Physics 110
Spring 2010
Exam #1
April 16, 2010
Name______________
Part
Multiple Choice
Problem #1
Problem #2
Problem #3
Total
/ 10
/ 27
/ 27
/ 36
/ 100
In keeping with the Union College policy on academic honesty, it is assumed that you will
neither accept nor provide unauthorized assistance in the completion of this work.
Part I: Free Response Problems
Please show all work in order to receive partial credit. If your solutions are illegible
no credit will be given. Please use the back of the page if necessary, but number the
problem you are working on. Each subpart of a problem is worth 9 points.
1. A Boeing F/A-18 Super Hornet with a mass of 30,000kg lands on the deck of an
aircraft carrier at a speed of 140 mi/hr (~63 m/s) and is brought to rest by an arresting
cable (shown at the very back of the aircraft) that is stretched across the deck of the
carrier as shown in the figure below.
a. If the jet needs to be brought to rest
in 1.5s, what are the minimum force
and acceleration exerted by the
arresting cable?
Arresting cable
zxfzdfasdfasdf
v fx = v ix + ax t → 0 = 63 − ax (1.5s) → ax = 42
m
s
m
s2
Fx = max = 30,000kg × 42 sm2 = 1.26 ×10 6 N
aircraft.
€
, both opposite the landing
b. From the time the jet hits the carrier deck, how far does it travel before coming to
rest?
(
)
2
Δx = v ix t − 12 ax t 2 = (63 ms ×1.5s) − 12 42 sm2 (1.5s) = 47.3m
€
c. Of course in order to land the jet, it must have taken
off from the carrier. A steam catapult located on the
front of the flight deck launches the planes. Suppose
that the plane needs to be launched from rest to 63
m/s. If the flight deck is 100m long, what is the time
it takes to launch this plane and what are the
magnitudes of the acceleration and force needed to
launch this plane?
2
v 2fx
63 ms )
(
v = v + 2ax Δx → ax =
=
= 19.9 sm2
2Δx 2 ×100m
in the direction of launch.
Fx = max = 30000kg ×19.9 sm2 = 5.95 ×10 5 N
2
fx
2
ix
Δx = 12 ax t 2 → t =
€
2Δx
2 ×100m
=
= 3.2s
ax
19.9 sm2
2. In a local bar, a customer slides an empty beer mug down the counter for a refill. The
bartender is momentarily distracted and does not see the mug, which slides off the
counter and strikes the floor 1.40m from the base of the counter. Suppose that the
height of the counter is 0.86m above the floor.
a. What was the speed of the mug when it leaves the counter?
b. What is the impact velocity of the mug just before the floor?
c.
How long does it take the mug to reach the floor?
3. Suppose that you are given the arrangement of masses shown below where the left
mass mL = 1kg, center mass mC = ½ kg and the right mass mR = 4kg. The masses are
on frictionless surfaces and are released from rest. You may assume that the center
mass moves to the right.
a. Draw a carefully labeled force diagram showing all of the forces that act on each
mass. You can draw these on the diagram below, but be sure to specify a
coordinate system for each mass. The triangular mass on the left is on an incline
that is oriented at 51o with respect to the horizontal.
F
NC
FTL
FTL
FTR
FNL
FWC
FTR
FWL
FWR
b. What is the magnitude of the acceleration of the system?
mC :
mR :
mL :
∑F
∑F
∑F
x
: FTR − FTL = mc a
x
: FTR − mR g = −mR a → FTR = mR g − mR a
x
: FTL − mL gsin θ = mL a → FTL = mL gsin θ + mL a
FTR − FTL = mR g − mR a − mL gsin θ − mL a = mc a
(
) (
)
4kg × 9.8 sm2 − 1kg × 9.8 sm2 sin51
mR g − mL gsin θ
∴a =
=
= 5.74 sm2
mL + mC + mR
5.5kg
€
c. What are the magnitudes of the tension forces that act on the left (FTL) and right
(FTR) sides of the center mass?
(
)
FTR = mR ( g − a) = 4kg 9.8 sm2 − 5.74 sm2 = 16.2N
(
)
FTL = mL ( gsin θ + a) = 1kg 9.8 sm2 sin51+ 5.74 sm2 = 13.4N
€
d. How long does it take the center mass to move 0.5m and how fast is the rightmost block moving after it has traveled 1.2m?
