LIST OF CONTENTS

Sections Page Number

Introduction & Objectives 2

Question 1 3-4

Question 2 5-6

Question 3 7-9

Question 4 10-11

Question 5 12

Conclusion & References 13

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INTRODUCTION

Based on the questions in Mathematics Problem Best-Learning (PBL) 1, the questions are asking
about some of the Physics concept that can be solve using Mathematics techniques by using “
”, “ ”, “ ” and other Mathematics signs in the formula. There are five questions in this PBL that
have to be solve by using the Mathematics signs. Each signs bring the different of functions that
could help Mathematicians to solve a problem. Basically, Mathematics can be apply in daily life
other than in Physics and Chemistry. Mathematics application can be useful to us as it can solve
our life problem such as to manage our financial, measure the height of building, calculate the
mass of an object and etc. We use Mathematics signs and knowledge as basic in the other
subjects and Mathematics should not be neglected. (Chris, 2018)

OBJECTIVES

1. To apply the subject of Mathematics in any situation and other subjects to solve the problem.

2. To identify which Mathematics signs can be use in some formula to solve the problem.

3. To understand the advantages of Mathematics application in daily life.

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QUESTION 1

Two light strings are attached to a block of mass 30kg. The block is in equilibrium on a
horizontal surface 𝐴� with the strings taut. The strings make angles of 60° and 30° with the
horizontal, on either side of the block, and the tensions in the strings are � N and 75N
respectively (see diagram).

(i) Given that the surface is smooth, find the value of � and the magnitude of the contact force
acting on the block.

From the question block is in equilibrium on horizontal surface AB. So,

ԐFx= 0

Force x-component y-component

T1 -T cos 60 T sin 60

75N 75 cos 30 75 sin 30

T3 0 294.3 N

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To find the T:

ԐFx= - T cos 60 + 75 cos 30 = 0

75 cos 30 = T cos 60

T= 75 cos 30 / cos 60

= 129.90 N

To find magnitude of contact force acting on the block:

ԐFy= 129.90 sin 60 + 75 sin 30 – (30kg×9.81ms -2) + (30kg×9.81ms-2)

= 149.90 N

Magnitude =

= 149.90 N

(ii) It is given instead that the surface is rough and that the block is on the point of slipping. The
frictional force on the block has magnitude 20N and acts towards 𝐴. Find the coefficient of
friction between the block and the surface.

Given Frictional force = 20 N

T cos 60 + 20 = 75 cos 30

T cos 60 = 75 cos 30 - 20

T = 44.95/ cos 60

= 89.90 N

89.90 sin 60 + 75 sin 30 + Fn= 294.30 N

Fn = 178.94 N

Thus, = 20 N/ 178.94 N

= 0.112
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QUESTION 2

A particle P of mass 0.5 kg is projected vertically upwards with speed 5.2 ms−1 from a point O
which is 6.2 m above the ground. Air resistance acts on P so that its deceleration is 10.4 ms−2
when P is moving upwards, and its acceleration is 9.2 ms−2 when P is moving downwards. Find

(i) The greatest height above the ground reached by P.

v = u + at

Where v is final speed, u is initial speed, a is acceleration and t is time taken. The greatest
height is reached when the final speed is zero, v = 0 and used deceleration value as a is
because P is moving upward. Hence, rearrange the formula (1) to get the t value.

t=v–

u = 5.2 ms-1

v = 0 ms-1

a = - 10.4 ms-2

t=0– = 0.5 s

After get the t value, to determine the height use the formula below and substitute the value
into this formula:

s = ut + at2

u = 5.2 ms-1

t = 0.5 s

a = - 10.4 ms-2

s = (5.2) (0.5) + (-10.4) (0.5)2 = 1.3 m

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 Since P is projected from point O which is 6.2 m above the ground, hence the greatest height

above the ground reached by P is, 6.2 m + 1.3 m = 7.5 m.

(ii) The speed with which P reaches the ground.

v2 = u2 + 2as

P reaches the ground when the initial speed is zero, u = 0 and used acceleration value as a is
because P is moving downward.

u = 0 ms-1

a = 9.2 ms-2

s = 7.5 m

v2 = 0 + 2 (9.2) (7.5) = 138, v = = 11.75 ms-1

(iii) The total work done on P by the air resistance.

Kinetic energy formula:

mv2

Potential energy formula:

mgh

Work done = Potential energy – Kinetic energy

g = 9.81 ms-2

m = 0.5 kg

v2 = (12)2 – (5.2)2, (final speed – initial speed)

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h = 6.2 m, (because at the starting point, P is static (potential energy), hence the height is 6.2
m)

Work done = (0.5) (9.81) (6.2) – (0.5) (122 – 5.22) = 30.411 – 29.24 = 1.171 J

QUESTION 3

Particles A and B are attached to the ends of a light inextensible string which passes over a
smooth pulley. The system is held at rest with the string taut and its straight parts vertical. Both
particles are at a height of 0.36 m above the floor (see diagram). The system is released and A
begins to fall, reaching the floor after 0.5 s.

(i) Find the acceleration of A as it falls.

The height given or displacement, S= 0.36 m

The time after A reaching the floor, t = 0.5 s

The formula to find acceleration of A as it falls, a is S= ut + ½ at 2

Since the system is at rest at first, the initial velocity = 0 and ut can be ignore in the formula.

