Download exam3_T113

Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 1
Q1.
A particle moves in one dimension such that its position x(t) as a function of time t is
given by x(t) = 2.0 + 27 t – t3, where t is in seconds and x(t) is in meters. At what position
does the particle change its direction of motion?
A)
B)
C)
D)
E)
56 m
54 m
27 m
81 m
100 m
Ans:
dx
= 0 = 27t − 3t 2
dt
⇒ t = 3.0 s
x(t) = 2.0 + 27 t − t 3 ⇒ x(3) = (2 + 27 × 3 − 27)m
x(3) = (2 + 81 − 27)m = 56.0 m
Q2.
What is your total displacement when you follow directions that tell you to walk 40.0 m
north then 30.0 m east?
A)
B)
C)
D)
E)
50.0 m, 53.1o North of East
50.0 m, 53.1o East of North
50.0 m, 45.0o East of North
45.0 m, 45.0o East of North
45.0 m, 30.1o North of East
Ans:
North
30 m
40 cm
West
��⃗ = �30 �i + 40 �j � m
D
��⃗� = 50.0 m
�D
Qx = cos−1 �
30
� = 53.1°
50
�⃗
D
θx
South
East
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 2
Q3.
A projectile is fired from the ground with an initial velocity
=
vo (30.00 iˆ + 19.62 ˆj ) m/s.
Find the horizontal distance the projectile travels before hitting the ground.
A)
B)
C)
D)
E)
120.0 m
60.00 m
78.48 m
156. 9 m
400.3 m
Ans:
v0x = 30.0 m/s ; v0y = 19.62 m/s
Time to reach ground:
0
∆y = 19.62 t −
Q4.
1 2
2 × 19.6
gt ⇒ t =
= 4.0 s
2
9.8
∆x = v0x xt = (30.0 × 4.0) = 120.0 m
There are two forces on the 3.00 kg box in the overhead view of Figure 1, but only one is
shown. For F 1 = 20.0 N, a = 12.0 m/s2, and θ = 30.0o, find the second force in unit vector
notation.
A) (−38.0 N ) iˆ + (−31.2 N ) ˆj
B) (−51.2 N ) iˆ − (+18.0 N ) ˆj
C) (−38.0 N ) iˆ + (+31.2 N ) ˆj
D) (+38.0 N ) iˆ + (+31.2 N ) ˆj
E) (−31.2 N ) iˆ + (−38.0 N ) ˆj
Ans:
�F⃗1 + �F⃗2 = ma�⃗
20.0 �i + �F⃗2 = 3.0 �−12.0 Sinθ �i − 12.0 cos θ �j �
= −36.0 �0.5 �i + 0.866 �j � N
= �− 18.0 �i − 31.2 �j � N
�⃗2 = �− 20.0 �i − 18.0 �i − 31.2 �j � N
∴F
�⃗2 = (−38.0 N) �i + (−31.2 N) �j
F
Fig# 1
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 3
Q5.
A block slides with a constant acceleration down a rough plane inclined at an angle of
40.0o with the horizontal. If the block starts from rest and slides a distance of 39.2 m
down the plane in 4.00 s, calculate the coefficient of kinetic friction between the block
and the plane.
A)
B)
C)
D)
E)
0.186
0.215
0.125
0.110
0.098
Ans:
fk
θ
mgsinθ − µk mgcosθ = ma
a = gsin400 − µk gcos40° = 9.8 × 0.643 − µk × 9.8 × 0.766
∴ a = 6.299 − 7.51µk
1
∆x = v0 t + 2 at 2
39.2 = 0 +
1
8
(6.299 − 7.51 µk ) × 16
2
39.2 = 50.39 − 60.08µk ⇒ µk = 0.186
Q6.
A 100-kg parachute falls at a constant speed of 1.00 m/s. At what rate is energy being
lost?
A) 980 W
B) 19.8 W
C) 89.0 W
D) 49.0 W
E) 490 W
Ans:
P = mgv
= 100 × 9.8 × 1.0 = 980 W
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 4
Q7.
The work done by a conservative force acting on a body
A)
B)
C)
D)
E)
does not change the total energy.
does not change the potential energy.
does not change the kinetic energy.
is always equal to zero.
is always equal to the sum of the changes in potential and kinetic energies.
