Download Final Solution-Phy 105-Fall2011-1

AIR UNIVERSITY
Final Exam Fall 2011
Subject: Physics 105
Maximum Marks: 45
Maximum Time: 3 Hrs
Class: BEE 1st Semester
Date: 06-01-2012
Instructors: A. Sadiq, M. Anwar, Mozaffar, H; M. Kashif and T. Zehra
Note: Attempt all the 9 questions on the two pages. Each question carries
equal marks. Encircle your answer
1. Fig 1 shows the sketch of displacement x in meters as a function of time t .in
seconds.
Fig 1
2.5 x
a) What is the average velocity for t  2s ?
(1)
b) What is the velocity for 3  t  6s ?
(1)
0.5
c) If x  t   x0  at 2 / 2 , for t  2s , find a .
(1)
d) Make a sketch of velocity as a function of time.
(2)
0
2
3
x2  x0 2.5  0.5

 1.0m / s
t 2  t0
2
_
x x
0.0  2.5
 0.83m / s
b) v26  v36  6 3 
t6  t3
3
6
_
Solution: a) v 02 
c) Given x  t   x0  at 2 / 2 , for t  2s , we have,
1.0
v
1
2
3
0
x  2   x  0   a  2  , or
2
2
0.83
a
2.5  0.5   4 , giving,
Fig 1a
2
2
a  1m / s ]
d) We have for t  2s , v  t   at  tm / s , for 2s  t  3s , v  t   0 and for 3s  t  6s ,
t
6
v  t   0.83m / s . Therefore the sketch of v  t  as a function of time is shown in Fig 1a.
2 a) Calculate the angle  between the two vectors A  3i  4 j and B  3i  4 j (2)
b) Fig 2 shows a particle of mass m  100 g tied to a string of length l  30cm
moving in a vertical circle at 100 revolutions per minute. Find the tension in the
string when the string makes an angle   300 with the vertical.
(3)
300
Solution: a) The angle  between two vectors A and B is given by,
Fig 2

.B 
1  9  16 
1  7 
0
  cos 
  cos    106.4
 9  16 
 25 
 AB 
  cos1  A
1
t
mg
0
100
 2  10.5rad / s
b) m  100 g  0.1kg ,r-l=30cm=0.3m, w 
Fig 2a Fc 30
60
2
Magnitude of centripetal force acting on m Fc  mrw2  0.1 0.3  10.5   3.29 N
This force Fc is directed towards the center of the circle and weight is directed
downwards as shown in the free b0dy diagram of Fig 2a.. Therefore, the component
of the weight along the string provides part of the centripetal force needed to keep
the object moving in the circle. The rest of the force is provided by the tension T in
the string given by,
3
T  Fc  mg cos 30  10.5  2.94 
 10.5  2.5  8 N
2
3 Fig 3 shows a block of mass M  10kg lying on a rough inclined surface with
s  0.6 and k  0.5 making an angle   300 with the horizontal. It is held in its
place by a light string passing over a light frictionless pulley. A mass m is
attached to the other end of the string..
Fig 3
a) Sketch a free body diagram of M when it just moves up the plane. (1)
b) Find the value of m that is just large enough to pull M up the plane. (2)
c) Find the acceleration of the block M when the string is cut.
(2)
M
m

