Phys101
Coordinator: Dr. A. A. Naqvi
First Major-131
Wednesday, September 25, 2013
Zero Version
Page: 1
Q1.
Consider two uniform solid spheres A and B made of the same material and having
radii r A and r B , respectively. Find the ratio r B / r A if the mass of sphere B is five times
the mass of sphere A.
A)
B)
C)
D)
E)
1.7
2.2
2.7
1.2
3.3
Ans:
mB = 5mA
sVB = 5sVA ⇒
1
R B = (5)3 R A
Q2.
4π 3
4π 3
RB = 5
R
3
3 A
1
RB
= 53 = 1.71
RA
The position x of a particle is given by
=
x R t3 +
H
R
t2
where x is in meters and t is in seconds. The dimension of H is
A)
B)
C)
D)
E)
Ans:
[H] =
L2T −5
L3T −2
L T −2
ML−3T −2
ML T −5
L × [R]
L
, [R] = 3 = LT −3
2
T
T
L × LT −3
=
= L2 T −5
T2
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-15-s-0-e-1-fg-1-fo-0
Phys101
Coordinator: Dr. A. A. Naqvi
First Major-131
Wednesday, September 25, 2013
Zero Version
Page: 2
Q3.
The velocity of a train is 80.0 km/h , due west. One and a half hour later its velocity
decreases to 65.0 km/h, due west. What is the train’s average acceleration ?
A)
B)
C)
D)
E)
Ans:
Q4.
aavg =
10.0 km/h2
10.0 km/h2
43.3 km/h2
43.3 km/h2
53.3 km/h2
due
due
due
due
due
east
west
west
east
east
65 − 80
= −10 km/h2 due west
1.5
A ball moves in a straight line along the x-axis and Figure 1 shows its velocity as a
function of time t. What is the ball average velocity and average speed, respectively,
over a period of
3.00 s.
Figure 1
A)
B)
C)
D)
E)
0.330 m/s, 2.33 m/s
2.33 m/s , 0.330 m/s
2.33 m/s , 2.33 m/s
1.66 m/s, 2.33 m/s
2.33 m/s, 1.66 m/s
Ans:
Average Velocity =
Average Speed =
=
∆x 2 × 2 − 1 × 3
=
∆t
3
4−3
= 0.330 m/s
3
∆s 2 × 2 + 1 × 3
=
∆t
3
=
7
= 2.33 m/s
3
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-15-s-0-e-1-fg-1-fo-0
Phys101
Coordinator: Dr. A. A. Naqvi
Q5.
Zero Version
Page: 3
The position of an object moving along the x-axis is given by x = 6.0 + 6.0 t − 3.0 t2,
where x is in meters and t in seconds. Which statement about this object is correct?
A)
B)
C)
D)
E)
Ans:
First Major-131
Wednesday, September 25, 2013
The object is momentarily at rest at t = 1.0 s.
The object position is negative at t = 0 s.
The acceleration of the object is zero at t = 0 s.
The acceleration of the object is positive at all times.
The object is momentarily at rest at t = 2.0 s.
x = 6 + 6t − 3t 2 m
v=
dx
= 6 − 6t m/s
df
For v = 0 , 6 − 6t = 0, Then v = 0, at t = 1 sec
a=
Q6.
dv
= −6 m/s2
dt
x(t = 0) = +6.0 m
A rock is thrown vertically upward from ground level at time t =0.0 s. At t =1.5 s it
passes the top of a tall tower, and then 1.0 s later it reaches its maximum height. What
is the height of the tower?
A)
B)
C)
D)
E)
Ans:
26 m
62 m
36 m
16 m
20 m
1
Tower Height H = viy t − gt 2
2
but vfy = viy − gt
for maximum heigh vfy = 0, and t = 1.5 + 1.0 = 2.5 sec
then viy = g × t = 9.8 × 2.5 = 24.5 m/s
1
H = viy t − gt 2
2
viy = 24.5 m/s, t = 1.5 sec
1
H = 24.5 × 1.5 − × 9.8 × (1.5)2
2
H = 25.7 m = 26 m
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-15-s-0-e-1-fg-1-fo-0
Phys101
Coordinator: Dr. A. A. Naqvi
First Major-131
Wednesday, September 25, 2013
Zero Version
Page: 4
Q7.
A man walks 50 m in a direction 37o north of east at 5.0 m/s, then 60 m
south at 4.0 m/s. How long would it take him to get back to his starting point
at 5.0 m/s by the shortest path?
A)
B)
C)
D)
E)
10 s
15 s
20 s
5.0 s
3.5 s
Ans:
∆x = 50cos37 = 39.9 m
∆y = 50 sin 37 − 60 = 30.09 − 60 = −29.9
Q8.
r = �(∆x)2 + (∆y)2 = 49.9 m, t =
r 49.9
=
= 9.98 sec
v
5
Vector �A⃗ has a magnitude of 35.0 m and makes an angle of 37.0o with the
�⃗ that is in the direction opposite to vector �A⃗
positive x axis. Find a vector B
�⃗.
and is one fifth the magnitude of A
A)
B)
C)
D)
E)
Ans:
�B⃗ = −
– (5.59 m) iˆ – (4.21m) jˆ
(5.59 m) iˆ + (4.21 m) jˆ
(0.798 m) iˆ – (0.602 m) jˆ
– ( 1.56 m) iˆ –(5.06 m) jˆ
– ( 0.798 m) iˆ + (0.602 m) jˆ
�A⃗
5
�A⃗ = 35cos37𝚤⃗ + 35sin37𝚥⃗
�A⃗ = 27.95𝚤⃗ + 21.06𝚥⃗
1
�B⃗ = − (27.95𝚤⃗ + 21.06𝚥⃗)
5
= −5.59𝚤⃗ − 4.21𝚥⃗
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-15-s-0-e-1-fg-1-fo-0
Phys101
Coordinator: Dr. A. A. Naqvi
First Major-131
Wednesday, September 25, 2013
Zero Version
Page: 5
Q9.