Ans:
Q8.
𝐀
The velocity of a given body is increased to such an extent that the kinetic energy of the
body is increased by a factor of 16. The momentum of the body is increased by a factor
of:
A)
B)
C)
D)
E)
4.0
16
8.0
2.0
1.0
Ans:
ki =
kf =
⇒
1
P1 2
mv1 2 =
2m
2
1
P2 2
mv2 2 =
2
2m
kf
P2 2
= � �
ki
P1
P2
4= � �
P1
P2 = 4.0 P1
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 5
Q9.
A 32.0-kg thin loop has a radius of 2.00 m and is rotating at 280 rev/min about its central
axis. What is the average power required to bring it to a stop in 20.0 s?
A)
B)
C)
D)
E)
2.75 × 103 W
1.32 × 103 W
3.15 × 103 W
1.32 × 102 W
9.38 × 101 W
Ans:
ωi = �280
rev
min
� �2π
rad
rev
min
� � 60 s� = 29.3 rad/s
I = MR2 = 32.0 × (2.0)2 = 128 kg. m2
1
1
2
2
× 128 × (29.3)2
W
∆k �2 I (ωf − ωi )�
2
P=
=
=
=
J/S
∆t
∆t
20.5
20
P = 2.75 × 103 W
Q10.
A uniform solid sphere rolls down an incline. What must be the incline angle if the linear
acceleration of the center of the sphere is to have a magnitude of 1.96 m/s2?
A)
B)
C)
D)
E)
16.3o
45.0o
30.0o
60.0o
25.4o
Ans:
a=
−g sin θ
−g sin θ
−gsinθ
=
=
2
I
7
2 MR
I+
1
+
2
2
5
MR
5 MR
∴a=
−5
5
gsinθ ⇒ −1.96 = − × 9.8 × sinθ
7
7
sinθ = 0.280
θ = 16.26° = 16.3°
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 6
Q11.
A uniform beam is held in a vertical position by a pin at its lower end and a cable at its
upper end (see Figure 2). The tension in the cable is 72.0 N. Find the horizontal force F
acting on this beam.
Fig# 2
A)
B)
C)
D)
E)
99.8 N
120 N
65.0 N
135 N
50.3 N
Ans:
� τ0 = 0
Q12.
5 F = 72 cos 30° × 8
72 cos 30° × 8
= 99.76 N = 99.8 N
F=
5
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.00 g
coins stacked over the 10.0 cm mark, the stick is found to balance at the 45.5 cm mark.
What is the mass of the meter stick?
A)
B)
C)
D)
E)
78.9 × 10−3 kg
78.9 kg
98.7 × 10−3 kg
50.0 × 10−3 kg
25.0 × 10−3 kg
10 cm
0
cm
100 cm
2 mg
45.5
Mg
Ans:
� τ0 = 0
2 mg (45.5 − 0.10) = Mg (50 − 45.5)
2 × 5.0 × 10−3 (35.5) = M (4.5) ⇒ M =
10 × 10−3 × (35.5)
= 78.9 × 10−3 kg
4.5
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 7
Q13.
A 12-gram bullet is fired into a 1.0 × 102-gram wooden block initially at rest on a
horizontal surface. After impact, the bullet is imbedded inside the wooden block which
slides 7.5 m before coming to rest. If the coefficient of friction between the block and the
surface is 0.65, what was the speed of the bullet immediately before impact?
A)
B)
C)
D)
E)
91.2 m/s
49.0 m/s
25.0 m/s
100 m/s
190 m/s
Ans:
→ vb
mb
VB = 0
100 g
MB
d = 7.5 m
µk = 0.65
mb vb = (mb + MB )v → (1)
∆k + ∆ug = Wnc → (2)
1
0 − 2 (mb + MB )v 2 = −µk (mb + MB )gd ⇒ v = �2µk gd = 9.77 m/s
Q14.
