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Physics 201 Midterm Exam 2
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The exam has 20 questions (questions 2-21) and you have 1 hour and 15 mintues to complete the exam.
Please use 9.8 m/s^2 for the acceleration of gravity on the Earth.
.
____
1. Please fill in your test ID code
a. A
b. B
c. C
d. D
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
2. An airplane travels 80.0 m/s as it makes a horizontal circular turn which has a 0.80-km radius. What is the
magnitude and direction of the resultant force on the 75-kg pilot of this airplane?
a. 0.60 kN forward in the direction the plane’s velocity at any moment
b. 600 kN forward in the direction the plane’s velocity at any moment
c. 0.74 kN upward
d. 0.60 kN inward toward the center of the circle
e. 0.95 kN upward and inward toward the center of the circle
à d 0.60 kN inward toward the center of the circle
Since the pilot is also moving in a circle they must also experience a centripetal acceleration caused by a force pointed
inward toward the center of the circle. In this case either a frictional force or normal force from the pilots
seat.
𝐹" = 𝑚
𝑣&
80.0&
= 75 ∗
= 600𝑁 = 0.60𝑘𝑁
𝑟
800
Note that the pilot also feels a force of gravity down and a normal force upward which cancel out.
____
3. A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of
the curve is half the radius for the outside. One car, car A, travels on the inside while another car of equal
mass, car B, travels at equal speed on the outside of the curve. Which statement about resultant forces on the
cars is correct?
a. The force on A is half the force on B.
b. The force on B is half the force on A.
c. The force on A is four times the force on B.
d. The force on B is four times the force on A.
e. There is no net resultant force on either as long as they stay on the road while turning.
à b The force on B is half the force on A.
Using:
𝑣&
𝐹" = 𝑚
𝑟
At twice the radius the force is half as much
____
4. The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is
0.800. The car is on a level circular track of 50.0 m radius on a planet where g = 2.45 m/s2 compared to
Earth's g = 9.80 m/s2. If the car is to be able to travel at the same maximum speed without slipping on the
planet as on Earth, the radius of the track on the planet must be ____ times as large as the radius of the track
on Earth.
a. 0.250
b. 0.500
c. 1.00
d. 2.00
e. 4.00
à e 4.00
The centripetal force is being provided by static friction
𝑣&
𝐹" = 𝑚
= 𝐹2 = 𝜇4 𝑚𝑔
𝑟
Where the g is four times smaller on the other planet. To have them still equal the radius has be four times bigger
____
5. A skydiver of 75 kg mass has a terminal velocity of 60.0 m/s. At what speed is the resistive force on the
skydiver half that when at terminal speed?
a. 15 m/s
b. 49 m/s
c. 30 m/s
d. 42 m/s
e. 36 m/s
à d 42 m/s
For a falling object in air:
1
𝐹6 = 𝑚𝑎 = 𝑚𝑔 − 𝐷𝜌𝐴𝑣 &
2
For the resistive force to be half as much the velocity must be, compared to the terminal velocity, 𝑣 =
____
>?
&
=
@A
&
= 42
6. If a dense 20.0-kg object is falling in air at half its terminal velocity, what is the drag force on the object at
this moment?
a. 24.5 N
b. 49.0 N
c. 69.3 N
d. 98.0 N
e. 139 N
à b 49.0 N
At terminal veloclty
1
𝐹6 = 𝑚𝑎 = 0 = 𝑚𝑔 − 𝐷𝜌𝐴𝑣 &
2
Thus
𝑚𝑔 = 20 ∗ 9.80 = 196 =
At half terminal velocity
1
𝑣D
𝐷𝜌𝐴
2
2
&
=
196
= 49
4
1
𝐷𝜌𝐴𝑣 &
2
____
7. A 2.0-kg projectile moves from its initial position to a point that is displaced 20 m horizontally and 15 m
above its initial position. How much work is done by the gravitational force on the projectile?
