NAROK UNIVERSITY COLLEGE
(A CONSTITUENT COLLEGE OF MOI UNIVERSITY)
UNIVERSITY EXAMINATIONS 2011/2012
FIRST YEAR FIRST SEMESTER EXAMINATION
SCHOOL OF SCIENCE
UNIVERSITY EXAMINATIONS FOR THE DEGREE OF BACHELOR OF
SCIENCE (COMPUTER, WILDLIFE MANAGEMENT, INFORMATION
SCIENCE), BACHELOR OF SCIENCE AND BACHELOR OF
EDUCATION (SCIENCE)
COURSE CODE:
PHY110
COURSE TITLE:
BASIC PHYSICS I
DATE: AUGUST, 2011
TIME: 3 HOURS
MARKING SCHEME AND SOLUTIONS
INSTRUCTIONS
-
Answer Question ONE and any other TWO Questions.
-
Use of sketch diagrams where necessary and brief illustrations are encouraged.
- Read the instructions on the answer booklet keenly and adhere to them.
PHYSICAL CONSTANTS
-
Acceleration due to gravity, g = 9.8 ms-2
1ft(foot) = 0.305 m
1lb (pound) = 0.4536 kg
1 cal (calorie) = 4.2 joules.
Initial intensity, Io = 10-12 W/m2
 This paper consists of __6___ printed pages.
Page 1 of 6
QUESTION ONE: 30 MARKS
a) A sound intensity of about 1.2 W/m2 can produce pain. To how many decibels is this
equivalent?(take Io=10-12W/m2)
[2]
I
1.2
= 10log
= 120.8dB
- n = 10log
-12
Io
10
b) Use mathematical analysis to determine which of the following equations is certainly
wrong:
m
mv
v2
λ = vt ;
F=
;
F=
;
h=
;
v = (2gh)1/2
a
t
2g
Where λ and h are lengths and  F =  MLT-2  and the other symbols have their usual


meanings.
[5]
-
Vt=[LT-1][T]=L but [λ]=[L] so eqn is correct
[m/a]=[M][T2L-1]=[ MT2L-1] but [F]=[ MT2L] hence F=m/a is incorrect.
[mv/t] = [M][LT-1][T-1]= [MLT-2] and F = mv/t is dimensionally correct.
[V2/2g] = [(L2/T2)/(L/T2)] =L and since h = L, h= V2/2g is dimensionally correct.
Since [v]=[LT-1] = , [(2gh)1/2] = [(L1/2T-1)L1/2] = [LT-1], v=2gh)1/2 is dimensionally
correct.
Note that pure numbers are dimensionless.
c) A force of 70 N gives an object of unknown mass an acceleration of 20 ft/s2. What is
the object’s mass?(take 1 ft = 0.305 m). [2]
F
70
70
- F = ma ® m = =
=
= 11.5kg
a 20× 0.305 6.10
d) A cord passing over a frictionless, massless pulley has a 4-kg block tied to one end
and a 12-kg block tied to the other. Compute the acceleration and the tension in the
cord.
[4]
- Make a drawing
- T- m1g = m1a and T- m2g = -m2a and eliminating T gives: a= (m2-m1)/ (m1+m2)
and T = (2m1m2g)/(m1+m2)
12 - 4
2× 4×12
- a=
(9.8) = 58.8 N
 9.8 = 4.9ms-2 and T =
(4 +12)
12 + 4 
e) A body is projected upwards from the level ground at an angle of 50o with the
horizontal has an initial speed of 40 m/s. How long will it be before it hits the
ground?
[4]
- Choose upwards as positive, and place the origin at the launch point.
u x = u o cos50o = 40 × 0.642 = 25.7m/s and u y = u o sin50o = 40 × 0.766 = 30.6m/s
a y = -g = -9.8m/s 2 and a x = 0 using y = u y t -1/2gt 2 and since y = 0 at end of flight;
0 = 30.6t - 4.9t 2 ® 4.9t 2 = 30.6t hence t = 0 or t = 6.24.
t = 0is for the starting point and t = 6.24s is the time needed on air.
f) While driving around a curve of 200 m radius, an engineer notes that a pendulum in
the car hangs at an angle of 15o to the vertical. What should the speedometer of the
car read(in km/hr)?
[2]
Page 2 of 6
mv 2
and Tcosθ = mg where T = tension.
- Tsinθ =
r
v2
-Thus tanθ =
or v = rgtan15o = 23m/s =82.5km/hr
rg
g) State Kepler’s three laws of planetary motion.
[3]
LAW 1: The orbit of a planet/comet about the Sun is an ellipse with the Sun's center of
mass at one focus
LAW 2: A line joining a planet/comet and the Sun sweeps out equal areas in equal
intervals of time
LAW 3: The squares of the periods of the planets are proportional to the cubes
of their semimajor axes:
h) At what temperature do the celcius and the Fahrenheit readings have the
same numerical value?