From equation (1): 12 × 10−3 vb = 112 × 10−3 × 9.77 ⇒ vb = 91.2 m/s
A certain wire stretches 1.00 cm when a force F is applied to it. The same force is applied
to a second wire of the same material with equal length but with twice the diameter. The
second wire stretches:
A)
B)
C)
D)
E)
0.25 cm
4.0 cm
2.5 cm
0.50 cm
2.0 cm
Ans:
∆L1
F1
=E
A1
L1
F2
∆L2
=E
A2
L2
F2 A1
∆L2 L1
⇒ � �� � = �
�� �
L
F1 A2
∆L1
2
A1
1
∴ ∆L2 = � � ∆L1 = ∆L1 = 0.25 cm
A2
4
Phys101
Coordinator: Dr. G. Khattak
Q15.
Final-113
Thursday, August 02, 2012
Zero Version
Page 8
A cube of volume 1.0 m3 is made of material with a bulk modulus of 2.0 × 109 N/m2.
When it is subjected to a pressure of 2.0 × 105 Pa, the magnitude of the change in its
volume is:
1.0 × 10−4
0.25 × 10−4
1.4 × 10−4
3.5 × 10−2
2.4 × 10−3
A)
B)
C)
D)
E)
m3
m3
m3
m3
m3
Ans:
∆P = −B
∆V = −
∆V
V
∆P×V
B
=
2×105 ×1.0
2.0×109
|∆V| = 1.0 × 10−4 m3
= −1.0 × 10−4 m3
Q16.
Let the acceleration due to gravity on the surface of Earth be g E . The acceleration due to
gravity (g p ) on the surface of a planet whose mass is equal to that of Earth but whose
radius is only 0.100 R E is given by (R E is Earth’s radius)
A)
B)
C)
D)
E)
gp
gp
gp
gp
gp
= 100 g E
= 10.0 g E
= 50.0 g E
= 25.0 g E
= 1.00 g E
Ans:
ge =
gp =
GMe
R2
GMp
Rp2
g p = 100 g e
2
gp
Re
Re 2
⇒
= � � = �
�
ge
Rp
0.1R e
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 9
Q17.
A solid uniform sphere has a mass of 1.00 × 104 kg and a radius of 1.00 m. What is the
magnitude of the gravitational force due to the sphere on a particle of mass m = 1.00 kg
located at a distance of 0.500 m from the center of the sphere
A)
B)
C)
D)
E)
3.34 × 10−7 N
1.17 × 10−5 N
6.68 × 10−3 N
1.50 × 10−5 N
1.67 × 10−8 N
m
r
GmM ′
r2
M′ =
Q18.
4
3
M
� πR3 �
4
× �3 πr 3 � =
R
M′
Ans:
F=
M
Mr3
R3
GmMr
6.67 × 10−11 × 1.0 × 1 × 104 × 0.5
F=
=
= 3.34 × 10−7 N
(1.0)3
R3
Figure 3 gives the potential energy function U(r) of a projectile, plotted outward from the
surface of a planet of radius R s . What kinetic energy is required of a projectile launched
at the surface if the projectile is to “escape” the planet?
Fig # 3
A)
B)
C)
D)
E)
+5.0 × 109 J
–5.0 × 109 J
+4.0 × 109 J
−4.0 × 109 J
−9.0 × 108 J
Ans:
∆K + ∆U = 0 ⇒ ∆K = − ∆U
k f − k i = −(Uf − Ui )
k i = (0 − Ui )
k i = −Ui
k = 5 × 109 J
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 10
Q19.
A satellite travels in a circular orbit around a certain planet. The orbit has a radius of
10 × 106 m and a period of 8.0 hours. What is the mass of the planet?
A)
B)
C)
D)
E)
7.1 × 1023 kg
9.1 × 1020 kg
9.8 × 1017 kg
4.9 × 1026 kg
8.5 × 1029 kg
Ans:
T2 =
4π2 3
r
GM
22 2
� × 1021
4π 3
7
r =
= 7.1 × 1023 kg
M=
GT 2
6.67 × 10−11 × (8 × 3600)2
2
(4) �
Q20.
A 250-kg spaceship is in a circular orbit of radius 3.00 R E about the earth. How much
energy is required to transfer the spaceship to a circular orbit of radius 4.00 R E ? (R E is
the radius of the earth)?