Mg cos 
Solution: a) Fig 3a shows the forces acting on the block.
T
Mg sin 
These are:
Fig 3a
Mg
its weight Mg acting vertically downward, with its components;
F
Mg sin  , acting downward along the plane
Mg cos  , acting upward perpendicular to the plane the plane
The
Force of friction F  s Mg cos  , acting downward along the plane, and
The tension T in the string acting upward along the plane
b) Since block M is at rest, we have,
T  Mg sin   F  Mg  sin   s cos   10  9.8  0.5  0.6  0.86  99.9 N
Also, since block m is at rest, we a e,
mg  T , giving,
T 99.9
m 
 10.2kg
g 9.8
c) When the string breaks, block M slides down then the plan under the component
of its weight opposed by the force of friction. We have
Ma  Mg sin   k Mg cos  , giving,
a  g sin   k g cos  9.8  sin 30  0.5  cos30  9.8  0.5 1.0  0.87   0.66ms 2
4 Fig 4 shows a 100kg block pushed by two forces on a rough horizontal surface
with s  0.5 and k  0.4 .
100kg
400N
a) Sketch its free body diagram.
(1)
b) Find the total work done in pushing the block through 10m . (1)
300
600N
600
2
Fy
1
c) Calculate the final kinetic energy of the block.
(3)
Solution: a) Fig 3a show the free body diagram of the block.
Fig 3a
Here Fx is the x-component of the applied forces, Fy is the y-component
N
F2
2
F1
F
mg
Fx
of the applied forces, N is the normal reaction of the floor on the block and F  k N
is the force of kinetic friction acting on the block.
N  mg  Fy  100  9.8   400sin 30  600sin 60  980  719.6  260.4N
b) The total work done on the block W  F N.s  Fx s , with
Fx  F1 cos   F2 cos 2  400cos30  600cos 60  646.4 N , s  10m
W  64644 10  6464J
c) Final kinetic energy of the block K is the difference of the total work done W by the
applied force and the work done against the force of friction W f . With,
N  mg  Fy  100  9.8   400sin 30  600sin 60  980  319.8  660.2 N , we get
W f  Fs  k Ns  0.4  660.2 10  2841.6 J , and,
K  W  W f  6464  2841.6  3622.5 J
5 a) Three identical particles each of mass m lie at the corners of an equilateral
triangle of side l .
a) Find the distance of their centre of mass from a corner of the triangle.
(2)
b) Two friends A and B , with mA  30kg and mB  40kg are stuck-up in the middle of
frictionless icy surface. To get out, A pushes away B , finding herself moving away
from B with a velocity vA  2m / s . Find the magnitude and direction of v B .
(2)
c) How much work did A do in pushing B ?
(1)
Solution: a) Let m be the mass of each particle and l be the length of the
m
y
side of the equilateral triangle. For finding the location of the center of
mass (CM) C of the particles we chases the origin O of our coordinated
system at one of the corners of the triangle, with x-axis along its base
l
m
and y-axis perpendicular to it as shown in Fig 4. Fig 4
C
The locations of the three particles are  0.0 ,  l.0 , and l / 2.l 3 / 2 .
O
m
If  xc , yc  are the coordinates of CM of the thee particles then,

3
 m  x  x   3ml / 2  3mx
i 1
i
i
c
c

 0 , giving,
xc  l / 2 , and
3
 m  y  y   ml
i 1
yc 
i
i
c
3 / 2  3mxc  0 , giving,
1
l
2 3
Therefore, distance fof CM from a corner of the triangle is given by,
l
OC  l 2 / 4  l 2 /12 
6
3
x
b) We have mA  30kg , mB  40kg and vA  2m / s
Since both persons A and B are at rest, their total initial linear momentum pi  0
Also since no external force act on them, especially when A pushed away B , their
total momentum is conserved during the push, giving their final total momentum,
p f  pi  0 ,
In terms of their individual final moments pA and pB w get,
p f  p A  pB  0 , giving,
pB  mB vB   pA  mAvA , or,
m
30
vB   A v A    2  1.5m / s
mB
40
c) Work done by A is the sum of the kinetic energies of A and B ;
1
1
9
WA   30  4   40   60  45  105 J
2
2
4
6 a) A particle of mass m  100 g moving with velocity v  5m / s collides elastically
head-on with a particle of mass M  200 g initially at rest. Calculate the magnitudes
and directions of the velocities of the two particles after their collision.
(4)
b) Impulse J is the change  p in linear momentum during the very short time of a
collision. Calculate the impulse imparted by M to m .
(1)
Solution: a) We have m  100 g  0.1kg , M  200 g  0.2kg , and v  5m / s ,
and V  0m / s
Since article m collides elastically head on with particle M , it is a one dimensional
problem in which their total linear momentum and their total kinetic energy are
conserved. If u and U are the velocities of the two particles after their collision, the
conservation of linear momentum gives,
mu  MU  mv  0.1 5  0.5kgms 1 , or,
(6.1)
2U  5  u
The conservation of kinetic energy gives,
1
1
1
1
mu 2  MU 2  mv 2  0.1 25 ,or,
2
2
2
2
2
2
2U  25  u
(6.2)
Substituting EQ (6.1) into EQ (6.2) we get,
1
1
1
2
2
 2U    5  u    25  u 2  10u   25  u 2 , giving,
2
2
2
2
3u  10u  25  0 , or,
5m / s
10  100  300 10  20
u


