Physics WAEC

Motion

Bayani Gaba-gaba

Welcome to the intriguing world of motion, where the interaction of matter, space, and time unfolds with captivating dynamics. In this comprehensive exploration, we delve into the fundamental concept of motion and its diverse manifestations in the realm of physics. Motion, the essence of change in position with respect to time, is a phenomenon that permeates every aspect of the universe, from the celestial bodies in space to the minutest particles on Earth.

Understanding the Various Types of Motion: Motion presents itself in a myriad of forms, ranging from the simple rectilinear motion to the complex orbital and oscillatory motions. Each type of motion exhibits distinct characteristics that contribute to the diverse dynamics observed in the physical world. Through qualitative treatments and illustrative examples, we aim to provide a holistic view of the various types of motion and their significance in the natural order.

Force as the Cause of Motion: At the heart of motion lies the concept of force, the agent responsible for initiating and altering the state of motion of objects. By exploring the fundamental principles of force and its effects on matter, we elucidate the intricate relationship between force and motion, shedding light on the underlying mechanisms that drive physical interactions.

Unveiling the Concept of Push and Pull: Push and pull, the ubiquitous forces that influence the direction and magnitude of motion, play a pivotal role in the dynamics of objects in space and time. Through insightful discussions and real-world examples, we unravel the essence of push and pull, elucidating their impact on the motion of bodies in various scenarios.

Deciphering the Role of Friction in Motion: Friction, the resistance encountered when two surfaces come into contact, exerts a profound influence on the dynamics of motion. By examining the principles of frictional force, coefficient determinations, and methods of friction reduction, we gain a deeper understanding of how friction shapes the behavior of objects in motion.

Exploring Fluid Friction and its Applications: Fluid friction, a unique form of resistance encountered in fluid mediums, plays a crucial role in lubrication and various industrial processes. Through qualitative analyses and practical examples, we illuminate the concept of fluid friction, underscoring its significance in enhancing efficiency and reducing wear in mechanical systems.

Delving into Terminal Velocity and Circular Motion: Terminal velocity, the maximum speed reached by a falling object, and circular motion, the circular path followed by objects in motion, offer fascinating insights into the dynamics of motion. By conducting experiments and illustrating concepts such as centripetal force and banking of roads, we unravel the intricate dynamics of terminal velocity and circular motion.

As we embark on this enlightening journey through the realm of motion, we invite you to immerse yourself in the captivating interplay of matter, space, and time, where the nuances of motion unfold with profound beauty and complexity.

Manufura

  1. Differentiate between angular speed and velocity
  2. Examine methods of reducing friction
  3. Discuss banking of roads in reducing sideways friction
  4. Demonstrate motion in vertical/horizontal circles
  5. Examine terminal velocity and its determination
  6. Describe centripetal force
  7. Explain the concept of push and pull
  8. Understand the various types of motion
  9. Illustrate fluid friction and its application in lubrication
  10. Analyze the disadvantages and advantages of friction
  11. Discuss the role of friction in motion
  12. Describe the force as the cause of motion

Takardar Darasi

Motion refers to the change in position of an object with respect to time. It is a fundamental concept in physics and occurs when an object changes its position relative to a reference point. Understanding motion is crucial for analyzing and predicting various physical phenomena.

Nazarin Darasi

Barka da kammala darasi akan Motion. Yanzu da kuka bincika mahimman ra'ayoyi da ra'ayoyi, lokaci yayi da zaku gwada ilimin ku. Wannan sashe yana ba da ayyuka iri-iri Tambayoyin da aka tsara don ƙarfafa fahimtar ku da kuma taimaka muku auna fahimtar ku game da kayan.

Za ka gamu da haɗe-haɗen nau'ikan tambayoyi, ciki har da tambayoyin zaɓi da yawa, tambayoyin gajeren amsa, da tambayoyin rubutu. Kowace tambaya an ƙirƙira ta da kyau don auna fannoni daban-daban na iliminka da ƙwarewar tunani mai zurfi.

Yi wannan ɓangaren na kimantawa a matsayin wata dama don ƙarfafa fahimtarka kan batun kuma don gano duk wani yanki da kake buƙatar ƙarin karatu. Kada ka yanke ƙauna da duk wani ƙalubale da ka fuskanta; maimakon haka, ka kallesu a matsayin damar haɓaka da ingantawa.