A) 6.52 × 108 J
B) 3.52 × 1011 J
C) 5.32 × 1014 J
D) 4.37 × 105 J
E) 8.97 × 101 J
Ans:
Ei = −
Gm ME
Gm ME
; Ef = −
2(3.00R E )
2(4.00R E )
∆E = −
∆E =
Gm ME 1 1
Gm ME
1
� − �= +
×
2R E 4 3
2R E
12
Gm ME
6.67 × 10−11 × 250 × 5.98 × 1024
=
J
24R E
24 × 6.37 × 106
⇒ ∆E = 6.52 × 108 J
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 11
Q21.
Find the pressure increase in the fluid in a syringe when a nurse applies a force of 31.4 N
to the syringe’s circular piston, which has a radius of 1.00 cm.
A)
B)
C)
D)
E)
1.00 × 105 Pa
1.50 × 104 Pa
2.00 × 103 Pa
3.00 × 106 Pa
5.00 × 103 Pa
Ans:
P=
F
31.4 N
=
= 9.99 × 104 N/m2
A
π(10−2 )2
Q22.
In Figure 4, a spring of spring constant 5.00 × 104 N/m is between a rigid beam and the
output piston of a hydraulic lever. An empty container with negligible mass sits on the
input piston. The input piston has area A, and the output piston has area 20 A. Initially
the spring is at its rest length. How many kilograms of sand must be (slowly) poured into
the container to compress the spring by 4.00 cm?
A)
B)
C)
D)
E)
10.2 kg
100 kg
9.80 kg
980 kg
250 kg
Fig # 4
Ans:
F2
kx
F1
=
=
A2
A1
A2
F1 = kx �
A1
�
A2
A
F1 = 5 × 104 × 4 × 10−2 × �20 A�
m=
F1
g
=
5×104 ×4×10−2
20×9.8
=
102
9.8
kg = 10.2 kg
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 12
Q23.
A hollow spherical iron shell floats almost completely submerged in water. The outer
diameter is 50.0 cm, and the density of iron is 7.87 g/cm3. Find the inner diameter.
A)
B)
C)
D)
E)
47.8 cm
12.5 cm
15.5 cm
7.87 cm
25.0 cm
𝑟𝑜𝑢𝑡 𝑟𝑖𝑛
Ans:
ρf Vsubmerged g = ρ0 V0 g
4
4
1.0 × 3 π(50)3 = 7.87 × 3 π(503 − rin 3 )
(50)3 = 7.87(503 − rin 3 ) ⇒ rin 3 = (50)3 −
(50)3
⇒ rin = 47.8 cm
7.87
Q24.
Two streams merge to form a river. One stream has a width of 8.0 m, depth of 4.0 m, and
current speed of 2.0 m/s. The other stream is 7.0 m wide and 3.0 m deep, and flows at 4.0
m/s. If the river has a width of 10.0 m and a speed of 4.0 m/s, what is its depth?
A)
B)
C)
D)
E)
3.7 m
4.0 m
3.0 m
8.0 m
7.0 m
Ans:
A1 v1 + A2 v2 = A3 v3
32 × 2 + 21 × 4 = 10 d × 4
64 + 84 = 40 d
d = 3.7 m
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 13
Q25.
A liquid of density 1.00 × 103 kg/m3 flows through a horizontal pipe that has a crosssectional area of 2.00 × 10 − 2 m2 in region A and a cross-sectional area of 10.0 × 10 − 2 m2
in region B. The pressure difference between the two regions is 10.0 × 103 Pa. What is
the mass flow rate between regions A and B?
A)
B)
C)
D)
E)
91.3 kg/s
87.3 kg/s
857 kg/s
576 kg/s
100 kg/s
Ans:
A1 v1 = A2 v2
2 × 10−2 v1 = 10 × 10−2 v2 ⇒ v1 = 5v2
0
1
1
P1 + 2 ρv1 2 + ρgy1 = P2 + 2 ρv2 2 + ρgy2
0
1
1
P2 − P1 = ρ(v1 2 − v2 2 ) = × 103 �25v2 2 − v2 2 �
2
2
∴ 10 × 103 =
v2 2 =
1
2
× 103 × 24v2 2
10
10
⇒ v2 = � = 0.9313 m/s
12
12
∴ Mass �low rate: = A2 v2 ρ = 91.3 kg/s
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 14
Q26.