a. +0.29 kJ
b. -0.29 kJ
c. +30 J
d. -30 J
e. -50 J
à b -0.29 kJ
𝑊 = 𝐹 ∙ ∆𝑥 = 𝐹∆𝑥𝑐𝑜𝑠𝜃 = mg*h*cos(180) = 2.0*9.8*15 = -294 J = -0.29 KJ
____
8. A block is pushed across a rough horizontal surface from point A to point B by a force (magnitude P = 5.4 N)
as shown in the figure. The magnitude of the force of friction acting on the block between A and B is 1.2 N
and points A and B are 0.50 m apart. If the kinetic energies of the block at A and B are 4.0 J and 5.6 J,
respectively, how much work is done on the block by the force P between A and B?
a.
b.
c.
d.
e.
2.8 J
1.0 J
2.2 J
1.6 J
3.2 J
à c 2.2 J
Using the work energy theorem
𝑊 = Δ𝐾 = 𝐾2 − 𝐾O
𝑊P + 𝑊2 = Δ𝐾 = 𝐾2 − 𝐾O
𝑊P − 1.2 ∗ 0.5 = Δ𝐾 = 5.6 − 4.0
𝑊P = 5.6 − 4.0 + 0.6 = 2.2
____
9. A 10.0-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m).
The block is initially at rest at its equilibrium position when a force (magnitude P = 80.0 N) acting parallel to
the surface is applied to the block, as shown. What is the speed of the block when it is 13 cm from its
equilibrium position?
a.
b.
c.
d.
e.
0.85 m/s
0.89 m/s
0.77 m/s
0.64 m/s
0.52 m/s
à a 0.85 m/s
Using the work energy theorem
𝑊 = Δ𝐾 = 𝐾2 − 𝐾O =
1
1
𝑚𝑣2& − 𝑚𝑣O&
2
2
1
1
𝑚𝑣2& − 𝑚𝑣O&
2
2
However the work of the spring is conservative so we can replace it with the negative of the change in potential energy.
1
1
𝑊R − 𝑈2 − 𝑈O = 𝑚𝑣2& − 𝑚𝑣O&
2
2
1 & 1 &
1
1
𝐹𝑑 − 𝑘𝑥2 − 𝑘𝑥O = 𝑚𝑣2& − 𝑚𝑣O&
2
2
2
2
1 &
1
&
𝐹𝑑 − 𝑘𝑥2 = 𝑚𝑣2
2
2
1
1
80.0 ∗ 0.13 − 800 ∗ 0.13& = 10.0𝑣2&
2
2
v = 0.85 m/s
𝑊R + 𝑊4 =
____ 10. When a crate of mass m is dragged a distance d along a surface with coefficient of kinetic friction µk, then
dragged back along the same path to its original position, the work done by friction is
a. 0.
b. -µkmgd.
c. +µkmgd.
d. -2µkmgd.
e. +2µkmgd.
à d -2µkmgd
Negative because the direction of movement and the frictional force are opposite. Twice rather than zero because the
work is negative in both directions.
____ 11. Two clowns are launched horizontally from the same spring-loaded circus cannon with the spring compressed
the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their
speeds at the instant of launch is
a. vA = (3/2) vB.
b. vA = sqrt (3/2) vB.
c. vA = vB.
d. vB = sqrt (3/2) vA.
e. vB = (3/2) vA.
à b vA = sqrt (3/2) vB
The same amount of work is done by the spring in each case. Using the work energy theorem
𝑊 = Δ𝐾 = 𝐾2 − 𝐾O =
𝑊=
1
𝑚𝑣O&
2
𝑣=
2𝑊
𝑚
1
1
𝑚𝑣2& − 𝑚𝑣O&
2
2
For clown A compared to B
𝑣=
2𝑊
=
2
𝑚
3
3 2𝑊
2 𝑚
____ 12. A spring (k = 600.0 N/m) is placed in a vertical position with its lower end supported by a horizontal surface.
The upper end is depressed 20.0 cm, and a 4.0-kg block is placed on top of the depressed spring, but not
attached to the spring. The system is then released from rest. How far above the point of release will the block
rise?
a. 27 cm
b. 36 cm
c. 41 cm
d. 31 cm
e. 20 cm
à d 31 cm
Using conservation of energy and converting spring energy to gravitational potential energy.
