Fall Multiple Choice Answers
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D Acceleration produces a change in velocity and/or direction.
B Fair = bv = ma  aair = bv/m  overall a = g – bv/m (down)
Fg = -kx  -(0.50 kg)(10 m/s2) = -k(0.25 m)
C  k = 20 N/m
Because the surface is rough, some energy is lost,
E potential energy after sliding is less and h' will be less.
2
2
2
C d = ½at  3.0 m = ½a(1 s) a = 6 m/s
Fg is proportional to mass on the Earth or Moon, but
C centripetal acceleration doesn't depend on mass (a = v2/r).
c
Momentum is conserved, The total momentum after release
E equal momentum before, which is zero,  momentums must
be equal but opposite in direction.
Gravitational force between spheres is equal, but opposite in
E direction.
At maximum displacement, the oscillating object is
A momentarily stopped while it changes direction, but the
spring restoring force is at a maximum (F = -kx).
Momentum is conserved (p = mv). Since the total mass is
D greater, then the combined velocity must be less.
Total momentum increases, so an external force (impulse)
B must have been applied, which could have come from a
loaded spring.
Velocity is horizontal at that instant and acceleration is
B upward (centripetal acceleration causes the car to climb up
the other side of the hill).
B Converging lenses are thicker in the center than the edges.
-1
D v = f = (5 s )(2 m) = 10 m/s
f
=
nf

5
s-1 = (2)f1  f1 = 2.5 s-1
1
B n
Radio has the longest wavelength, lowest frequency (c = f)
B and energy (E = hf) and gamma is opposite.
A Sound waves in air are longitudinal.
Image position corresponds, but on the other side of the
E mirror.
Photoelectric effect correlates each released electron with the
C absorption of a quanta of energy (photon as particle).
C Interference is a wave phenomenon.
Maximum positive acceleration occurs when the object is at
A the lowest point in its motion (F = -kx = ma).
E P = W/t = Fd/t = mgh/t
B F = ma  mg = (3 m + m)a  a = g/4
The package slides because of inertia. There wasn't enough
E friction to generate the centripetal acceleration inward.
2
2
2
2
½
D K = Us  ½mv = ½kx  (m)vo = kx  x = vo(m/k)
2
2
2
2
Fg = mg = GMm/r  G = gr /M  gErE /ME = gMrM /MM
D 10 m/s2(1)2/(1) = g (½)2/1/  g = 4 m/s2
M
10
M
8
10 -1
E c = f  3 x 10 m/s = f(0.03 m)  f = 1 x 10 s
Angular momentum deals with radius of orbit and velocity,
A answer deals with orbital values.
When balanced counter clockwise torque = clockwise torgue
D m r = m r  6x = 8(70 - x)  x = 40
1 1
2 2
C f' = f(vw ± vo)/(vw ± vs)
The left sides of p and Q combine to make an extra high crest
A and the right sides of p and Q cancel each other.
I and II are both interference phenomena where  is similar to
C the bubble thickness and spacing between grating lines.
Bending of light is independent of  (n1sin1 = n2sin2).
2
D Fg = mg  m = 20 N/10 m/s = 2 kg
C Fn = mgcos37 = (20 N)(4/5) = 16 N
B W = F||d = mgh = (20 N)(3 m) = 60 J
vx is constant, but vy decreases with elevation.
D v = (v 2 + v 2)
x
y
E Acceleration is due to gravity, which is downward.
71
A The only force acting on the ball is gravity,  downward.
do is more than 2f, let's say 3f.
D 1/d + 1/d = 1/f  1/d = 1/f – 1/3f = 2/3f  d = 1.5f
o
i
i
i
B All objects accelerate at the same rate,  speeds are equal.
72
C P = W/t = mgh/t = (700 N)(8 m)/10 s = 560 W
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96
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108
109
Inelastic collision
E m v + m v = (m + m )v'  v' = (mv + 0)/(m + M)
A A
B B
A
B
Momentum is conserved,  with an increase in mass there is
C a corresponding decrease in speed.
