Physical Science 101
EXAM 1
Summer 2010
1. An object’s weight would be different on the Earth and on the Moon. Which of the following would also be different?
a. density
b. mass
c. volume
d. None of these
2. What is the mass of 25 cm 3 of pure water?
a. None of these
b. 25 kg
c. 25 g
d. 25 mL
Recall that the density of water is 1 g/cm 3
3. The speed of light in a vacuum is 300,000,000 m/s. Represented in powers-of-10 notation, this is
a. 30 x 108m/s.
b. 3 x 107m/s.
c. 300 x 106m/s.
d. All of these
Move the decimal point 6 places.
4. If an object moves with constant velocity,
a. its speed is constant.
b. its direction is constant.
c. its average speed is constant.
5. An automobile’s acceleration may be changed by using the
a. windshield wipers.
b. steering wheel.
c. radio.
d. All of these
d. horn.
6. An object with a mass of 2.0 kg is dropped and takes 3.0 s to hit the ground. Then another object with a mass of 4.0
kg is dropped from the same point. How long does it take to hit the ground?
a. 3.0 s
b. 6.0 s
c. 1.5 s
d. 9.0 s
Both masses have an acceleration due to gravity, which we all know is CONSTANT here.
7. Ideally, a projectile’s horizontal velocity component
a. does not change.
Clearly, ideally means no friction so there are no forces in the horizontal direction.
b. changes most rapidly near the bottom of its trajectory.
c. changes at a constant rate.
d. changes most rapidly near the top of its trajectory.
8. A golfer hits a ball on a level fairway at an angle of 40 relative to the horizontal, and it lands on the green. Another
golfer hits a ball with the same speed but at another angle, and it lands on the green by the other ball. What was the
launch angle for the second golfer?
a. 30
b. 60
c. 45
d. 50
Use symmetry wrt 45
initial angle is 5 below so other angle is 5 above.
9. The natural tendency of an object to remain at rest or in motion with a constant velocity is called
a. an external force.
b. an unbalanced force.
c. inertia
d. All of these
10. The quantity mv is an object’s
a. inertia.
b. torque.
c. linear momentum.
d. centripetal force.
11. If a single force is exerted on an object will its momentum change?
a. not always
b. yes
c. sometimes
e. angular momentum.
d. not enough information
12. Work is defined as force times
a. distance.
b. time.
c. parallel distance.
d. perpendicular distance.
13. Which of the following is a vector quantity?
a. energy
b. power
c. work
d. None of these
14. The kinetic energy of a pendulum is greatest
a. when its potential energy is greatest.
b. at the top of its swing.
c. at the bottom of its swing.
d. when its total energy is greatest.
1
Physical Science 101
EXAM 1
15. Which of the following four objects has the greatest kinetic energy?
a. An object of mass m moving with speed 4v
b. An object of mass 3m moving with speed 2v
c. None of these; all four objects have equal kinetic energy
d. An object of mass 2m moving with speed 3v
e. An object of mass 4m moving with speed v
Summer 2010
Because KE = 1/2mv2
new KE is 1 X 42 = 16 times larger
new KE is 3 X 22 = 12 times bigger
new KE is 2 X 32 = 18 times larger
new KE is 4 X 12 = 4 times larger
16. We define the total mechanical energy of an object as
a. potential plus kinetic energy.
b. inertial plus kinetic energy.
c. gravitational plus potential energy.
d. mental plus physical energy.
17. A system may
a. be isolated.
b. have real boundaries.
18. What is bought from an electric company?
a. electrical energy
b. electric charge
c. have imaginary boundaries.
c. electrical power
d. All of these
d. none of these
19. A kilowatt-hour is a unit that measures
a. work per unit of time.
b. energy.
c. power.
d. work x time.
Note kilowatt is power and hour is time so kilowatt-hour is Power X T which is Energy because P = E/T so PT = E
20. The average adult radiates heat energy at the same rate as
a. a 10-W light bulb.
b. None of these
c. a 100-W light bulb.
d. a 100-kW light bulb.
F-I-B
1. The scientific method holds that no theory is valid unless its predictions are in accord with experiment or data
2. The ratio of mass to volume is known as density.
3. What two quantities are constant in uniform circular motion? speed and radius
4. The centripetal acceleration of an automobile in uniform circular motion on a circular track is supplied by force of
friction. Note: To just say force doesn't really do it. There is also the force of gravity, which ain't the answer.
5. Because of air resistance, an object falling a great distance may reach a terminal velocity.
6. An acceleration is evidence of an unbalanced (net) force.
7. The natural tendency of an object to remain at rest or in uniform straight-line motion is called inertia.
8. When a moving object’s velocity doubles, its kinetic energy increases by a factor of 4.
Because KE = = 1/2mv2 , if v  2v0 then KE  22 KE0 = 4 KE0
9. When the height of an object is doubled, its gravitational potential energy is increased by a factor of 2.
Because PE = mgh, if h  2h0 then PE  2 PE0
10. A 2-hp motor can do the same amount of work as a 1-hp motor in half the time.
Because P = W/t or W = Pt, if P  2 P0 then t must go as t  1/2 t0 in order to keep W constant.