The potential energy the spring has compared at the starting point is.
𝑈=
1 & 1
𝑘𝑥 = 600 ∗ 0.20& = 12𝐽
2 O
2
The gravitational potential energy to raise to the springs zero point is
𝑈 = 𝑚𝑔ℎ = 4.0 ∗ 9.8 ∗ 0.2 = 7.84𝐽
There is still energy to go higher! Equating the extra energy to the gravitational and spring energy to go higher.
12 − 7.84 = 4.16 = 𝑚𝑔𝑥 = 39.2𝑥
x = 0.0692 m = 10.5 cm, total adding the 20 cm of the original displacement down = 31 cm
____ 13. A champion athlete can produce one horsepower (746 W) for a short period of time. If a 70.0-kg athlete were
to bicycle to the summit of a 0.50-km high mountain while expending power at this rate, she would reach the
summit in ____ seconds.
a. 1
b. 460
c. 500
d. 1 000
e. 35 000
à b 460 s
P = W/t
t = W/P = mgh/P = (70*9.8*500)/746= 460 s
____ 14. As an object moves from point A to point B only two forces act on it: one force is nonconservative and does
-30 J of work, the other force is conservative and does +50 J of work. Between A and B,
a. the kinetic energy of object increases, mechanical energy decreases.
b. the kinetic energy of object decreases, mechanical energy decreases.
c. the kinetic energy of object decreases, mechanical energy increases.
d. the kinetic energy of object increases, mechanical energy increases.
e. None of the above.
à a the kinetic energy of object increases, mechanical energy decreases.
Kinetic energy increase because the overall work is positive. Mechanical energy decreases because the nonconservative
force removes energy
____ 15. A 3.0-kg object moving in the positive x direction has a one-dimensional elastic collision with a 5.0-kg object
initially at rest. After the collision the 5.0-kg object has a velocity of 6.0 m/s in the positive x direction. What
was the initial speed of the 3.0 kg object?
a. 6.0 m/s
b. 7.0 m/s
c. 4.5 m/s
d. 8.0 m/s
e. 5.5 m/s
à d 8.0 m/s
Using conservation of momentum and kinetic energy
𝑣X2 =
𝑚X − 𝑚&
2𝑚&
𝑣XO +
𝑣
𝑚X + 𝑚&
𝑚X + 𝑚& &O
𝑣&2 =
2𝑚X
𝑚& − 𝑚X
𝑣XO +
𝑣
𝑚X + 𝑚&
𝑚X + 𝑚& &O
𝑣X2 =
3.0 − 5.0
2𝑚&
𝑣XO +
0 = −0.25𝑣XO
3.0 + 5.0
𝑚X + 𝑚&
6.0 =
2 ∗ 3.0
𝑚& − 𝑚X
𝑣XO +
0 = 0.75𝑣XO
3.0 + 5.0
𝑚X + 𝑚&
then: 𝑣XO = 8.0
____ 16. A ball of mass mB is released from rest and acquires velocity of magnitude vB before hitting the ground. The
ratio of the kinetic energy the Earth acquires to the kinetic energy the ball acquires is
a. 0.
b. (mB/mE)2
c. mB/mE
d. 1
e. mE/mB
à c mB/mE
Using conservation of momentum the Earth must acquire equal magnitude and opposite direction momentum.
comparison the magnitudes
mEvE=mBvB
vE=(mB/mE)vB
and the kinetic energy
K = ½mE(mB/mE)2vB2
= ½(mE/mB)(mB/mE)2 mB vB2
= (mB/mE) ½ mB vB2
____ 17. A 3.0-kg mass moving in the positive x direction with a speed of 10 m/s collides with a 6.0-kg mass initially
at rest. After the collision, the speed of the 3.0-kg mass is 8.0 m/s, and its velocity vector makes an angle of
35° with the positive x axis. What is the magnitude of the velocity of the 6.0-kg mass after the collision?