  1. What is the term for a type of motion where an object moves along a straight line? A. Rectilinear motion B. Circular motion C. Rotational motion D. Oscillatory motion Answer: A. Rectilinear motion
  2. Which of the following forces is responsible for keeping an object in circular motion? A. Frictional force B. Tension force C. Centripetal force D. Gravitational force Answer: C. Centripetal force
  3. What is the force responsible for causing an object to start moving from rest? A. Tension force B. Frictional force C. Gravitational force D. Applied force Answer: D. Applied force
  4. What is the term used to describe the force that opposes the motion of an object sliding on a surface? A. Tension force B. Normal force C. Frictional force D. Gravitational force Answer: C. Frictional force
  5. Which of the following is NOT an advantage of friction in motion? A. It helps in locomotion B. It reduces efficiency C. It aids in using belts and grindstones D. It causes wear and tear of machines Answer: B. It reduces efficiency

Littattafan da ake ba da shawarar karantawa

Concepts of Physics
Concepts of Physics
Sunaƙa: Understanding the Principles of Physics
Mai wallafa: HC Verma
Shekara: 2000
ISBN: 9788177091878
Fundamentals of Physics
Fundamentals of Physics
Sunaƙa: Introduction to Motion and Forces
Mai wallafa: David Halliday, Robert Resnick, Jearl Walker
Shekara: 2013
ISBN: 9781118230718

Ƙarin koyo daga tushe na buɗe

Wasu ƙungiyoyi ne suka wallafa su ƙarƙashin lasisin buɗe, kuma muna ambaton su yadda suka nema. Hanyoyin suna buɗe shafukansu.

Tambayoyin Sake Nazari

Kana ka na mamaki yadda tambayoyin baya na wannan batu suke? Ga wasu tambayoyi da suka shafi Motion daga shekarun baya.

JAMB UTME - Physics - 2024

Tambaya 1 Rahoto

Two points on a velocity-time graph have coordinates (2s, 5m/s) and (4s, 15m/s). Calculate the mean acceleration
Bayanin Amsa

The mean acceleration of an object is determined by the change in velocity over the change in time. This is given by the formula:


Mean Acceleration (a) = (Final Velocity - Initial Velocity) / (Final Time - Initial Time)


From the velocity-time graph, we have the following points:

Initial Point: (2s, 5m/s)

Final Point: (4s, 15m/s)


Here, the Initial Velocity is 5m/s, the Final Velocity is 15m/s, the Initial Time is 2s, and the Final Time is 4s.


Plug these values into the formula:

Mean Acceleration (a) = (15m/s - 5m/s) / (4s - 2s)


Simplifying this, we get:

Mean Acceleration (a) = 10m/s / 2s = 5m/s²


The mean acceleration is therefore 5.0 m/s².

Duba Amsa
NECO SSCE - Physics - 2005

Tambaya 1 Rahoto

A body moves along a circular path with uniform angular speed of 0.6 rad s-1 and at a constant speed of 3.0 ms-1.

Calculate the acceleration of the body towards the centre of the circle.
Duba Amsa
WAEC SSCE - Physics - 2007

Tambaya 1 Rahoto

(a) State the conditions of equilibrium for a number of coplanar parallel forces.

(b) A metre rule is found to balance horizontally at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate the;

(i) mass of the metre rule;

(ii) distance of the balance point from the zero end, if the body were moved to the 13 cm mark.

(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;

(i) draw a diagram to illustrate all the forces acting on the box while in motion;

(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.

Bayanin Amsa

(a) Conditions of equilibrium for coplanar parallel forces

A number of coplanar parallel forces are in equilibrium when:

  1. The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
  2. The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).

(b) Balancing a metre rule

The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.

(i) Mass of the metre rule

With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:

\[ 60\times(30-6)=m\times(48-30) \]\[ 60\times24=m\times18 \]\[ m=\frac{60\times24}{18}=\frac{1440}{18}=80\,\text{g} \]

The mass of the metre rule is 80 g.

(ii) New balance point with the body at the 13 cm mark

Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):

\[ 60\times(x-13)=80\times(48-x) \]\[ 60x-780=3840-80x \]\[ 140x=4620 \]\[ x=\frac{4620}{140}=33\,\text{cm} \]

The new balance point is 33 cm from the zero end.

(c) Pulling a box up an inclined plane

Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).

(i) Diagram of the forces acting on the box while in motion

The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.

figure
Free-body diagram of the box on the 30° inclined plane: weight W acts vertically down, normal reaction N perpendicular to the plane, effort E up the plane, and friction F down the plane opposing motion.

(ii) Calculations

I. Minimum effort applied in pulling up the box

Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:

\[ E=W\sin\theta+F=700\sin30^{\circ}+1000 \]\[ E=700\times0.5+1000=350+1000=1350\,\text{N} \]

The minimum effort is 1350 N.

II. Velocity ratio of the plane

\[ V.R.=\frac{1}{\sin\theta}=\frac{1}{\sin30^{\circ}}=\frac{1}{0.5}=2 \]

(This agrees with \(\dfrac{\text{length}}{\text{height}}=\dfrac{5}{2.5}=2\).) The velocity ratio is 2.

III. Force ratio (mechanical advantage) of the plane

\[ \text{Force ratio}=M.A.=\frac{\text{load}}{\text{effort}}=\frac{700}{1350}=0.52 \]

The force ratio of the plane is 0.52.

Duba Amsa