speed of the cannon ball when the cannon is
held fixed. Then,
1
1
m v 02 = (m v 2 + M V 2 ) .
2
2
r
M 2
V
v2 +
m
r
m
=v 1+
M
s
⇒ v0 =
= (140 m/s)
1+
18.8 kg
2170 kg
= 140.605 m/s .
Thus, the velocity difference is
v 0 − v = 140.605 m/s − 140 m/s
→
7. VEarth = − v man .
µ ¶
M →
8. VEarth = −
v man .
m
Explanation:
The momentum is conserved. We have
→
So
→
V Earth = −
³m´
M
→
v man .
017 (part 1 of 1) 10 points
A child bounces a 55 g superball on the sidewalk. The velocity change of the superball is
from 23 m/s downward to 17 m/s upward.
If the contact time with the sidewalk is
1/800 s, what is the magnitude of the force
exerted on the superball by the sidewalk?
1. 1414.4 N
2. 1452 N
3. 1480 N
4. 1545.6 N
= 0.605144 m/s .
5. 1579.2 N
016 (part 1 of 1) 10 points
Bill (mass m) plants both feet solidly on the
ground and then jumps straight up with ve→
locity v .
The earth (mass M ) then has velocity
³m´ →
v man .
1. VEarth = +
M
r
m →
2. VEarth = +
v
.
M man
→
m v man + M V Earth = 0
6. 1640 N
7. 1680 N
8. 1760 N correct
9. 1788.8 N
10. 1856 N
Explanation:
Version 074 – Midterm 2 – OConnor (05141)
Let :
m = 55 g = 0.055 kg ,
vu = 17 m/s ,
vd = 23 m/s , and
∆t = 0.00125 s
Choose the upward direction as positive.
Then the impulse is
I = F ∆t = ∆P
= m vu − m (−vd )
= m (vu + vd )
m (vu + vd )
F =
∆t
(55 g) (17 m/s + 23 m/s)
=
0.00125 s
= 1760 N .
018 (part 1 of 1) 10 points
As shown in the top view above, a disc of mass
m is moving horizontally to the right with
speed v on a table with negligible friction
when it collides with a second disc of mass
9 m. The second disc is moving horizontally
v
at the moment of
to the right with speed
8
impact. The two discs stick together upon
impact.
v
8
v
before
m
9m
vf
after
10 m
The speed of the composite body immediately after the collision is
1. vf =
2. vf =
3. vf =
4. vf =
5. vf =
6. vf =
7. vf =
8. vf =
9
11
v.
20
5
v.
9
17
v. correct
80
3
v.
10
1
v.
4
11
v.
35
7
v.
15
7
v.
27
9. None of these are correct.
1
v.
2
Explanation:
The total momentum of the system is conserved because there is no exterior force. So
what we have is
v
(m + 9 m) vf = m v + 9 m
8¶
µ
9
10 m vf = m v 1 +
8
µ
¶
8 9
+
10 vf =
v
8 8
17
v
10 vf =
8
(17)
vf =
v
(8) (10)
17
=
v , so
80
17
v .
=
80
10. vf =