C P = W/t = J/s = W (kWhr is a measure of energy).
2
E L = rp = rmv = (4 m)(2 kg)(3 m/s) = 24 kg•m /s
A With no acceleration, the vector sum of the forces equal 0.
D K + U is a constant,  the sum of the two graphs = Vo.
A P = Ffv = mgv
1/do + 1/di = 1/f  1/ + 1/di = 1/f  0 + 1/di = 1/f  di = f
B f = ½R = ½(1.0 m) = 0.5 m
E nwater > nair, vn = c/n and n = /n
1/do + 1/di = 1/f
D 1/6 + 1/di = 1/9  1/di = 2/18 – 3/18 = -1/18  di = -18
M = -di/do = -(-18)/6 = -3 (virtual, upright, 3 x larger)
Transverse wave reflects off of a more dense surface
B undergoes phase shift (v and A remain the same).
D Real, inverted magnified image is formed for f < di < 2f.
n2 > n1 because light bend toward normal in medium 2.
A n = n because rays are parallel.
1
3
4 kg will stretch the spring 16 cm to reach equilibrium
D position, then it will stretch an additional 16 cm before turning
around,  32 cm.
2
1
E W = Fg = GMm/r ,  when r  4r, W  /16W
2
2
2
110 B Fc = Fg  mv /r = GMm/r  K = ½mv = ½GMm/R
Since there was no initial y momentum, then the total y
111 A momentum = 0: (0.2 kg)(1 m/s) = (0.1 kg)v,  v = 2 m/s 
Refraction angle depends on the difference in nprism and
112 E nmedium. Since the difference decreases, the angle of
separation will decrease, but the order remains the same.
