PHYS151: General Physics I Sectional Test I - Summer 1 2018

This document presents the complete solutions for the PHYS151 General Physics I Sectional Test I, administered during the Summer 1, 2018 session. The test is divided into two sections: Section A consists of fill-in-the-blank questions designed to assess the student's understanding of fundamental physics concepts such as collinear vectors, base units, scalar quantities, and vector diagrams. Section B provides detailed solutions to problem-solving questions, including the identification of correct kinematic equations, dimensional analysis of a given formula, unit conversions, and calculations involving vector forces and their resultant magnitudes and directions. The solutions demonstrate step-by-step approaches, utilizing relevant formulas and principles to arrive at the correct answers for each problem, including force vector components and resultant calculations.

PHYS151: GENERAL PHYSICS I Summer 1 2018
SECTIONAL TEST I
Name…………………………………………………… ID………………………………………………. Date……………..
SECTION A. FILL IN THE BLANK SPACES OF EACH STATEMENT BELOW
1. Collinear vectors travelling in opposite directions will be zero.
2. Dimensions use quantities such as: mass, length, and time in contrast to units.
3. Three base units are: meters, Kilograms, and seconds.
4. Derived units are also called supplementary units.
5. Temperature, speed, and distance are all scalar quantities.
6. Models can be either linear or non-linear in nature.
7. The conversion of units is either from one system to another or from one level to a next.
8. The normal force in a free body diagram is always perpendicular to the mating surface of the
body.
9. The direction of a vector is always taken with respect to some point.
10. A diagram of a vector starts from the origin and moves in the direction shown by the vector
arrow.

PHYS151: GENERAL PHYSICS I Summer 1 2018
SECTIONAL TEST I
SECTION B. ANSWER THE FOLLOWING QUESTIONS.
1. Three students while studying forgot a kinematic equation of motion. One student says that the
proper equation is: x = x0 + v0 + ½ t2.
Another student says that the equation is x = x0 + v0t – ½ at2.
The third is confident that the equation is x = x2 +a t2
2 + v.
Which of the students is correct?
The second student is correct. x = x0 + v0t – ½ at2.
2. A certain formula states that: F = Km2 t2
x3 ,
where K is an experimentally derived constant. Using dimensional analysis, what would the units
for the constant be?
(x, x0 are distances (unit: m). v0 is velocity (unit: m/s). a is acceleration (unit: m/s2). t is time
(unit: s). m is mass (unit: kg))
Solution
Let M represent mass, T represent time while L represent length.
K= F x3
m2 t2
F=Ma
a= m
s2 =L . T−2
F=M . L .T−2
K= M . L . T−2 . L3
M 2 T 2
K= L4
M .T 4
= m4/Kg.s4
3. Convert 3 lbs/inch to kg/mm given that: 1 inch = 2.54 cm, 1cm = 10mm, 1 lb = 0.453 kg.
Solution
3 × 0.453 = 1.359 Kgs/inch
1.359 ÷ (1×2.54×10) = 0.0535 Kg/mm

PHYS151: GENERAL PHYSICS I Summer 1 2018
SECTIONAL TEST I
4. Two force vectors are travelling on the same line and in the same direction. One vector has a
magnitude of 12N, while the other has a magnitude of 12 lbf. What will the resultant of the two
vectors be in N (Newton)? (1 lbf = 4.45N)
Solution
12×4.45= 53.4 N
Since the forces are travelling in the same line and direction, the resultant vector force(R) will be
the sum of the two vectors.
R= 12 + 53.4
R= 65.4 N
5. Two boats are pulling a third boat as shown below. What is the magnitude and direction of the
resultant force on the third boat due to the forces of the other two?
Solution
Since all the forces are pulling diagollany, both forces will have a vertical and horizontal component.
Ship 1
20o
The diagonal force is 980 N therefore,
The vertical component will be

PHYS151: GENERAL PHYSICS I Summer 1 2018
SECTIONAL TEST I
= 980 Cos 200
= 920.90 N
The vertical component will be
= 980 Sin 20o
= 335.18 N
Ship 2
58O
The diagonal force is 1200 N. Therefore,
The vertical component will be
= 1200 Cos 58o
= 635.90 N
The horizontal force will be
= 1200 Sin 58O
=1017.66 N
The combined force diagram is as shown below
920 N
(335+1017.66) N
635.9 N
Therefore the resultant forces will be

PHYS151: GENERAL PHYSICS I Summer 1 2018
SECTIONAL TEST I
920 – 635.9 = 284.1 N upwards.
335+1017.66 = 1352.66 N Horizontal ( to the right)
The resultant horizontal force will be
= √[(284.1)2+(1352.66)2]
= 1382.17 N
Angle of the resultant from the vertical
= tan-1(1382.17/284.1)
= 78.38o
78.38o