PHYS 1443-003, Fall 2002, 2nd Term Exam, Wednesday, Oct. 30, 2002
Name:
ID:
Problems 1 – 26 are 3 points each. The points of problems 27 and 28 vary.
1. What force is needed to make an object move in a circle?
a) kinetic friction
b) static friction
c) centripetal force
d) weight
2. How many revolutions per minute must a circular, rotating space station of radius 1000 m rotate to
produce an artificial gravity of 9.80 m/s2?
a) 0.65 rpm
b) 0.75 rpm
c) 0.85 rpm
d) 0.95 rpm
Solution: In order for the station to produce an artificial gravity, its centripetal acceleration must be the
2
same as the gravitational acceleration. Since the radial acceleration is ar = g = v . Since speed is v = rϖ ,
r
we obtain rϖ
R=
2
= g . Therefore ϖ = g = 9.8 = 9.9 × 10 −2 rad / sec . Thus, it needs to rotate at a rate of
r
1000
ϖ ⋅ 60 60 × 9.9 × 10
=
2π
2π
−2
= 0.95rpm .
3. A car traveling 20 m/s rounds an 80-m radius horizontal curve with the tires on the verge of slipping.
How fast can this car round a second curve of radius 320 m? (Assume the same coefficient of
friction between the car's tires and each road surface.)
a) 20 m/s
b) 40 m/s
c) 80 m/s
d) 160 m/s
2
2
r1
r2
Solution: The centripetal force is the same for both the case, therefore we can obtain mar = m v1 = m v2 .
From this equation, the speed for the second curve is v2 = r2 v1 = 2 ⋅ 20m / s = 40 m / s .
r1
4. A motorcycle has a mass of 250 kg. It goes around a 13.7 m radius turn at 96.5 km/h. What is the
centripetal force on the motorcycle?
d) 4.31x104 N
2
Solution: Radial acceleration of the motorcycle in this situation is ar = v = (26 .8m / s) = 52 .4m / s2 . Thus
r
13.7 m
2
v
the centripetal force on the motorcycle is F = mar = = 250kg × 52.4m / s2 = 1.31 ×10 4 N .
r
a) 719 N
b) 2.95x103 N
c) 1.31x104 N
2
5. A stone, of mass m, is attached to a strong string and whirled in a vertical circle of radiusr. At the exact
top of the path the tension in the string is 3 times the stone's weight. The stone's speed at this point is
given by
a)
2
gr
b)
c)
2 gr
gr
d) gr
Solution: Since the total force that provides centripetal force is tension and the weight, we obtain
v2
mar = m = mg + 3mg = 4mg . Therefore the stone’s speed at this point is v = 4 gr = 2 gr .
r
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PHYS 1443-003, Fall 2002, 2nd Term Exam, Wednesday, Oct. 30, 2002
6. An object moves in a circular path at a constant speed. Consider the direction of the object's velocity
and acceleration vectors.
a) Both vectors point in the same direction.
b) The vectors point in opposite directions.
c) The vectors are perpendicular.
d) The question is meaningless, since the acceleration is zero.
Solution: Since the tangential velocity is always perpendicular to the centripetal acceleration, the two
vectors perpendicular to each other.
7. Two objects attract each other gravitationally. If the distance between their centers doubles, the
gravitational force
a) quadruples.
b) doubles.
c) is cut in half.
d) is cut to a fourth.
Solution: Since the gravitational force is proportional to the inverse square of the distance between the two
centers, if the distance between the two increases by a factor of two, the force cuts to a fourth of the orginal.