Solution:

0.36 m= (0 x t) + ½ a (0.5 s) 2

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a = 0.36 m x 2

(0.25 s2)

a = 2.88 ms-2

The mass of A is 0.55 kg. Find

(ii) the tension in the string while A is falling,

Using Newton’s Second Law formula to find the tension in string while A is falling:

mg – T = ma

The mass of A, m = 0.55 kg

The acceleration of gravity, g = 9.81 ms-2

The acceleration of A, a = 2.88 ms-2

Solution:

mg – T = ma

[(0.55 kg) x (9.81 ms-2)] –T = (0.55 kg) x (2.88 ms-2)

T = 5.40 kgms-2 – 1.58 kgms-2

T = 3.82 kgms-2 or T = 3.82 N

The tension in the string is 3.82 N

(iii) the mass of B,

T – mg = ma

T = ma + mg

T = m (a + g)

Where,

T = 3.82 N or 3.82 kgms-2

m = mass of B, kg

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The acceleration, a = 2.88 ms-2

The gravity acceleration, g = 9.81 ms-2

Solution:

T – mg = ma

T = ma + mg

T = m (a + g)

3.82 kgms-2 = m (2.88 ms-2 + 9.81 ms-2)

3.82 kgms-2 = m (12.69 ms-2)

m = 3.82 kgms-2

12.69 ms-2

m = 0.30 kg

The mass of B is 0.30 kg.

(iv) the maximum height above the floor reached by B.

Using the formula of final velocity to find the maximum height of B which is

V2 = u2 – 2gs

Where,

V = 0 ms-2

u = (0.5 s x 2.88 ms-2) = 1.44 ms-1

g = 9.81 ms-2

s = maximum height of B

Solution:

V2 = u2 – 2gs

0 = (1.44 ms-1) 2 – [2 x 9.81 ms-2 x s]

19.62 s = 2.074

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s = 2.074 m2s-2

19.62 ms-2

s = 0.106 m

Maximum height above the floor reached by B is 0.106 m.

QUESTION 4

A M B

2m
A particle P of mass 1.6 kg is attached to one end of each of two light elastic strings. The other
ends of the strings are attached to fixed points A and B which are 2 m apart on a smooth
horizontal table. The string attached to A has natural length 0.2 m and modulus of elasticity 4 N,
and the string attached to B has natural length 0.2 m and modulus of elasticity 8 N. The particle
is held at the mid-point M of AB (see diagram).

(i) find the tensions in the strings.

Solution:
With using T = λ x/L,
 [TA = 4 x 0.8/0.2 and TB = 8 x 0.8/0.2]
 Tensions are 16 N and 32 N

(ii) Show that the total elastic potential energy in the two strings is 19.2 J.

Solution:
With using T = λ x2/2L,
 [Total EE = 4 x 0.82 / (2 x 0.2) + 8 x 0.82 / (2 x 0.2)]
 Total EE = 19.2

P is released from rest and in the subsequent motion both strings remain taut. The
displacement of P from M is denoted by x m. Find

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(iii) The initial acceleration of P,

Solution:

With using Newton’s second law,

(iv) The non-zero value of x at which the speed of P is zero.

Solution:

 4 (0.8+x)2 / (2 x 0.2) + 8 (0.8 – x)2 / (2 x 0.2) = 19.2

 [–2x (8–15x) = 0 ⇒ x = 0, ]

 x is ≈ 0.5333
 Value of x is : 0.533

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QUESTION 5

A uniform rigid wire AB is in the form of a circular arc of radius 1.5m with center O. The angle
AOB is a right angle. The wire is in equilibrium, freely suspended from the end A. The chord AB
makes an angle of θ° ◦ with the vertical (see diagram).

(i) Show that the distance of the center of mass of the arc from O is 1.35m, correct to 3
significant figures.

Solution:

OG = 1.5 sin 45° / (π / 4)

(ii) Find the value of θ.

Solution:

tan θ = (1.35 – 1.5 cos 45°) / 1.5 cos 45°

θ = 15.3o

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CONCLUSIONS

As a summary, Mathematics can be in our daily life to solve problem and to make our life
become easier. Mathematicians are the most person that needed in any occupation such as
accountant, banker, engineering, and others. So, the subject of Mathematics is fun if we learn
this subject by apply it in our life instead just follow the principle of Mathematics which is during
our learning in Mathematics. (Stunning Uses of Mathematics in Daily Life You Never Thought
About, 2018)

REFERENCES

Chris. (2018, August 6). Practical Applications of Mathematics in Everyday Life. Retrieved from

https://owlcation.com/stem/Some-Practical-Applications-of-Mathematics-in-Our-Everyday-Life

Shilpa, R. (2007, March 4). Advantages of Mathematics. Retrieved from

http://ezinearticles.com/?Advantages-of-Mathematics&id=476015

Stunning Uses of Mathematics in Daily Life You Never Thought About. (2018, March 7).

Retrieved from https://sciencestruck.com/mathematics-in-daily-life

TEN REASONS TO LEARN MATH. (2018, October 9). Retrieved from

http://thesoni.com/WhyLearnMath.php

What Is the Importance of Mathematics in Mechanical Engineering? (2018, October 9).

Retrieved from https://www.reference.com/math/importance-mathematics-mechanical

engineering-4308fc353291a5c3.

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