If �A⃗ = 2 iˆ
A)
B)
C)
D)
E)
Ans:
Q10.
�⃗ xB
�⃗). �C⃗.
+ 3 jˆ , �B⃗= iˆ – jˆ and �C⃗= iˆ + jˆ , find (A
0
–6
+6
–3 kˆ
+2 iˆ
�⃗
�⃗ = 𝐴⃗ × 𝐵
�⃗ = 𝑘
C = 𝚤⃗ + 𝚥⃗ ; 𝐷
�⃗. 𝐶⃗ = 0
𝐷


The scalar product of vectors A and B is 6.00 and the magnitude of their vector
product is 9.00. Find the angle between these two vectors.
A)
B)
C)
D)
E)
Ans:
ABcosθ = A. B = 6 , |A × B| = 9.0 = ABsinθ
tanθ =
Q11.
56.3o
43.0o
23.4o
37.5o
90.0o
A. B
9
= = 1.5
|A × B| 6
θ = tan−1 (1.5) = 56.3°


The position of a particle is given by r = (4t – t2) iˆ + t3 jˆ , where r is in meters and t
in seconds. Find the average acceleration (in m/s2) of the particle in the time interval
between t = 2 s and t = 4 s.
A)
B)
C)
D)
E)
Ans:
–2 iˆ +18 jˆ
–4 iˆ – 6 jˆ
–5 iˆ – 10 jˆ
–7 iˆ – 12 jˆ
–10 iˆ – 6 jˆ
dr
= (4 − 2t)ı̂ + 3t 2 ȷ̂
dt
v(t = 2s) = (4 − 4)ı̂ + 12ȷ̂ = 12ȷ̂
v(t = 4s) = (4 − 8)ı̂ + 48ȷ̂ = −4ı̂ + 48ȷ̂
∆v −4ı̂ + 48ȷ̂ − 12ȷ̂ −4ı̂ + 36ȷ̂
aavg =
=
=
= −2ı̂ + 18ȷ̂
∆t
2
2
v=
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-15-s-0-e-1-fg-1-fo-0
Phys101
Coordinator: Dr. A. A. Naqvi
First Major-131
Wednesday, September 25, 2013
Zero Version
Page: 6
Q12.
A projectile is thrown from the ground into the air with an initial speed v 0 . Its
velocity, 1.50 s after it was thrown, is 42.3 m/s making an angle 30.40 above
the horizontal. Determine the initial velocity v 0 of the projectile.
A)
B)
C)
D)
E)
Ans:
51.3 m/s at 44.70 above the horizontal
43.1 m/s at 34.20 above the horizontal
21.6 m/s at 49.20 above the horizontal
32.5 m/s at 23.50 above the horizontal
12.2 m/s at 54.50 above the horizontal
vy (t = 1.5 sec) = vsinθ = 42.3 sin(30.4) = 21.4 m/s
v0x = vcosθ = 42.3 cos(30.4) = 36.48 m/s
v0y = vy + gt = 21.41 + 9.8 × 1.5 = 36.11m/s
v0 = �v0x 2 + v0y 2 = �36.112 + 36.482 = (51.3 m)/s,
Q13.
v0y
36.11
θ = tan−1 � � = tan−1 �
� = 44.7°
v0x
36.48
A 0.150 kg ball, attached to the end of a string, is revolving uniformly in a horizontal
circle of radius 0.600 m. The ball makes 10.0 revolutions in 5.00 seconds. Calculate
the centripetal acceleration of the ball?
A)
B)
C)
D)
E)
Ans:
94.8 m/s2
25.7 m/s2
12.6 m/s2
9.81 m/s2
zero
v2
2πR
5 sec
1
;v =
;T =
= sec
R
T
10 rev
2
2π × 0.6
v=
= 7.54 m/s
0.5
a=
(7.54)2
a=
= 94.8 m/s
0.6
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-15-s-0-e-1-fg-1-fo-0
Phys101
Coordinator: Dr. A. A. Naqvi
First Major-131
Wednesday, September 25, 2013
Zero Version
Page: 7
Q14.
A boat is to travel from point A to point B directly across a river. The water in the
river flows with a velocity of 1.20 m/s toward the west, as shown in Figure 3. If the
speed of the boat in still water is 1.85 m/s, at what angle from the north must the boat
head?
Figure # 3
A)
B)
C)
D)
E)
Ans:
40.40 east of north
30.20 west of north
10.50 east of north
90.00 west of north
55.00 west of north
1.2
θ = sin−1 �
�
1.85
= 40.44° of north
V=1.20 m/s
θ
1.85 m/s
Q15.
Which one of the curves shown in Figure 2 best represents the vertical component of
the velocity v y versus time t for a projectile fired at an angle of 450 above the
horizontal?
Figure 2
A)
B)
C)
D)
E)
AE
AB
OC
DE
AF
Ans:
vy = viy − gt
∂y
= −g , line with − ve slope
∂t
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-15-s-0-e-1-fg-1-fo-0