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AIR UNIVERSITY
Final Exam Fall 2013
Subject: Physics 105 Maximum Marks: 45 Maximum Time: 3 Hrs
Class: BEE 1st Semester
Date: 06-01-2014
Instructors: A. Sadiq, M. Anwar, M. Atif, R. Nasir, Z. Ali and K. Ali
Note: Attempt all the 9 questions. Each question carries equal marks. Use of Scientific calculators is allowed.
2
1
1 A particle with an initial speed v0  6ms and constant acceleration a  4ms starts from an initial
position x0
 5m .
a) Find its velocity and acceleration after t  2s
b) When do its speed and displacement become zero?
c) Sketch its speed and displacement as a function of t
Solution: a) For constant acceleration a we have,
(2)
(1)
(2)
vt   v0  at , and,
1
xt   x0  v0t  at 2
2
Therefore for t  2s we get,
vt  2  6  4  2  2ms1
1
xt  2  5  6  2  4  4  1m
2
b) The speed of the particle becomes zero when,
 6  4t  0 , or t  3 / 2  1.5s
The displacement of the particle becomes zero when,
5  6t  2t 2  0 , or, t 
x(t)
0
t
6  36  40 3  i

,
4
2
This means that in this case the displacement never becomes zero.
c) Fig shows sketches of v (t ) , straight line, and x(t ) , curved line, as functions of
2 Vector
v(t)
A of magnitude 4 makes an angle   30
0
with the x axis and vector
B
A.
of
magnitude 3 makes an angle   60 with the x axis.
a) Calculate their sum and the magnitude and direction of their cross product.
0
(2.5)
1
b) You are to go across a channel with water flowing at an average speed v  1ms in a boat
1
at v  2ms . In which direction would you go relative to the flow?
c) How long will it take you to cross a 100m wide channel?
Solution: a) The components of the two given vectors
A and B
(1.5)
(1)
are;
Ax  4 cos 300  2 3 , Ay  4 sin 300  2 ; Bx  3 cos 600  3 / 2 , By  34 sin 600  3 3 / 2 ,
This gives the components of their sum,
Cx  Ax  Bx  2 3  3 / 2 , and C y  Ay  By  2  3 3 / 2
Therefore the magnitude of the sum of the two vectors is;
2
C  Cx  C y 
2
2
 
2

2
3  3 / 2  2  3 3 / 2  25  12 3  6.77
Making an angle  with the x -axis given by,
1




Cy 
1 2  3 3 / 2
1 4  3 3
1
0
  tan 
  tan 0.93  45
  tan 
 Cx 
2 3  3/ 2
4 3  3 
  tan 1 
The magnitude of the cross product of the two given vectors
AB sin 300  12 
A and B
B
is,
1
6
2

A
Its direction is perpendicular to plane of these two vectors pointing out of the paper as shown by the
black dot in the figure.
b) Here we also have a problem of vector addition with the average velocity of river flow as one vector
and the velocity of the boat as the other vector. We need to find the direction of the boat velocity vb
that makes the resultant vector perpendicular to the river flow. This means that the component
boat velocity along the river flow must be equal and opposite to average velocity of the river flow.
From the figure we it follows that,
vr  vb sin  , or,
  sin 1

vr
vr
1
 sin 1  300
vb
2
c) Time taken to cross the river t

100
100

 57.7 s
2 cos 30
3
3 Fig 1 shows a block of mass M  1kg lying on a rough inclined surface with s
 0.5 , k  0.4 .
2
Take g  10ms
a) With the help of a free body diagram find the angle when the block just starts sliding down.
(2)
b) Find its downward acceleration for an angle of inclination   30 .
(1.5)
c) Starting from rest how long does the block in part b) takes to slide down by 1m and how much
velocity does it acquire?
(1.5)
Solution: a) The forces acting on the block, shown by arrows in the following free body diagram, are;
Component of its weight along the plane Mg sin 
0
Component of its weight perpendicular to the plane Mg cos 