5 m/s
6
6

3
4
The first of these choices gives us the velocities before collision.
5
Choosing u   m / s , EQ (6.1) gives,
3
5  u 5  5 / 3 10
U

 m/s
2
2
3
b) Impulse J imparted by M to m is the change in the linear momentum of m .
Therefore,
2
 5 
1
J  p f  pi  m  v f  vi   0.1  3  5    3 kgms ,
As expected from Newton’s 3rd law of motion, this is equal and opposite to the
impulse imparted by m to M .
7 a) Derive the expression for the moment of inertia I c of a thin rigid ring of mass m
and radius R about an axis passing through its CM and perpendicular to its plane.
Find its value for m  250 g and R  30cm.
(2)
2
b) Using the parallel axis theorem I P  I c  md , calculate its moment of inertia I P of
the ring about a parallel axis through point P on its circumference.
(1)
c) Calculate the work done to spin the ring at w  50rpm about its diameter. 2)
Solution: a) All parts of the ring are at the same distance R from the axis of
rotation passing through its CM at its center and perpendicular to its plane.
We consider a small part of the ring of mass  mi located between angles 
and   d as shown in Fig 7. Its moment of inertia about the axis of rotation
is,
 I i   mi R 2
Fig 7
The moment of inertia of the ring about the given axis of rotation is the sum of the
moments of inertias of ll such parts of the ring,


I c    I    mi R 2     mi  R 2  mR 2
i


i
For m  250 g  0.25kg and R  30cm  0.3m
I c  mR 2  0.25   0.3  0.0225kgm 2
2
b) Moment of inertia I P about an axis through a point P on the ring that is
perpendicular to the plane of the ring is given by,
I  I c  mR 2  2mR 2  0.045kgm2
50
c) Spin angular velocity of the ring w   2  5.24rad / s
60
Work done W to spin the ring about one of is diameter is the kinetic energy of rotation
K r about its diameter. If I d is the moment of inertia of the ring about its diameter then,
1
W  K r  I d w2
2
From perpendicular axis theorem,
5
  d

I d  I c / 2  0.0112kgm2
Therefore,
1
2
W  K r  I d w2  0.5  0.0112   5.24   0.161J
2
8 A uniform thin rigid rod of mass m  50 g , length l  30cm and I c  ml 2 /12 , is
vertically hung from one of its ends. It oscillates when slightly disturbed from its
mean position.
a) Find its time period of oscillations
b) What is its kinetic energy K r when the velocity of its CM is vc  5cm / s ?
Solution: a) We have m  50 g  0.05kg , l  30cm  0.3m and
(3)
(2)
I c  ml 2 /12  0.05   0.3 /12  0.00375kgm 2
2
We find the time period of small oscillations of a thin rigid rod about an axis passing
though its end and perpendicular to it by the following equation,
d 2
I P 2  mgl sin   mgl , or,
dt
2
d  mgl
d 2



 w2  0 , giving,
2
2
dt
IP
dt
w
mgl
, with I P  I c  ml 2 / 4  ml 2 / 3 , so that,
IP
IP
2
ml 2 / 3
l
0.3
T
 2
 2
 2
 6.28
 0.63s
w
mgl
mgl
3g
3  9.8
b) Since the CM of the rod is at a distance of l / 2 from the axis of rotation, the
angular velocity of the rod w is related to the linear velocity of CM vc by the following
relation
3
w  lvc / 2  0.3  0.05 / 2 
rad / s
400
Therefore rotational kinetic energy of the rod when its CM has a velocity
vc  0.05m / s is given by,
0.05   0.3  3 
1
8
I P w2 

  4.22 10 J
2
6
 400 
9 The following equation describes the free oscillations of an object having a time
period T  0.25s ,
d 2x
 w2 x  0
2
dt
a) Verify that its displacement at any time t is x  t   x0 cos  wt    .
(2)
2
2
Kr 
b) Find the values of its amplitude x0 and the phase angel  if the object has the
maximum velocity v0  5cm / s at t  0s .
Solution: a) We have,
(3)
6
x  t   x0 cos  wt   
Differentiation the above equation twice with respect to time we get,
dx  t 
  wx0 sin  wt    ,
dt
d 2 x t 
  w2 x0 cos  wt      w2 x  t  , giving,
2
dt
2
d x t 
 w2 x  t   0
2
dt
2 6.28

 25.1rad / s
b) For T  0.25s , w 
T
0.25
We are also the maximum velocity of the oscillator, v0  5cm / s  0.05m / s at t  0s ,
which gives,
dx  t 
 0.05   wx0 sin  wt     25.1 x0 sin  
dt
Since v0 is the maximum velocity of the oscillator, we have,
   / 2 , and
0.05
x0 
 98.2 105 m  982 m
25.1
7