Physics 120
Spring 2007
Exam #2
May 14, 2007
Name__________________
Part
Multiple Choice
Problem #1
Problem #2
Problem #3
Total
/ 10
/ 30
/ 30
/ 30
/ 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.
1
Multiple choice questions are worth 2 points each. Please circle the best answer to each
question.
1. After colliding, two bodies move away with different velocities. In the space below,
explain how you would determine if the collision was elastic or inelastic?
Measure the kinetic energy of both objects before and after the collision. If they
are equal then the collision was elastic, if not it was inelastic.
2. A Ping-Pong ball is thrown at a stationary bowling ball hanging from a wire. The
Ping-Pong ball makes a one-dimensional elastic collision and bounces back along the
same line. After the collision, the Ping-Pong ball has, compared with the bowling
ball,
a.
b.
c.
d.
a larger magnitude of momentum and more kinetic energy.
a smaller magnitude of momentum and more kinetic energy.
a larger magnitude of momentum and less kinetic energy.
a smaller magnitude of momentum and less kinetic energy.
r
r
r
Since the momentum of the two-ball system is conserved, p Pi + 0 = p Pf + p B .
Because the Ping-Pong ball bounces back from the much more massive bowling ball
r
r
r
r
with approximately the same speed, p Pf = −p Pi . As a consequence, p B = 2p Pi .
Kinetic energy can be expressed as K = p2/2m. Because of the much larger mass of
the bowling ball, its kinetic energy is much smaller than that of the Ping-Pong ball.
3. A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a
distance d. For the next loading, the spring is compressed a distance 2d. How much
work is required to load the second dart compared to that required to load the first?
a.
b.
c.
d.
It takes four times as much work.
It takes two times as much work.
It takes the same amount of work.
It takes half as much work.
Since the work done in compressing a spring is proportional to the square of the
compression distance x, doubling the value of x causes the work to increase fourfold.
2
4. Suppose you are talking by interplanetary telephone to your friend, who lives on the
Moon. He tells you that he has just won a Newton of gold in a contest. Excitedly, you
tell him that you entered the Earth version of the same contest and also won a Newton
of gold! Who is richer?
a.
b.
c.
d.
You are.
Your friend is.
You are equally rich.
There is no way to determine the answer from the information given.
Since the value of g is smaller on the Moon than on the Earth, more mass of gold
would be required to represent 1 N of weight on the Moon. Therefore, your friend on
the Moon is richer, by about a factor of 6!
5. Three identical point masses are lying along the x axis at the following locations;
x = -2m, 3m and 4m. What is the x-coordinate of the center of mass?
a. 5/3 m
xcom =
b. 5 m
∑m x
∑m
i i
i
=
c. 3 m
d. 3/5 m
m(−2m) + m(3m) + m(4m) 5
= m
3m
3
3
Part II: 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. The numbers in parentheses following the question
correspond to the point values for each part.
1. You are to design a roller coaster in which the cars of total mass 500kg are
accelerated from rest by a spring of spring constant 850 N/m toward a downhill ramp
of height 30m above the ground. The cars travel through a valley and then up a
mountain.
A
C
B
a. What is the speed of the cars at the top of the incline, at point A if the spring is
compressed by 4m? (6)
850 Nm
k
(4m ) = 5.22 ms
Ei = E f → 12 kx 2 = 12 mv 2 → v =
x=
m
500kg
b. What is the speed of the cars at the bottom of the incline, at point B? (8)
E A = E B → 12 mv A2 + mghA = 12 mvB2
vB = v A2 + 2 ghA =
(5.22 ms )2 + 2 × 9.8 sm × 30m = 24.8 ms
2
c. If the passengers are to feel 8g’s at the bottom of the valley, what must the radius
R of the arc of the circle that fits the bottom of the valley? (8)
v2
v 2 (24.8 ms )
ac = B → R = B =
= 7.85m
R
ac 8 × 9.8 sm2
2
d. The top of the next mountain, point C, is an arc of a circle of the same radius R.
If the passengers are to feel 0g’s (zero-g’s) at the top of this mountain, what must
be its height h’? (8)
FN
E B = E C → 12 mv B2 = 12 mvC2 + mghC → hC =
v B2 − v C2
2g
v2
From a Free − Body diagram : ∑ F y ,C : FN − mg = −m C → v C = Rg ,
R
since zero − g' s corresponds to feeling " weightless" at the top which means FN = 0.
(
m
FW
)
m
m
v B2 − Rg (24.8 s ) − 7.85m × 9.8 s 2
∴ hC =
=
= 27.5m
2g
2 × 9.8 sm2
2
4
5
2.
For any aircraft in flight there are four main forces that act that aircraft. These
four forces are: the lift, FLift which acts perpendicular to the aircraft’s wings and
makes the aircraft rise, the thrust, FThrust which propels the aircraft forward and
points in the direction the aircraft’s nose is pointing, the drag, FDrag which is a
frictional force due to air resistance and opposes the thrust, and the weight of the
aircraft, FWeight. Suppose that the US Air-Force Thunderbird’s F-16 jet shown
below is climbing out 30o angle with respect to the runway.
a. On the jet draw a carefully labeled free body diagram labeling the four forces
above. Please use the coordinate system shown below. (10)
y
x
FLift
FThrust
FDrag
FWeight
30o
b. The engines of the aircraft can produce a thrust of magnitude FThrust =
108,000 N. If the weight of the aircraft is FWeight = 150,000N and climbs out
at constant velocity, what is the magnitude of Flift? (10)
ΣFx : FThrust − FDrag − FWeight sin θ = max = 0
ΣFy : FLift − FWeight cosθ = ma y = 0
(1)
(2)
From (2 ) we have that :
FLift = FWeight cosθ = 150,000 N cos 30 = 129,904 N .