Normal Reaction of the plane on the block N  Mg cos , and
Force of static friction acting up along the plane Fs
 s N  s Mg cos
Just before the block starts sliding we have,
Fs  s N  s Mg coss  Mg sin  s , or,
s  tan 1  s  tan 1 0.5  270
b) The block will start sliding down the plane for an angle of inclination   30 with acceleration,
0
a  g sin 30  k cos 30  9.80.5  .04  0.87   1.5ms2
c)
t  2as  2  0.15  1  0.3  0.55s
4 Now consider pushing the block in Fig 1 that is initially at rest up the same rough inclined plane with
a force F  10N . Its angle of inclination remains   30 .
a) Find the work done in pushing the block by 1m up the plane.
b) Find the change in its potential energy.
c) Calculate the change in its kinetic energy.
d) Using the law of conservation of energy find the work done against friction.
0
(1)
(1)
(2)
(1)
2
 Fd  10J
0
b) Change in the potential energy of the block U  Mgh  1  9.8  sin 30  4.9 J
Solution: a) Work done on the block W
c) For calculating the change in the kinetic energy of the block we need to find its acceleration up
along the plane. This is given by,
Fn
 10  g sin 30  k cos 30  10  9.80.5  .04  0.87   2.2ms 2
M
v 2  2aS  4.4m 2 s 2
1
K  Mv 2  2.2 J
2
a
d) Since the total energy of the block is conserved, the difference of the work done and the total
change in its kinetic and potential energies must be the work done against the force of kinetic friction;
W f  Ff d  W  U  K   10  7.1  2.9 J
This can also be directly calculated by first calculating the force of kinetic friction.
5 Fig 2 shows a mass m  200 g lying on a smooth horizontal surfaced attached to a spring of spring
1
constant k  100 Nm .
a) Find the magnitude and direction of the force acting on m when it is displaced by x0
 2mm
d 2x
  w2 x describes the motion of m after it is released.
2
dt
c) Find the amplitude and angular frequency w of oscillations of m ?
d) Show that xt   x0 coswt    is the solution of the above equation of motion of m .
e) Find the value of  when xt  0  0
2
 0.2 N
Solution: a) Magnitude of force: F  kx  100 
1000
b) Using Newton’s 2nd law show that
(1)
(1)
(1)
(1)
(1)
Its direction will be opposite to the displacement.
k
d 2x
d 2x
k
  x , with w2 
,
or
we get the desired relation.
2
2
m
dt
dt
m
1 k
1 100
c) Amplitude x0  2mm , frequency f  w / 2 

 7.1Hz
2 m 2 0.2
dxt 
d 2 xt 
  wx0 sin wt    ,
d)
  w2 x0 coswt      w2 x
2
dt
dt
e) xt  0  x0 cos   0 , gives    / 2
b) F
 kx  ma  m
1
6 A m  100 g ball moving with v  6ms elastically collides head-on with an identical ball at rest.
a) What are the linear momentum and kinetic energies of the two balls before their collision?
(1)
b) Using the conservation of kinetic energy and momentum find their velocities after the collision. (2)
c) What is the impulse imparted to the ball at rest during the collision?
(1)
d) For a collision time   0.1s find the average force exerted by the moving ball on the ball at rest. (1)
Solution: a) The linear momentum and kinetic energies of the two balls before collision,
p1  0.1 6  0.6kgms1 , and p2  0kgms1
K1  0.5  0.1 36  1.8J , and K2  0 J
3
b) Let v1 , v2 be the velocities of the two balls just before and v1 ' , v2 ' be their velocities just after their
collision.
Since v2
 0 , the conservation of their total linear momentum gives,
v1  v  v , or
`
v1  v  v
'
1
'
1
'
2
'
2
(1)
The conservation of their total kinetic energy gives,
2
2
2
2
v 21  v1'  v2' , or
v1  v1'  v2'
2
2)
Dividing EQ (2) by EQ (1) we get,
v1  v1'  v2'
(3)
Adding EQs (1) and (3) we get,
v2'  v1  6ms1
'
Substituting this value of v2 in EQ (3) gives,
v2'  v1
Therefore when a particle collides with another particle of the same mass that is at rest then the first
particle comes to a complete stop and the second particle takes off with its velocity.
c) Impulse imparted to the second particle during the collision is the change in its momentum during
1
the collision J 2  p2  p  0.6kgms
d) Average force exerted during the collision by the first particle on the second particle,

F  J 2 / 
0.6
 6N
0.1
7 Two particles m1 and m2 are at a distance l from each other as shown in Fig 3.
n
n
i 1
i 1
 mi xi /  mi find the distance of their CM from m1 .
a) Using the definition of CM xc 
(2)
(1)
 m2  m and for m1  m2 ?
c) Calculate the distance of the CM of uniform thin rod of mass m and length l from one of its ends. (2)
b) What are the values of xc for m1
8 a) Using the definition I 
n
m x
i 1
i i
2
show that the moment of inertia of a uniform thin ring of
2
mass m and radius R about an axis through its CM and perpendicular to its plane is mR .
(1.5)
b) Find its moment of inertia I d about one of its diameters for m  250 g and R  30cm.
(2)
c) Find the kinetic energy stored in the ring when it is rotating at 100rpm about its diameter.
(1.5)
9 Fig 4 shows the ring in question 8 b) suspended from a nail through its rim perpendicular to its
plane.
a) How much work is done in rotating this ring through an angle of   15 ?
b) What is its moment of inertia about the nail?
(1.5)
(1)
c) Write the expression for its mechanical energy when 0    15
d) Find the time period of its oscillations.
(1)
(1.5)
0
0
0
P
m
R
4
Fig 4
m1 
Fig 1
Fig 2
l
 m2
Fig 3
5