Momentum increases when velocity increases, which
113 B produces a curved d vs. t graph.
Fnet = 0 means no acceleration, where v = 0 or constant (graph
114 C is straight).
Frequency is greatest where the wavelength is shortest (v =
119 C f ). This occurs where the wave fronts are closest together.
120 D sinc = n/d  d = n/sinc = (1)(0.12 m)/(3/5) = 0.20 m
121 B Image formed is the "mirror" image of the object.
From midpoint to one extreme position = ¼T,
126 D  T = 4(0.1 s) = 0.4 s.
127 C Impulse, J = FT = mv = (0.4 kg)(5.0 m/s) = 2 N•s
Counterclockwise torque = clockwise torque
128 C Rm + ½RM = R(2M)  m + ½M = 2 M,  m = 3M/2
2
½
½
130 E d = ½at  t = (2d/a) = (2h/g)
2
131 D K' = K + Ug = ½mvo + mgh
Momentum is conserved,  their momenta must be equal in
134 C magnitude and opposite in direction.
px: 3mvx = mV,  vx = V/3, py: 3mvy = mV,  vy = V/3
137 D v = (v 2 + v 2)½ = ((V/3)2 + (V/3)2)½ = 2V/3
x
y
138 A torque,  = r x F = r x Fsin = rsin x F
Only non-accelerated motion is constant straight line motion,
141 E but straight line can also be accelerated.
2
142 B ma = F – Fg = 50 N – 30 N = 20 N  a = 20 N/3 kg = 6.7 m/s
Velocity is vector sum of compressed spring () and the
143 D rotation ()  
144 E Velocity is zero, but acceleration is downward  K = 0.
Maximum K occurs at the equilibrium position and maximum
146 A U occurs at the release point.
½
½
147 E Ts = Tp  2(m/k) = 2(L/g)  m1/k = L/g  k = m1g/L
T = 2(m/k)½  period T depends on mass m and force
148 E constant k.
The order will not change the final velocity, since it only
149 C depends on the total mass at the end.
2
150 B Fg = GMm/r , where M and r are 2x  Fg = ½ as much (250 N).
151 D vav = d/t = (8 m – 2 m)/1 s = 6 m/s
Center of curvature is twice as far from the mirror as the focal
164 B point.
165 B fn = nf1: f2 = 2f1  f1 = f2/2 = f/2
When the object is placed inside the focal length, then the
166 A image is virtual, upright and larger.
8
6 -1
167 C c = f  3 x 10 m/s = (100 x 10 s )   = 3 m
All waves show interference. Sound is longitudinal and light is
168 E transverse. Light is 106 x faster.
169 A slope = y/x = 20 N/4 s = 5 N/s
170 C Ft = p. Ft = area = ½(4)(20) + (4)(5) = 60 N•s
The air mattress lengthens the stopping time, which reduces
171 E the decelerating force on the stunt person.
172 B TL + TR = Fg  TL + 250 N = 125 N + 500 N  TL = 375 N
173 B  =   (250 N)(4 m) = (500 N)x + (125 N)(2 m)  x = 1.5 m
180 B 0 = mAvA + mBvB  mA/mB = -vB/vA = 2/5
F = ma  10 N – 30cos60 N = (10 kg)a
181 A net
a = -5 N/10 kg = -0.5 m/s2
182 B ac = v2/r = 42r2/T2r = 42R/T2
In order for the ball to stay in circular motion, there must be a
184 D
net force Fc toward the center of the circle.
Left interface, light bends toward normal (D-E).
190 E
Right interface, light bends away from normal (E).
Left interface, light bends away from normal (A-B). Right
191 A
interface, light bends toward normal (A).
f/f  v/vw.
192 A
If frequency from X > frequency from Y, then vx > vy.
193 E nf is extreme with bright reflection  T = ¼
194 E 1/do + 1/di = 1/f  1/8 cm + 1/di = 1/2cm  di = 8/3 cm
195 D Less light reaches the screen  the image is dimmer.
Mv + 2M(-v) = 3Mv'  v' = -v/3
201 D
K = ½[(3M)(-v/3)2 – (3M)v2] = ½[-8/3Mv2] = -4/3Mv2
202 E The first force is , the second force is   the velocity is .
Each force causes kinetic energy to increase by the square of
203 B
time 1/2mv2 = 1/2m(vo + at)2  parabolic shape increases.
207 C Fg = Fc  GMm/r2 = mv2/r  r = r = GM/v2
208 E P = Fv = (900 N)(4 m/s) = 3600 W
Since the ball is thrown straight up, it is farthest when it
211 C
reaches its highest point.
Speed is a scalar value  it reaches its lowest speed when it
212 C
momentarily stops at the top of its flight.
Acceleration is constant and downward during the entire
213 A
flight.
214 A v = vo + at. At 3 s, 0 = vo + (-10 m/s2)(3 s)  vo = 30 m/s
F = ma regardless of any relative velocity the more massive
215 B
box would have the lesser acceleration.
216 D K = ½mv2 and v2 = 2ad  K  d.
At ½H, half of the kinetic energy is converted to Ug  it kinetic
217 B
energy is half the initial kinetic energy.
mbvb = mgvg  mvb = 2mv  vb = 2v
218 E
W = Kb + Kg = ½m(2v)2 + ½(2m)v2 = 3mv2
For elastic collisions both momentum and energy are
219 A
conserved  p = Mv and K = ½Mv2.
The area under the graph equals impulse, J = mv = 6.
220 C
6 kg•m/s = (2 kg)v  v = 3 m/s
221 A Displacement is  to force W = F||d = 0.
222 B Tp = Ts  2(L/g)½ = 2(m/k)½  L/g = m/k L = mg/k
F = GMm/r2. The planet has a greater M, but without knowing
223 D g
r, you can't compare Fg.
The upper string supports both masses,
224 C
 Ft = Fg = mg = (4kg)(10 m/s2) = 40 N
The lower string supports the lower mass and pulls down with
225 E
a force Ft = Fg = ma = (2 kg)(10 m/s2) = 20 N
Amplitude is the change in vertical displacement measured
244 A
from the midpoint  4 cm.
245 C v = /T = 10 cm/0.2 s = 50 cm/s
All are characteristic of waves, which fit both sound and
246 E
electromagnetic waves.