8. Which of the following is the correct unit of work expressed in SI units?
a) kg m/s 2
b) kg m2/s
c) kg m2/s 2
d) kg 2 m/s 2
Solution: Work is defined as W = F ⋅ d and is the same as energy, the unit of energy is kg m2/s 2
9. A 4.00-kg box of fruit slides 8.0 m down a ramp, inclined at 30.0o from the horizontal. If the box slides
at a constant velocity of 5.00 m/s, what is the work done by the weight of the box?
a) 157 J
b) -157 J+78.4 J
c) 78.4 J
d) -78.4 J
Solution: Potential energy of the box at the top of the hill is
PE = mgh = mgl sinθ = 4.00 × 9.80 × 8.0 × sin 30o = 157 J . This is the work done by the weight of the box
of which 50J was spent to change the box’s kinetic energy, and the remainder was spent to
overcome friction.
10. If you push twice as hard against a stationary brick wall, the amount of work you do
a) doubles
b) is cut in half
c) remains constant but non-zero
d) remains constant at zero
Solution: Since the force did not contribute at all to move any object, it did not do any work.
11. A driver, traveling at 22 m/s, slows down her 2000 kg car to stop for a red light. What is the work done
by the friction force against the wheels?
a) -2.2X104 J
b) -4.4X104 J
c) -4.84X105 J
d) -9.68X105 J
Solution: In order for the car to stop, all its kinetic energy should be taken away. Therefore the friction
must do all the work to take away the car’s kinetic energy Wf = Ff d = − 1 mv2 = − 1 ×2000kg×( 22m/ s)2 = −4.84×105J .
2
2
12. The quantity 1/2 mv 2 is
a) the kinetic energy of the object.
c) the work done on the object by the force.
b) the potential energy of the object.
d) the power supplied to the object by the force.
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PHYS 1443-003, Fall 2002, 2nd Term Exam, Wednesday, Oct. 30, 2002
13. An 800-N box is pushed up an inclined plane. The plane is 4.0 m long and rises 2.0 m. It requires
3200 J of work to get the box to the top of the plane. What was the magnitude of the average friction
force on the box?
a) 0 N
b) Non-zero, but less than 400 N
c) 400 N
d) Greater than 400 N
Solution: Total work needed to push the box to the top of the hill was 3200J but the potential energy of the
box at the top of the hill is PE = mgh = 800 × 2 .0 = 1600 J . Therefore the remaining 1600J was used to
overcome friction along the 4.0m length of the hill, W f = Ff l = Ff × 4. 0m = 1600 J . Therefore, the
friction is 400N.
14. A 50-kg pitching machine (excluding the baseball) is placed on a frozen pond. The machine fires a
0.40-kg baseball with a speed of 35 m/s in the horizontal direction. What is the recoil speed of the
pitching machine? (Assume negligible friction.)
a) 0.14 m/s
b) 0.28 m/s
c) 0.70 m/s
d) 4.4x10 3 m/s
Solution: A linear momentum conservation principle problem. Since the pitching machine fires the baseball
on a frozen pond, we neglect friction in this problem. From the momentum conservation, we obtain
0 = mBvB − mpv p . Thus the speed of the machine after firing the ball is v p = mB v B = 0.4 × 35 = 0.28m / s .
mp
50
15. A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block. The bullet emerges (the bullet does not
embedded in the block) with half of its original speed. What is the velocity of the block right after
the collision?
a) 1.50 m/s
b) 2.97 m/s
c) 3.00 m/s
d) 273 m/s
Solution: A linear momentum conservation problem. Since the momentum before and after the collision
need to be conserved, we obtain mbvb = mbv'b + mB v'B . Solving the equation for the block’s final velocity, we
'
−2
obtain vB' = mbvb − mbvb = mb (v b − vb' ) = 1.00 ×10 150 = 1.50 m / s .
mB
mB
1.00
16. A constant 6.0-N net force acts for 4.0 s on a 12 kg object. What is the object's change of velocity?
a) 2.0 m/s
b) 12 m/s
c) 18 m/s
d) 288 m/s
Solution: The 6.0N net force on a 12kg objects provides acceleration to the object: a = F = 6.0 = 0.5m / s 2
m 12
Using kinematic equation of motion, we obtain: ∆v = at = 0.5 ⋅ 4. 0 = 2.0m / s .