Δx = v ix t + 12 ax t 2 → t =
2Δx
2 × 0.5m
=
= 0.42s
ax
5.74 sm2
v 2fx = v ix2 + 2aΔx → v fx = 2aΔx = 2 × 5.74 sm2 ×1.2m = 3.7 ms
€
Part II: Multiple-Choice
Circle the best answer to each question. Any other marks will not be given credit.
Each multiple-choice question is worth 2 points for a total of 10 points.
1. In the diagram below, the block of mass m is pulled to the right by an applied force
FA oriented at an angle θ above the horizontal and the block experiences an
acceleration a. The magnitude of the frictional force is given by
a. FAsinθ − ma
b. FAcosθ − ma
FA
c. FAcosθ + ma
d. FN
m
2. Suppose that you stand up quickly and accidentally bump your head on heavy object
located above your head. The greater force of impact will be
a. on the your head.
b. on the heavy object.
c. the same for both.
d. unable to be determined with the information given.
3. Suppose that an object travels in three steps. In the first step the object covers 100m
in 10 seconds in the negative x-direction then the second step has the object
decelerates to rest from velocity of 10 m/s in 10 seconds and third step has the object
traveling 150m in the positive x-direction for 25s. For the entire trip, the average
velocity of the object is
a. -4.8 m/s
b. 0 m/s
c. 9.8 m/s
d. 12.7 m/s
4. Suppose that a 10,000kg block is sitting on a horizontal frictionless surface. A
0.01kg washer is connected to the block by a string passing over a pulley that is
attached to the block. If the washer is released from rest, the acceleration of the block
is
a. equal to 0 m/s2
b. 0 m/s2< a < 9.8 m/s2
c. greater than 9.8 m/s2
d. unable to be determined.
5. Two marbles sitting on a tabletop. One is dropped from rest vertically
(marble A) and the other shot horizontally off the table (marble B). Compared to
marble B, the acceleration of marble A’s is
a. the same.
b. greater.
c. less.
d. unable to be determined.
Useful formulas:
Motion in the r = x, y or z-directions
1
2
rf = r0 + v 0r t + ar t
2
v fr = v 0r + ar t
v fr 2 = v 0r 2 + 2arΔr
Uniform Circular Motion
Geometry /Algebra
v2
ar =
r
Circles
Triangles
Spheres
v 2 C = 2πr
A = 12 bh
A = 4 πr 2
Fr = mar = m
r
A = πr 2
V = 43 πr 3
2πr
v=
Quadratic equation : ax 2 + bx + c = 0,
T
FG = G
m1m2
r2
Vectors
−b ± b 2 − 4ac
2a
whose solutions are given by : x =
Useful Constants
€
€
G = 6.67×10−11 Nm
g = 9.8 m s2
N A = 6.02×1023 atoms mole
→
→
→
→
p f = p i + F Δt
→
→
F = ma
→
vsound = 343 m s
Work/Energy
Heat
€ 1 mv 2
Kt =
2
p = mv
→
kg 2
kB = 1.38×10−23 J K
σ = 5.67×10−8 W m 2 K 4
Linear Momentum/Forces
2
K r = 12 Iω 2
U g = mgh
→
Fs = −k x
U S = 12 kx 2
F f = µFN
WT = FdCosθ = ΔE T
W R = τθ = ΔE R
W net = W R + WT = ΔE R + ΔE T
ΔE R + ΔE T + ΔU g + ΔU S = 0
ΔE R + ΔE T + ΔU g + ΔU S = −ΔE diss
€
Rotational Motion
Fluids
Simple Harmonic Motion/Waves
2π
ω = 2πf =
T
TS = 2π
m
k
TP = 2π
l
g
1
v =±
k  x 2 2
A1− 
m  A2 
x ( t ) = Asin( 2Tπt )
k
cos( 2Tπt )
m
k
a( t ) = −A sin( 2Tπt )
m
FT
v = fλ =
µ
v(t) = A
Sound
v
2L
I = 2π 2 f 2 ρvA 2
f n = nf1 = n
€