Frequency is proportional to n and red has a lower n. vn = c/n
247 B
 red is faster than violet.
There is less light  bright bands are darker. There is unequal
248 E
interference  dark bands are not as dark.
Angle of reflection is always equal to incident angle. Angle of
249 A
refraction increases, n1sin1 = n2sin2.
250 A 1/do + 1/di = 1/f  1/di = 1/0.4 – 1/0.3 = -1.2 m (left of the lens)
251 D -di  virtual and upright. M = di/do = 1.2/.4 > 1  large.
255 D
256 C
257 C
258 E
259 C
260 B
265 C
266 C
268 D
269 B
270 D
271 A
274 C
276 E
277 E
278 C
280 D
Terms in an equation must have the same units in order to
combine them.
It takes half of the time to travel from ship to shore.
d = vt = (340 m/s)(6 s/2) = 1020 m
Fx = Fy  mxax = myay
mx(2.2 m/s2) = 2mxay  ay = 1.1 m/s2
v2 = vo2 + 2ad. Since vo = 0 and 2a is constant, then the graph
of d vs. v2 is a straight line.
Since py = 0 initially, then py after the collision must also equal
zero, which is impossible for C.
 = r x F. Since the masses cause opposite torques, then net =
L – R = (.6 m)(20 N) – (.4 m)(10 N) = 8 N•m
Speed increases with increased tension (vw = (Ft/m/L)½ vw = f
and  is unchanged f also increases.
sinc = nair/nprism  sin60 = 1/nprism
 nprism = 1/sin60 = 1/3/2 = 2/33
P = K/t = ½mv2/t = ½(900 kg)(20 m/s)2/5 s = 36,000 W
Ff = Fg  Fn = mg  ma = mg  a = g/
T = 2(m/k)½  T6/T3 = (m6/m3)½
ma = m6-m3=(6 s/3 s)2(2 kg) -2 kg = 6 kg
Fg = mg = GMm/r2  g = GM/R2
Adjacent nodal points are ½ apart = 1.0 m
v = f  340 m/s = f(1 m)  f = 340 Hz
py is unchanged. px = -mvcos – mvcos = -2mvcos.
E = hf = qV, when V is doubled, f is also doubled.
tan = x/L. Since tan is constant, then x/L is constant.
10/100 = x/30  x = 3 cm.
tan = Fc/Fg =( mv2/r)/mg = v2/rg
Fall Free Response Answers
1.
a.
b.
c.
d.
2.
a.
b.
c.
d.
e.
m wv w = m s v s
vs = mwvw/ms = (70 kg)(0.55 m/s)/35 kg = 1.1 m/s
Ft = mv
F = mwvw/t = (70 kg)(0.55 m/s)/0.60 s = 64 N
The average forces are equal, but in the opposite
directions. This is the result of Newton's third law,
which states that for every action force there is an
equal but opposite reaction force.
K = Wf
½mv2 = mgd
 = v2/2gd  If  is the same and the son's velocity is
2 x the woman's, then the distance the son travels
must be 22 times farther than the women's, which is
4 x 7 m = 28 m.
d = vot + ½at2
55 m = (25 m/s)(3.0 s) + ½a(3.0 s)2
a = -4.4 m/s2
Fn: , Fg: , Ffr = 
(1) F = Ffr = Fn
ma = mg
 = a/g = 4.4 m/s2/10 m/s2 = 0.44
(2)  is static
a = (v – vo)/t = 25 m/s/10 s = 2.5 m/s2
F = Fs = kx
ma = kx
x = ma/k = (900 kg)(2.5 m/s2)/(9200 N/m) = 0.24 m
Less, since the truck and crate are no longer
accelerating, there is no spring force needed to keep
the crate on the truck bed. With out force acting on
the spring, it will return to its unstretched length.
3.
a.
d.
9.
a.
b.
c.
b.
c.
d.
5.
a.
b.
Virtual because the image forms on the same side of
the lens as the object as a result of the apparent
convergence of rays as seen through the lens.
