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[Ended]Physics Tutoring Session #9 "Motion" on Wed. 2nd March, 2016 @ 3:00 - 4:00pm
Physics
Chima Sunday (Tutor)
02-03-2016 13:12:00 +0000Today we are going to take a new dimension. Basically, we will be discussing on Motion, but our priority will be on solving motion related problems often tested in JAMB and other exams. So sit tight and have a nice tutoring hour. Don't forget to come along with your questions, and were you need more explanations, and i promise to do the best within my power to make the time worth.
Welcome and enjoy your time.
This section has ended Don't forget to fill our the survey after the session:
Survey Feedback Link - http://goo.gl/forms/Tasr40Ck1Zmotion
If you had to think consciously in order to move your body, you would be severely disabled. Even walking, which we consider to be no great feat, requires an intricate series of motions that your cerebrum would be utterly incapable of coordinating. The task of putting one foot in front of the other is controlled by the more prim- itive parts of your brain, the ones that have not changed much since the mammals and reptiles went their separate evolutionary ways. The thinking part of your brain limits itself to general directives such as “walk faster,” or “don’t step on her toes,” rather than mi- cromanaging every contraction and relaxation of the hundred or so muscles of your hips, legs, and feet.
Physics is all about the conscious understanding of motion, but we’re obviously not immediately prepared to understand the most complicated types of motion. Instead, we’ll use the divide-and- conquer technique. We’ll first classify the various types of motion, and then begin our campaign with an attack on the simplest cases. To make it clear what we are and are not ready to consider, we need to examine and define carefully what types of motion can exist.
Motion involves a change of position of a body with time. The study of motion without involving the cause of it is call KinematicsTypes of Motion
a) Random Motion: Here body moves in a haphazard or disorderly manner. Example is the Brownian motion, motion of smoke particles etc.
b) Translational Motion: here body moves from one point to another without rotating. E.g is the motion of a boy running from one goal post to another, a ball rolling vertically upward. etc
c) Rotational Motion: It involves motion of a circle about a center or axis. E.g rotation of the fan blade, motion of the wheel of a moving car. etc
d) Vibratory Motion: In this kind of motion the path of a moving body is repeated at a succesive equal intervals of time. E.g is the motion of devil as described in the christain bible, motion of a simple pendulum. etc.Parameters of Motion
i) Displacement is defined as the distance travelled in a specified direction
ii)Speed v is the rate of change of distance with time. v= s/t. Speed is said to be uniform if s/t is constant through out the journey.
Velocity is speed calculated in a given direction[recall that distance is for speed and rate of change of displacement is for velocity]
iii)Acceleration a is the rate of change of velocity increase with time While De acceleration is the rate of change of velocity decrease with time. a = v/t.
A motion is said to be a uniform acceleration if v/t is constant through out the journey e.g is the acceleration due to gravity.Equations of Motion
Examples
If a body starts with initial velocity u, accelerates uniformly along a straight line with acceleration a and covers a distance s in a time t when it velocity reaches a final velocity v. then
s = (v+u/2)t ...............................(1)
a = v-u/t; v = u + at ......................(2)
Eliminating t in (1) and (2) to obtain
v2 = u2 + 2as ........................... (3)
Eliminating v in (1) and (2) to have
s = ut + (1/2)at2
These four equations are used in solving problems associated with uniformly accelerated motion.
(Jamb 2015 Question 4) What force has to be exerted on a mass 60kg to give it an acceleration of 10ms-2 vertically upwards? [g =10ms-2]
Option A 600N
Option B 1200N
Option C 400N
Option D 300N
Solution
Here we need the force that to find the resultant R
let the upward force be F, and down ward force be W(weight of the body i.e mg). So since the two are opposite, then, m = 60kg, a = 10m/s^2 and g = 10m/s^2
R = F - W
so F = R + W = ma + mg = m(a + g) = 60(10 + 10) = 1200N.
(Jamb 2015 Question 5) A satellite revolving around the earth is kept on its orbit by
Option A centrifugal forces only
Option B centripetal forces only
Option C centripetal and frictional forces
Option D centripetal and centrifungi forces
Solution
When a body is undergoing a rotational motion, two forces are invited for a play, one is centripetal force, this force is trying to keep this body in motion in its circulating orbit while the other force centrifugal, just like its name "fugal" it is trying to force the body out from its orbit. But since this are equal and opposite. its resultant effect is zero. Note the question ask for the one keeping the body in its orbit, so correct answer is Option B.
Question 46 A train with an initial velocity of 20ms-1 is subjected to a uniform deceleration of 2ms-2 the time required to bring the train to a complete halt is
Option A 20s
Option B 40s
Option C 10s
Option D 5s
Solution recall from equations of motion that v = u + at, were symbols have there usual meaning.
so t = (v - u )/a. since a is decelerating a = -a, final velocity v = 0
then t = -u/-a = u/a = 20/2 = 10s
So Answer is Option C
(Jamb 2013 Question 7) Calculate the apparent weight loss of a man weighing 70kg in an elevator moving downwards with an acceleration of 1.5ms-2
Option A 686N
Option B 595N
Option C 581N
Option D 105N
Solution
When a body is in an elevator moving downward its resultant force R is given below
R = F + W
, F = W - R = mg + ma = m(g - a).
a = 1.5m/s^2, g = 10m/s^2
F = 70(10 - 1.5) = 70(8.5) = 595N
So apparent loss in weight = the man weight in air - his weight in the elavator = 700 - 595 = 105N
Answer Option D
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Chima Sunday (Tutor)
once again i welcome you all to our today's tutorial class. i still remain my humble self, your one and the only tutor in this forum. Don't think my name has actually changed still my self. You can post your question or go through the script above to see if you really understand, then in absent of any questions i will bombard you guys with my own questions once again welcome @Oluwaseun Babatope, Francis David and co.
0 02-03-2016 14:07:00 +0000
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Chima Sunday (Tutor)
Can any one try this
? (Jamb 2013 Question 6) A train with an initial velocity of 10ms-1 is subjected to a uniform deceleration of 2ms-2. The time required to bring the train to a complete halt is
Option A 5s
Option B 10s
Option C 20s
Option D 40s
solution
recall from equations of motion that v = u + at, were symbols have there usual meaning. by subject formular
so t = (v - u )/a. since a is decelerating a = -a, final velocity v = 0
then t = -u/-a = u/a = 10/2 = 5s.
answer is option A.0 02-03-2016 14:14:00 +0000
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Chima Sunday (Tutor)
hey hope i got some student in here were have they gone to @Oluwaseun Babatope Francis David etc
0 02-03-2016 14:26:00 +0000
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Chima Sunday (Tutor)
Ok this is another one here?
Question 9 An object is moving with a velocity of 5m/s. At what height must a similar body be situated to have a potential energy equal in value with kinetic energy of the moving body?
Option A 25.0m
Option B 20.0m
Option C 1.3 m
Option D 1.0m
updated solution for this to happen, we equate the p.e =k.e
i.e kinetic energy = potential energy
(1/2)mv^2 = mgh. observe that m cancels
so h = (1/2)v^2/g = 25/20 = 1.3
so answer is option C.0 02-03-2016 14:30:00 +0000
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Admin Admin
Thought you will miss todays class you must be a very serious student @Jibo Khadijah. Anway you are welcome ok. Yah which question number are you answering dear?
0 02-03-2016 14:37:00 +0000
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