From (1) we have that :
FDrag = FThrust − FWeight sin θ = 108,000 N − 150,000 N sin 30 = 33,000 N
c. What is the magnitude of FDrag? (10)
FDrag = 33,000N
Two comments: 1. This jet is actually landing, since the speed brakes are
deployed, and 2: The landing angle (and the take off angle) are not usually 30o.
Rather the take off and landing angles are usually more like 10o to the horizontal.
6
7
3. Suppose that you shoot a bullet through a piece of wood sitting on a table. The piece
of wood acquires a speed of 8.0 m/s, and the bullet emerges with a reduced speed.
The mass of the bullet is 15g, its initial speed is 600 m/s, and the mass of the piece of
wood is 300g.
a. What is the velocity of the bullet as it exits the block? (8)
r r
r
r
r
pi = p f → mbullet vi ,bullet = mbullet v f ,bullet + mblock v f ,block
r
r
mbullet vi ,bullet − mblock v f ,block ⎡ (0.015kg × 600 ms ) − (0.3kg × 8 ms )⎤
r
m ˆ
ˆ
=⎢
v f ,bullet =
⎥i = 440 s i
mbullet
kg
0
.
015
⎣
⎦
b. What is the change in kinetic energy of the bullet? (7)
(
∆KEbullet = 12 mbullet v 2f ,bullet − 12 mbullet vi2,bullet = 12 × 0.015kg × (440 ms ) − (600 ms )
2
2
)
∆KEbullet = −1248 J
c. What is the change in kinetic energy of the block? (7)
(
∆KEblock = 12 mblock v 2f ,block − 12 mblock vi2,block = 12 × 0.300kg × (8 ms ) − (0 ms )
2
2
)
∆KEblock = 9.6 J
d. How much work is done by the block on the bullet? How does this missing
energy manifest itself? (8)
W = ∆KEbullet = 1248 J and the energy lost by the bullet shows up as heat and
friction in the block as the block does work to slow the bullet down.
8
Useful formulas:
Motion in the x, y or z-directions
Uniform Circular Motion
Geometry /Algebra
9
r f = r0 + v0 r t + 12 a r t 2
ar =
v2
Fr = ma r = m
r
2πr
v=
T
mm
FG = G 1 2 2
r
v fr = vir + a r t
v fr = vir + 2 a r ∆r
2
2
Vectors
magnitude of a vector = vx + v
2
Triangles
Spheres
A = 12 bh
A = 4πr 2
A = πr
V = 43 πr 3
2
Quadratic equation : ax + bx + c = 0,
2
whose solutions are given by : x =
Linear Momentum/Forces
− b ± b 2 − 4ac
2a
N A = 6.02 × 10 23 atoms mole
σ = 5.67 × 10− 8 W m
2
K4
2
kg 2
k B = 1.38 × 10 − 23 J K
vsound = 343 m s
Work/Energy
→
p=mv
K t = mv
1
2
→
G = 6.67 × 10−11 Nm
g = 9.8 m s 2
y
⎛ vy ⎞
direction of a vector → φ = tan −1 ⎜⎜ ⎟⎟
⎝ vx ⎠
→
Circles
C = 2πr
Useful Constants
2
→
v2
r
→
Pair = 1.013 × 105 N / m 2
Rotational Motion
2
θ f = θ i + ωi t 12 αt 2
ω f = ωi + αt
p f = p i + F ∆t
K r = 12 Iω 2
→
ω 2 f = ω 2 i + 2α∆θ
τ = Iα = rF
U S = 12 kx 2
L = Iω
r r
WT = ∫ F ⋅ d r = ∫ Fdr cos θ = ∆ ET ∆s = r∆θ : v = rω : at = rα
→
F =ma
→
U g = mgh
→
Fs = − k x
F f = µ FN
W R = ∫ τd θ = ∆ E R
a r = rω 2
W net = W R + WT = ∆ E R + ∆ ET
P=
Simple Harmonic Motion/Waves
ω = 2π f =
TS = 2π
2π
T
m
k
k
cos
m
k
sin
a (t ) = − A
m
FT
v = fλ =
⎛ N r⎞
⎜ ∑ mi ri ⎟
r
i =1
⎟ = rrdm = rr
rcom = N
⇒ lim ⎜ i =N1
com
N → ∞⎜
⎟ ∫
m
m
⎜ ∑ i ⎟
∑
i
i =1
⎠
⎝ i =1
m − m2
2m2
v1 f =
v2 i + 1
v1i
m1 + m2
m1 + m2
r
∑ mi ri
⎛
x2
k
A ⎜⎜ 1 −
v = ±
A2
m
⎝
x (t ) = A sin ( 2Tπ t )
v (t ) = A
Other Misc. Formulas
N
l
g
T P = 2π
∆E
∆t
⎞
⎟⎟
⎠
1
2
( 2Tπ t )
v2 f =
m − m1
2m1
v1i + 2
v2 i
m1 + m2
m1 + m2
( 2Tπ t )
µ
f n = nf
P =
1
2
ω
= n
1
2
µ vA
v
2L
2
10
11