[2]
9
9
4
t F = t C + 32. Let t F = t C = x. Then x = x + 32 or x = -32 and x = -40
5
5
5
Thus t F = -40o F and t C = -40o Care the same temperatures
i) Explain a simple harmonic motion (SHM).
[2]
 Simple harmonic motion (SHM) is the motion of a simple harmonic oscillator,
a motion that is neither driven nor damped.
 A body in simple harmonic motion experiences a single force which is given by
Hooke's law; that is, the force is directly proportional to the displacement x and
points in the opposite direction.
 The motion is periodic: the body oscillates about an equilibrium position in a
sinusoidal pattern.
 Each oscillation is identical, and thus the period, frequency, and amplitude of the
motion are constant.
 If the equilibrium position is taken to be zero, the displacement x of the body at
any time t is given by x(t) = Acos(2πft + Φ) Where A is the amplitude, f is the
frequency, and Φ is the phase.
j) A body describing a SHM has a maximum acceleration of 8π radians/s2 and a
maximum speed of 1.6m/s. Find the period T and the amplitude A.
[4]
-
-
We express amax and vmax in terms of T and A.
4π2
2π
a max =
A  = 8πm/s2
and vmax =

 A  = 1.6m/s
T
T2
a max 2π 8π
2π 2πA
0.4×1.6
=
=
® T = 0.4s Then
=
= 1.6 ® A =
= 0.102m
vmax
T 1.6
T
0.4
2π
QUESTION TWO: 20 MARKS
a) Discuss thermal resistance of a material for heat conduction. [6]
- The thermal resistance R is characteristic not only of the material but also of the
geometry involved, specifically the total length which supports the temperature
L
difference. R =
where L is length of material and K is the htermoconductivity
K
Page 3 of 6
-
The thermal resistance is large for thick matter with low thermal conductivity. In
terms of the thermal resistance we can calculate the rate of heat flow from
ΔQ 1
= AΔT where symbols have their usual meanings
Δt R
b) A refrigerator door is 150 cm high, 80 cm wide, and 6 cm thick. If the coefficient of
conductivity is 0.0005 cal/cm.s.oC, and the inner and outer surfaces are at 0 and
30oC, respectively, what is the heat loss per minute through the door , in calories?
(take 1 cal = 4.2 joules).
[5]
c) (i)
(ii)
KA(t h - t c )t 0.0005(150×80)(30 - 0)(60)
=
= 1800cal
d
6
Find the heat resistance of 1m2 of window glass that is 0.5 cm thick. [5]
Specially insulated window pane consists of two 0.5-cm thick sheets of glass
separated by a 0.15-cm thick layer of dry air. Calculate the resistance of 1m2
of this pane. (take Kg= 0.60W/m.oC and Ka = 0.025 W/m.oC). [4]
Q=
a) From the definition we have that the resistance Rg of a single sheet of glass is
d
0.5×10-2
Rg =
=
= 0.0083o C/W
KA
0.60 ×1
b) The resistance Ra of the layer of air between the two sheets of glass is
d
0.15×10-2
=
= 0.060o C/W
- Ra =
KA
0.025×1
-
Since the pane consists of two layers of glass and one layer of air, its total resistance is
-
R = Rg + Ra + Rg = 0.077oC
The resistance of the double-sheeted pane is more than 9 times greater than that of the
single-sheeted pane and most of its increased resistance comes from the layer of air
between the glass sheets.
QUESTION THREE: 20 MARKS
a) Define the terms Density and pressure. [4]
- Density(ρ) is mass per unit volume of a material while
- Pressure is the force acting normally per unit cross-sectional area.
b) A liquid can not withstand a shear stress. How does this imply that the surface of a
liquid at rest must be level, that is, normal to the gravitational force? [4]
-
If the surface is not horizontal a fluid element at the surface will experience a
component of gravitational force parallel to the surface. Since the adjacent liquid can
not exert a shear stress, this element will flow until the surface is horizontal.
c) If the blood vessel in a human being acted as simple pipes, what would be the
difference in blood pressure between the blood in a 1.8-m-tall woman’s feet and her
head when she is standing? (assume the specific gravity of blood is 1.06).
[4]
- Δp = ρgh = 1060(9.8)(1.8) = 18.7kPa
d) A cubical copper block is 1.50cm on each edge.
(i)
What is the buoyant force on it when it is submerged in oil for which
ρ=820kgm-3?
[4]
Page 4 of 6
(ii)
What is the tension in the string that is supporting the block when
submerged? (take ρcu = 8920kgm-3)
[4]
(i) The buoyant force equals the weight of the displaced liquid.