18. A 3.0-kg object moves to the right with a speed of 4.0 m/s. It collides in a perfectly elastic collision
with a 6.0-kg object moving to the left at 2.0 m/s. What is the total kinetic energy after the collision?
a) 72 J
b) 36 J
c) 24 J
d) 0 J
Solution: A linear momentum and kinetic energy conservation problem. Let’s define right-hand side
positive direction. Since the collision is a elastic collision, it’s kinetic energy is conserved. Therefore
kinetic energy after the collision is the same as that before the collision. Thus the kinetic energy after the
collision is KE = 1 m1v12 + 1 m2v22 = 1 (3.0 × 4.0 2 + 6.0 × 2 .02 ) = 36 J .
2
2
2
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PHYS 1443-003, Fall 2002, 2nd Term Exam, Wednesday, Oct. 30, 2002
19. Three masses are positioned as follows: 2.0 kg at (0, 0), 2.0 kg at (2.0, 0), and 4.0 kg at (2.0, 1.0).
Determine the coordinates of the center of mass.
a) (0.50, 1.5)
b) (1.5, 0.50)
c) (2.5, 1.5)
d) (2.5, 0.50)
Solution: Using definition of center of mass for each axis, one obtains
m x 2.0 ⋅ 0 + 2.0 ⋅ 2 .0 + 4.0 ⋅ 2.0 12
∑ mi yi = 2.0 ⋅ 0 + 2.0 ⋅ 0 + 4.0 ⋅1. 0 = 4.0 = 0.50
xCM = ∑ i i =
=
= 1.5 and yCM =
2 .0 + 2.0 + 4.0
8.0
2 .0 + 2.0 + 4.0
8 .0
∑ mi
∑ mi
20. Which of the following is a false statement?
a) For a uniform symmetric object, the center of mass is at the center of symmetry.
b) For an object on the surface of the earth, the center of gravity and the center of mass are the same
point.
c) The center of mass of an object must lie within the object.
d) The center of gravity of an object may be thought of as the "balance point."
Solution: Depending on the shape of the object, the center of mass of the object could be outside of the
object’s body. Thus C) above is a false statement.
21. A plane, flying horizontally, releases a bomb, which explodes before hitting the ground. Neglecting air
resistance, the center of mass of the bomb fragments, just after the explosion
a) is zero.
b) moves horizontally.
c) moves vertically.
d) moves along a parabolic path.
Solution: The center of mass of the bomb will continue the motion the bomb has been performing, moving
along the parabolic path the bomb was following before the explosion.
22. A wheel of radius 1.0 m is rotating with a constant angular speed of 2.0 rad/s. What is the linear speed
of a point on the wheel's rim?
a) 0.50 m/s
b) 1.0 m/s
c) 2.0 m/s
d) 4.0 m/s
Solution: Since linear speed is express in terms of angular speed and radius, v = rϖ . Thus the linear
speed of a point on the wheel’s rim is v = rϖ = 1 .0 ⋅ 2.0 = 2.0m / s .
23. Consider a rigid body that is rotating. Which of the following is an accurate statement?
a) Its center of rotation is its center of gravity.
b) All points on the body are moving with the same angular velocity.
c) All points on the body are moving with the same linear velocity.
d) Its center of rotation is at rest, i.e., not moving.
24. Consider two uniform solid spheres where both have the same diameter, but one has twice the mass of
the other. The ratio of the larger moment of inertia to that of the smaller moment of inertia is
a) 2
b) 8
c) 10
d) 4
Solution: Since moment of inertia is proportional to the mass, if the two objects have the same shape but
one has twice the mass, the moment of inertia of the heavier object is twice that of the lighter one.
25. What is the quantity used to measure an object's resistance to changes in rotation?
a) mass
b) moment of inertia
c) torque
d) angular velocity
Solution: Moment of inertia is equivalent to mass in linear motion which resists to the change.