Exercises
1. On a trip, a family travels 200 km in 2.5 hr (or 80km/hr) on the first day, 300 km in 4.0 hr (or 75km/hr) on the second
day, and 250 km in 4.0 hr (or 62.5km/hr)on the third day. What was the average speed in km/hr for the total trip?
average speed = total distance / time. Total dist = 200 km + 300km + 250 km = 750 km
Time = 2.5h + 4.0h + 4.0h = 10.5h
avg speed = 750km/10.5h = 71.4 km/h
2
Physical Science 101
EXAM 1
Summer 2010
2. An object is dropped and falls for 2.00 s before hitting the ground. Use g = 10.0 m/s2 and ignore air resistance to
find: a. What is the final velocity? b. What is the average speed? c. How far has the object fallen? d. What is the
final acceleration?
a. What is the final velocity?
b. What is the average speed?
a. How far has the object fallen?
d. What is the final acceleration?
with constant acceleration, vf = gt = 10.0m/s2 x 2.00s = 20.0 m/s, down
because a is constant use <v> = ½(vi + vf) = ½(0 + 20m/s) = 10.0m/s
From <v> = total dist / time find d = <v>t = 10.0m/s x 2.00s = 20.0m
OR use d = 1/2gt2 = 1/2 x 10.00m/s2 x (2.00s)2 = 20.0 m
hey, its gravity! a = g = 10.0m/s2, down
3. Calculate the centripetal force on a 3.0-kg object moving in a horizontal circular path of 20-m radius with a speed of
8.0 m/s.
We have Fc = mac = m v2/r = 3.0kg x (8.0m/s)2 / 20m = 9.6 N
4. To increase the force of gravity between two masses by a factor of 9, how would the objects have to be moved?
From Fg = G m1 m2 / r2 and Fnew = 1/9 Foriginal we hust have r2new = 1/9 r2original or rnew = 1/3 roriginal or 3 times closer
5. Two carts (A’s mass = 4.5 kg and B’s mass = 1.5 kg) are held against a strong, compressed spring. The carts are
released simultaneously, and B moves off to the right with a speed of 2.0 m/s. What is A’s initial velocity?
For this question we consider conservation of momentum of the system consisting of both carts. Before release
there is zero momentum (p0 = mAvA0 + mBvB0 = 0) so after the release the total momentum pf = mAvAf + mBvBf = 0.
Solving for vAf we have vAf = - mBvBf / mA = -1.5kg x 2.0m/s / 4.5kg = -0.67m/s (left)
6. A constant force of 7.2 N acting on a 1.0-kg object initially at rest moves it through a parallel distance of 2.5 m.
Neglecting friction, (a.) how much work is done on the object and (b.) using the results for part a., what is the speed of
the object when it is moved 2.5 m?
a. From the definition of work we have W = Fd = 7.2N x 2.5m = 18 Nm = 18 J.
b. OK, let's do a common student thing. How do we find d? We could ignore the requirement to use the results of
part a and think: Well, use d = vt ... but NO! Can't do that because v is not constant - after all there is a force
acting ain't there? I know, I know, use d = 1/2 at2... I sorta remember that equation. But what good does that
do, I don't know a or t. But wait, I can use a = 10m/s2 just like we did so many times already. Wait a minute,
this is the a for gravity and the question doesn't say any thing about gravity. Well if I just use it maybe he won't
notice or give me credit anyway. No, wait! I can use THE LAW!! I think F = ma so maybe a = F/m is ok???
But I still don't know what to use for t. I know, I will just use 1 second and he will think I figured it out....
Tell you what, why don't we think just a bit more about physics and recall there is a relationship between work and
change in kinetic energy and kinetic energy is after all the energy of motion and the object is certainly moving.
From the relationship W = ΔKE, and because the initial KE = 0, we have 1/2 mv 2 = 18J or v2 = 2 x 18J/1.0kg =
36 J/kg. And, to get to fundamental units, we have v 2 = 36 J/kg = 36 Nm/kg = 36 kg m/s2 m/kg = 36 m2/s2
Thus v = (36 m2/s2)1/2 = 6.0 m/s
3
Physical Science 101
EXAM 1
Summer 2010
Special questions
1. An automobile with a mass of 100 kg is moving north on a level, straight highway with a constant velocity of 80
km/h. What is the unbalanced force acting on the automobile?
Because velocity is constant there is no acceleration so Fnet = 0
2. A 6.0-kg object exerts a force of 8.0 N on a 2.0-kg object. What is the magnitude of the force on the 6.0-kg object?
By the third law F6kg on 8kg
= F8kg on 6kg so F = 8.0 N
Special Bonus Question
3. A comet is traveling at a speed of 50,000 mi/h when it is 100 million miles from the Sun. Compute its speed when it
is 25 million miles from the Sun.
For this question we consider conservation of angular momentum. L0 = Lf or mv0r0 = mvf rf or v0r0 = vf rf.
Solve for vf to get vf = v0r0 / rf = 50,000 mi/h x 100 x 106 / 25 x 106 = 2 x 105 mi/h = 200, 000 mi/h
Or, without the algebra, since the product vr must be constant and r got to be four times smaller then v must be
four times larger and fifty thousand becomes two hundred thousand.
4