BF = ρoil Vcu g = 820×3.38×10-6 ×9.8 = 0.027N
(ii) The forces acting on the block are tension T up, BF up, and the block weight
down= Weight = ρcu Vcu g = 8920×3.38×10-6 ×9.8 = 0.295N
Then T = 0.295 - 0.027 = 0.268 N
QUESTION FOUR: 20 MARKS
a) Define the terms weight density and specific gravity. Discuss each with respect to
water.
[5]
- Weight density (D) of a material is the weight per unit volume of the material.
weight of body g
D=
= = ρg its units is N/m3 D for water is 62.4lb/ft3
volume of body V
- The specific gravity of a substance is the ratio of the density of an object to the density
of some standard substance. The standard is usually water at 4oC for liquids and solids.
density of body
ρ
sp gr =
=
its unitless
density of water ρ water
b) Battery acid has sp gr = 1.285 and is 38% sulfuric acid by weight. What mass of
sulphuric acid is contained in 1 litre of battery acid?
[3]
-
ρbattery acid = 1.285ρ w = 1.285×1 = 1.285kg/L
of the1.285kg, 0.38×1.285 = 0.488kg issulphuric acid.
c) A load of 100 lb is applied to the lower end of steel rod 3 ft long and 0.20 inch
diameter. How much will the rod stretch?(take Y=3.3 X 107lb/inch2) [ 1ft= 0.305
m and 1lb =0.4536kg]
[6]
F/A
LF
3×100
Y=
or ΔL =
=
= 2.9 ×10-4 ft = 3.5×10-3 inch = 8.845×10-5 m
7
ΔL/L
AY π × 0.1× 3.3×10
d) Compute the compressibility of glycerin if a pressure of 290lb/inch2 causes a volume
of 3 x 10-3inch3
[6]
- The compressibility is the reciprocalof the bulk modulus. Thus
1 ΔV 3×10-3
κ==
= 1.62 ×10-7 inch 2 /lb
Δp Vo 290 × 64
QUESTION FIVE: 20 MARKS
a) Show that the linear momentum p and the kinetic energy K of a mass m are related
p2
through K =
.
[4]
2m
since p = mv, multiply the numerator and denomerator of K = mv 2 /2 by m to get
K=
(mv)2 p2
=
2m
2m
b) Discuss ‘coefficient of restitution’ in mechanics.
Page 5 of 6
[6]
-
-
-
-
-
The coefficient of restitution, e , is a measure of the elasticity of the collision between
ball and racquet. Elasticity is a measure of how much bounce there is, or in other
words, how much of the kinetic energy of the colliding objects before the collision
remains as kinetic energy of the objects after the collision.
With an inelastic collision, some kinetic energy is transformed into deformation of the
material, heat, sound, and other forms of energy, and is therefore unavailable for use in
moving.
A perfectly elastic collision has a coefficient of restitution of 1. Example: two
diamonds bouncing off each other. A perfectly plastic, or inelastic, collision has e = 0.
Example: two lumps of clay that don't bounces at all, but stick together. So the
coefficient of restitution will always be between zero and one.
The coefficient of restitution is the ratio of the differences in velocities before and after
the collision. In other words, the difference in the velocities of the two colliding
objects after the collision, divided by the difference in their velocities before the
collision. In symbolic language:
v -v
Relative speed after inpact
e= 2 1 =
u1 - u 2 Relativespeedbeforeimpact
e = coefficient of restitution
u1 = linear velocity of the racquet mass center before impact
v1 = linear velocity of the ball before impact (will be negative according to our
convention that away from the player is positive)
u2 = linear velocity of the racquet mass center after impact
v2 = linear velocity of the ball after impact
c) A proton of mass 1.66 x 10-27 kg collides head-on with a helium atom at rest. The
helium atom has a mass of 6.64 x 10-27 kg and recoils with a speed of 5 x 10 5 m/s. if
the collision is elastic, what are the initial and final speeds of the proton and the
fraction of its initial energy transferred to the helium atom/?
[6]
- Mass of helium = 4(mass of proton) and from conservation of momentum,
mp vi + 0 = mp vf + (4m p )Vf
and vi = vf + 4Vf
For elastic collision,
velocity of approach = - velocity of separation or vi = -vf + Vf
Addingthe equations, we get 2vi = 5Vf = 5(5 x105 m/s) or vi = 1.25 x106 m/s.
and from momentum equation, - vf = 4Vf - vi so vf = 1.25 x106 - 2.00 x 106 = -7.5 x105 m/s
5 2
ke of helium 1/2(4mp )(5x10 )
energy fraction =
=
= 0.64
initial ke
1/2mp (1.25x105 )2
d) A 16-g mass is moving in the +x-direction at 30m/s while a 4-gmass is moving in the –
x-direction at 50 cm/s. they collide head-on.and stick together. Find their velocity
after collision.
[4]
- Apply conservation of momentum.
(0.016 × 0.30) + (0.004 × -0.50) = 0.020v
solving gives v = 0.14m/s
The End
Page 6 of 6