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PHYS 1443-003, Fall 2002, 2nd Term Exam, Wednesday, Oct. 30, 2002
26. A triatomic molecule is modeled as follows: mass m is at the origin, mass 2m is at x = a, and, mass 3m
is at x = 2a. What is the mo ment of inertia about the origin?
a) 2 m a2
b) 3 m a2
c) 12 m a2
d) 14 ma2
Solution: Using the definition of moment of inertia, one can obtain I = ∑ mi ri2 = 2ma 2 + 3m ⋅ (2 a) 2 = 14ma 2
i
27. A billiard ball with a speed of v1 collides head-on elastically with an identical ball moving with a speed
v 2 in the opposite direction.
a) What are the quantities conserved in this collision? (4 points)
Answer: Linear momentum and kinetic energy
b) Determine the speeds of both balls after the collision. (7 points)
Answer:
From momentum conservation
'
'
mv1 − mv 2 = − mv1 + mv2 (1)
therefore,
'
'
v1 + v1 = v 2 + v 2 (2)
Now from the kinetic energy conservation:
1
1 2 1
1
2
'2
'2
mv1 + mv 2 = mv1 + mv 2 (3)
2
2
2
2
therefore one can obtain the following:
2
'2
'2
2
v1 − v1 = v 2 − v 2 (4),
and by factorizing the terms we obtain
'
'
'
'
v1 − v1 v1 + v1 = v 2 − v 2 v 2 + v 2 (5)
Using eq. 2 above, eq. 5 becomes:
'
'
v1 − v 1 = v 2 − v 2 (6)
Using eq. 2 and eq. 6 one can deduce
(
)(
) (
)(
)
'
v1 = v 2
and
'
v 2 = v1 .
Therefore, perfectly elastic collisions between the two objects with identical masses exchange
their speeds.
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PHYS 1443-003, Fall 2002, 2nd Term Exam, Wednesday, Oct. 30, 2002
28. A very thin aluminum rod of length L has a density that varies as a function of the position in the
rod, λ = αx , where α is a constant and x is the position from the origin where the rod’s axis of
rotation is.
a) What is the total mass of the rod? (3 points)
Answer: Since the mass is integration of density over the entire length,
L
L
L
1

1
M = ∫0 λdx = ∫0 αxdx =  αx 2  = αL2
2
0 2
b) What is the moment of inertia of this rod when it rotates about the axis located at one end (x=0)?
(5 points)
Answer: Using the definition of moment of inertia,
L
L
L
L
1

1
1
I = ∫0 x 2 dm = ∫0 λx 2 dx = ∫0 αx 3 dx =  αx 3  = αL4 = ML2
2
4
0 4
c) What is the rotational kinetic energy of the rod when its angular speed is ω in the rotation above?
(3 points)
Answer: Using the definition of rotational kinetic energy,
K=
1 2 11

1
Iϖ =  ML2 ϖ 2 = ML2ϖ 2
2
22
4

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PHYS 1443-003, Fall 2002, 2nd Term Exam, Wednesday, Oct. 30, 2002
Useful Formulae and information
Velocity: v f = vi + at
1 2
at
2
Angular velocity: ϖ f = ϖ i + αt
Position: r f = ri + vi t +
Angular Displacement: θ f = θ i + ϖ i t +
∑m r
=
∑m
i
Center of Mass: r CM
1 2
αt
2
i
i
i
i
Moment of Inertia: I =
∑m r
i i
f
2
or I =
i
∫x
2
dm
i
Newton’s universal law of gravitation: The gravitational force between the two objects of masses m1 and
m2 is proportional to their masses and inversely proportional to the square of the distance between them.
FG =
Gm1m 2
r2
The solutions for a 2-dimensional equation:
ax 2 + bx + c = 0
are:
x=
− b ± b 2 − 4ac
2a
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