AIR UNIVERSITY
Mid-Term Exam with Solution (Physics-103)
Sections: BEMTS(A,B,C,D) AND BEMEC-A,B,C.
MAX.MARKS: 20
Instructors: Dr.Pervaiz Akhtar, Ch Ibrahim, Mr.Mozaffar.
Q # 1 a)consider three vectors a ,b c with magnitudes 3m 4m and 10m respectively
are shown in figure .The angle between a nd b is 300 , where as between b and c is
900
(i) find x- and y-components of all the three vectors individually.
Solution:
(ii) find the resultant of all the x-components and y-components.
Solution:
Resultant of all x components = 3+3.46-5 = 1.46
Resultant of all y components = 0+2+8.66= 10.66
(iii) if c = La + Mb , then find the value of L and M.
Solution:
In terms of components (first x and then y), we must have
-5.00 m = L(3.00 m)+M(3.46 m)
8.66 m = L(0)+M (2.00 m).
Solving these equations, we find L = –6.67. and M = 4.33
(b) If a = 3.0i – 4.0j and b = -2,0i + 3.0k
(i) find the magnitude of cross product of two vectors.
Solution:
a×b = (3i -4j)×(-2i + 3k) = -12i -9j-8k
(ii) also find the angle between the two vectors.
Solution:
a.b = (3i -4j).(-2i + 3k) = -6
Also a.b = ab cosθ
a  32  4 2 = 5,
b  4  9 = 13
 6 
 = 109o
Therefore (5)( 13 ) cosθ = -6, Therefore θ = cos 1 
 5 13 
Q #2 (a) A particle moves so that its position (in meters) as a function of time (in seconds) is
r =1.0i + 4t2j + t k. Write the expression for its velocity and acceleration as a function of
time.
Solution:
(b) A plane, diving with constant speed at an angle of 530 with the vertical, releases a
projectile at an altitude of 730m. The projectile hits the ground 5.00s after release
(i) What is the speed of the plane?
Solution:
Which give vo = 202 m/s
(ii) How for does the projectile travel horizontally during its flight?
Solution:
Q # 3 (a) A 2.0 Kg object is subjected to three forces that give it an acceleration
a = -(8.00m/s2)i+ (6.00m/s2 )j. If two of the three forces are
F1 = (30.0N) i+ (16.0N)j and F2 = -(12.0N)i+(8.00N)j, find the third force.
Solution:
m = 2 kg
Therefore

F3  7iˆ  12 ˆj N
(b) Consider a block of mass m = 3.0kgslides along a floor while a force F of
magnitude 12.0N is applied to it .at an upward angle θ = 300.The co-efficient of
kinetic friction between the block and the floor is μk 0.40.
Solution:
m = 3 kg
F = (12 N , 30o)
Resolving F in x and y components
Fx= 12 cos 30o = 10.4N,
Fy = 12 sin30o = 6 N
Net force along y-axis is
Net force along x-axis is
(i) Calculate the magnitude of the normal force acting on the block.
FN = mg- Fsinθ = 29.4-6= 23.4N
(ii) Calculate the magnitude of the force of kinetic friction.
= (0.4)(23.4) = 9.36N
(iii) find the value of acceleration a produced in the block.
ma = 12cos30 – 9.36
Therefore a = 0.35 m/s2
Q#4(a) the angular position of a point on a rotating wheel is given by
θ = 2.0+ 4.0t2 + 2.0t3 , where θ is in radians and t is in seconds. At t=0 what
are
i)
the points angular position
ii)
the points angular velocity
iii)
What is its angular velocity at t=4.0s.
iv)
What is its angular acceleration at t= 2.0s.
Solution:
Angular position is given by
(i)
(ii)
(iii)
At t = 4 s, the angular velocity is
(iv)
Angular acceleration is given by:
(b) (i) Calculate the rotational inertia of a meter stick about an axis perpendicular
to the stick and located at 20cm mark. The mass of the meter stick is 0.56Kg.
( I =1/12ML2
(ii) If the angular velocity of the meter stick is 20rev/mint, then find the
rotational kinetic energy of the meter stick.
Solution:
(i) Using parallel axis theorem, rotational inertia at a given point is
Therefore
(ii) The rotational K.E is given by
K.E = ½ Iω2
Where ω is
ω = 20 rev/min = (20)(2π) rad/min
so ω = 40π/60 = (2/3)π rad /s.
Also I = 9.7×10-2kg.m2 as calculated in part (i)
so K.E = (½)(9.7×10-2kg.m2) ((2/3)π rad /s)2.
Q#5 (a) Which of the following relationships between the force F on a particle and
the particles position x implies simple harmonic motion.
i)
F = -5x
ii)
F = -400x2
iii)
F = 10x
2
iv)
F = 3x
Solution:
(a)
A particle execute S.H.M if F = -kx, where F is the net force acting on a particle
at particular instant and x gives the position of particle at that instant. This condition is
satisfied for case (i) only.
(b) Consider a meter stick that swings about a pivot point O at a distance h from the
sticks centre of mass. Treat mater stick as thin meter rod and apply the
parallel axis theorem to find
o
h
com
(i) its rotational inertia about an axis passing through the pivot.
Solution:
Using parallel axis theorem, rotational inertia at a given point is
Where Icom = (1/12)ML2 and h = L/2
so I = (1/12)ML2 + M (L/2)2
I = (1/3)M L2.
As L = 1 m so I = (1/3)M
(ii) The time period of oscillation T?
Solution:
The time period of physical pendulum is given by
I
T  2
Putting the value of I and h,
Mgh
T = 1.64 s.
(c) In an electric shaver, the blade moves back and forth over a distance of 2.0mm in
simple harmonic motion. Find
i)
the amplitude
ii)
the maximum blade speed
iii)
the magnitude of the maximum blades acceleration.
Solution:
(i) As blade covers 2 mm in moving from one extreme position to other extreme position.
So amplitude is half of this value. Amplitude = 1 mm
(ii) vmax = xm ω
and ω = 2πf = 2π(100)
vmax = (1mm)2π(100) = 0.2 π m/s.
(iii) the maximum value of the acceleration is given by
amax = ω2 xm = (2π(100))2(1mm) = 20π m/s2