a. 2.2 m/s
b. 2.9 m/s
c. 4.2 m/s
d. 3.5 m/s
e. 4.7 m/s
à b 2.9 m/s
Using conservation of momentum
x: 3.0*10 + 6.0*0 = 3.0*8.0*cos(35) + 6.0*vx
y: 0 = 3.0*8.0*sin(35) + 6.0*vy
vx = 1.7234
vy = -2.2943
|v| = 2.8695 = 2.9 m/s
____ 18. A 4.2-kg object, initially at rest, "explodes" into three objects of equal mass. Two of these are determined to
have velocities of equal magnitudes (5.0 m/s) with directions that differ by 90.0°. How much kinetic energy
was released in the explosion?
a. 70 J
b. 53 J
c. 60 J
d. 64 J
e. 35 J
à a 70 J
Using conservation of momentum to find the momentum of the third object. Assume the object go in the x direction but
45 degrees up and down from x axis. Then the y components of the momentum cancel out
p = 1.4*5.0*cos(45) + 1.4*5.0*cos(45) = 9.8995
Using K = p2/(2m) = ½ mv2
Total K = 9.8995^2/(2*1.4) + 2*0.5*1.4*5^2 = 70
____ 19. Three particles are placed in the xy plane. A 40-g particle is located at (3, 4) m, and a 50-g particle is
positioned at (-2, -6) m. Where must a 20-g particle be placed so that the center of mass of this three-particle
system is located at the origin?
a. (-1, -3) m
b. (-1, 2) m
c. (-1, 12) m
d. (-1, 7) m
e. (-1, 3) m
à d (-1, 7) m
consider the two dimensions separately
x: (40*3 + 50*(-2) + 20*x)/110 = 0, x = -1
y: (40*4 + 50*(-6) + 20*y)/110 = 0, y = -7
____ 20. Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the
collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v
before the collision. After the collision
a. each car has half the momentum.
b. car A stops and car B has momentum mv.
c. car A stops and car B has momentum 2mv.
d. the momentum of car B is four times as great in magnitude as that of car A.
e. each car has half of the kinetic energy.
à d the momentum of car B is four times as great in magnitude as that of car A.
Using conservation of momentum and kinetic energy
𝑣X2 =
𝑚X − 𝑚&
2𝑚&
𝑣XO +
𝑣
𝑚X + 𝑚&
𝑚X + 𝑚& &O
2𝑚X
𝑚& − 𝑚X
𝑣XO +
𝑣
𝑚X + 𝑚&
𝑚X + 𝑚& &O
𝑚 − 2𝑚
2𝑚&
1
=
𝑣XO +
0 = − 𝑣XO
𝑚 + 2𝑚
𝑚X + 𝑚&
3
𝑣&2 =
𝑣X2
𝑣&2 =
2𝑚
𝑚& − 𝑚X
2
𝑣XO +
0 = 𝑣XO
𝑚 + 2𝑚
𝑚X + 𝑚&
3
Since the second car (B) is twice and heavy it has 4 times the momentum
____ 21. Consider a physical situation like that in lab M-5, Projectile Motion. If the projectile has a mass of 0.010 kg
is moving at 90.0 m/s, and is caught by a catcher of mass 0.50 kg, how far does the catcher rise?
a. 0.16
b. 0.51
c. 0.32
d. 0.64
e. 0.080
à a 0.16 m/s
Using conservation of momentum
mvi = (m+M)vf
0.010*90.0 = (0.010+0.50)vf
vf = 1.7647
Using conservation of Energy
½ (M+m)vf2 = (M+m)gh
h = ½ vf2/g = 0.5*((0.010*90.0/(0.010+0.50))^2)/9.80 = 0.15889 = 0.16m/s
Physics 201 Midterm Exam 2
Answer Section
MULTIPLE RESPONSE
1. ANS: A, B, C, D
PTS: 1
MULTIPLE CHOICE
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D
B
E
D
B
B
C
A
D
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D
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A
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C
B
A
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D
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Average
Easy
Challenging
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Challenging
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Challenging