What is the phase constant for the harmonic oscillator with the velocity function v(t)
given in Figure 5 if the position function x(t) has the form x(t) = x m cos(ωt + φ)? The
vertical axis scale is set by 𝑣 s = 4.00 cm/s.
R
Fig# 5
A)
B)
C)
D)
E)
−53.1o
+53.1o
−5.24o
+5.24o
−40.0o
Ans:
From Fig. vs = 4, v =
v=
4
× 5 = 5 cm/s
4
5
v
4 s
x = xm cos(ωt + ϕ)
dx
= −xm ωsin(ωt + ϕ)
dt
vs = −vsin(ωt + ϕ)
4 = −5sin(ϕ); t = 0
sin(ϕ) = −
4
⇒ ϕ = sin−1 (−4/5) = −53.1°
5
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 15
Q27.
If the phase angle for a block-spring system in SHM is π/6 rad and the block’s position is
given by x(t) = x m sin(ωt + φ), what is the ratio of the kinetic energy to the potential
energy at time t = 0?
A)
B)
C)
D)
E)
3
1/3
9
1/9
5
Ans:
Q28.
x (t) = xm sin(ωt + ϕ) ; v = ωxm cos(ωt + ϕ)
1
1
U = kx 2 = kxm 2 sin2(ϕ)
2
2
1
1
1
k = mv 2 = mω2 xm 2 cos(ϕ) = k xm 2 cos 2 (ϕ)
2
2
2
2
π
cos �6�
cosϕ
k
= �
∴ =
π �
sinϕ
U
sin �6�
k
∴ =3
U
Find the mechanical energy of a block-spring system having a spring constant of 2.0
N/cm and an oscillation amplitude of 3.0 cm.
A)
B)
C)
D)
E)
9.0 × 10−2 J
7.8 × 10−1 J
12 × 10−3 J
13 × 10+2 J
5.0 × 10−3 J
Ans:
E=
1
2
kxm 2 =
1
�2.0
2
N
102 cm
��
cm
1m
−2
= 102 × 9 × 10−4 = 9 × 10
E = 9 × 10−2 J
� × (3 × 10−2 m)2
J
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 16
Q29.
In Figure 6, a stick of length L = 1.73 m oscillates as a physical pendulum. What value of
𝑥 between the stick’s center of mass and its pivot point O gives the least period?
Fig# 6
A)
B)
C)
D)
E)
0.50 m
0.87 m
1.73 m
0.58 m
0.43 m
Ans:
T = 2π�
I=
I
;d = x
mgd
1
mL2 + mx 2
12
1
1
2
2 2
mL
+
mx
�
T = 2π �12
mgx
1
mL2 + mx 2
12
2
2
T = 4π �
�
mgx
4π2 1 2 −1
T =
� L x + x�
g 12
2
2T
dt
4π2
1
=
�− L2 x −1 + 1� = 0
dx
g
12
(1.73)2
L2
L2
2
=
1
⇒
x
=
=
12x 2
12
12
or x 2 = 0.249 ⇒ x = 0.499 m
Phys101
Coordinator: Dr. G. Khattak
Final-113
Thursday, August 02, 2012
Zero Version
Page 17
Q30.
Figure 7 shows the kinetic energy K of a simple pendulum versus its angle θ from the
vertical. The vertical axis scale is set by 𝐾𝑠 = 20.0 mJ. The pendulum bob has mass 0.30
kg. What is the length of the pendulum?
Fig# 7
A)
B)
C)
D)
E)
2.04 m
1.25 m
2.25 m
0.500 m
3.05 m
Ans:
θ = 100 × 10−3 rad = 5.73°
k max = Umax = mgL(1 − cosθ)
∴ 30 × 10−3 = (0.30)(9.8)(L)(1 − Cos 573°)
30 × 10−3 = (0.30)(9.8)(L)(1 − 0.995)
30 × 10−3
L=
(0.30)(9.8)(0.00499)
L = 2044 × 10−3 m = 2.04 m