1/do + 1/di = 1/f
1/6 cm + 1/di = 1/10 cm  di = -15 cm
The image is higher than the previous image.
1/di = 1/f – 1/do = 1/10 cm – 1/9 cm di = -90 cm
M = di/do = -(-15 cm)/6 cm = 2.5 for the first image
and –(-90 cm)/9 cm = 10 for the second image. The
greater magnification produces a larger image.
Fc = mv2/r = m2g  v =( m2gr/m1)½
v = 2r/P
P = 2r/v = 2r(m1/m2gr)½ = 2(m1r/m2g)
P2 = 42m1r/m2g = (42m1r/g)(1/m2)
The quantities that should be graphed are P2 and
1/m2.
d.
e.
K = Ug
½mv2 = mgh
½(2.4 m/s)2 = (10 m/s2)(hmin – 1.20 m)  hmin = 1.5 m
Stopwatch only
Pull the pendulum to one side and let it go at the
same time you start the stopwatch. Time how long it
takes the pendulum to swing back and forth 10
times. Divide the total time by 10; this equals the
period of motion. Frequency = 1/Period.
Length of string. Attach the bob to different lengths
of string. Measure the distance from the string
attachment to the center of the bob and the time it
takes for the bob to swing back and forth 10 times.
Calculate frequency (f = 1/T) and graph the
frequency, f, (y-axis) vs. length, L, (y-axis); also
graph f2 vs. L and 1/f vs. L to see if you get a straight
line.
The temperature would slightly increase because of
the conservation of energy. The kinetic energy of the
pendulum becomes heat as a result of the collision
of the bob with air molecules and friction between
the string and attachment point.
The period of the pendulum would increase. Period
is related to the length of the pendulum: T = 2(L/g)½.
10. a/b.
c.
f
d.
6.
a.
b.
c.
d.
slope = (1.96 – 0.56 s2)/(50 – 12.5 kg-1) = 0.037 kg•s2
slope = 42m1r/g
g = 42m1r/slope = 42(0.012 kg)(0.80 m)/0.037 kg•s2 =
10 m/s2
vn = c/n = 3 x 108 m/s/1.7 = 1.8 x 108 m/s
n = 1/n = 5.2 x 10-7 m/1.7 = 3.1 x 10-7 m
T = ¼n = ¼(3.1 x 10-7 m) = 7.8 x 10-6 m
c.
d.
12. a.
sin = m/d and tan = y/L, but since D >> b, then
ymax/L = m/d   = by3/3D
The fringe spacing would decrease If the index of
refraction increases, the wavelength in that region
decreases. From the relationship in part (c), one can
see that that means a decrease in fringe spacing.
Fn
F
Ff
Fg
8.
a.
K = Ug
½mv2 = mgh
½v2 = (10 m/s2)(2.0 m – 0.50 m)  v = 5.5 m/s
b.
F g Fn
c.
Fc = Fg (Fn = zero for minimum speed)
mv2/r = mg
v2 = rg = (0.60 m)(10 m/s2)  v = 2.4 m/s
b.
c.
d.
e.
f.
Fn = Fgcos20 = (50 kg)(10 m/s2)cos20 = 470 N
F|| = Fgsin20 = (50 kg)(10 m/s2)sin20 = 171 N
Ff = Fn = (0.30)(470 N) = 141 N
F = Fg|| + Ff = 171 N + 141 N = 312 N
W = F•d = (312 N)(3.0 m/sin20) = 2737 J
13. a.
b.
c.
d.
e.
f.
f = c/ = 3 x 108 m/s/2.4 x 10-2 m = 1.25 x 1010 s-1
Xm = mL/d = (1)(2.4 x 10-2 m)(2.5 m)/(0.20 m) = 0.30 m
1 x = 0.00 m
3 x = 0.15 m
2 x = 0.30 m
The center maximum (x = 0 m) has the greatest
intensity, where there is constructive interference.
The minimum at x = 0.15 m is the least because of
destructive inteference. The maxima at 0.30 m is
constuctive interference, but less than the center.