Secondary 4 Preliminary Examination Physics (18th September 2008) II

From: MARIS STELLA HIGH SCHOOL, Secondary 4 Preliminary Examination 2, PAPER 1 MULTIPLE CHOICE, NO. 22, 18th September 2008

A girl of height 1.6m stands 0.5m in front of a vertical plane mirror. What is the minimum length of the mirror which will the girl to see the whole of herself?
A) 0.5m
B) 0.8m
C) 1.3m
D) 1.6m

Solutions:

The minimum length of the mirror is obtained based on the following assumptions.

• The eye is right at the top of the head.
• The mirror is placed with its top at eye level, that is, same level as the top of the girl's head.
• The girl has normal good eyesight.
• Sufficient light intensity.

PICTURE: http://faculty.etsu.edu/gardnerr/einstein/e-reflection.jpg


According to Laws of reflection

The incident ray, the reflected ray and the normal at the point of incidence all lie on the same plane.
The angle of incidence, i, is equal to the angle of reflection, r.

• The angle of incidence, i, is the angle between the incident ray and the normal.
• The angle of reflection, r, is the angle between the reflected ray and the normal.
• The normal is the perpendicular line to the reflecting plane.

Using the laws of reflection, a mirror of half the height of the girl is sufficient for the girl to see the whole of herself. The incidence ray of the lowest part of the girl, the toe, reaches the bottom edge of the mirror and is reflected to the eye of the girl.


PICTURE: https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9Auue76yBpsaqU6-kEIqCYK5FKpdvgJFrj1rkwHdV-17G7tB4GQHUr43MrC0oOY-m9DduhAhHS-acG3xgtdxROlR3UIUYqpYk6qwhjz6BXsV1sz0vX8kivfJiP4uwC1s67D9gN6ZKtcc/s320/reflection.bmp

If the length of the mirror is shorter, then the angle of incidence ray of the toe will be too great causing the reflected ray to be above the eye and thus not seen by the girl.

The answer is B) 0.8m




Source: http://www.physicsclassroom.com/class/refln/u13l2d.cfm

What Portion of a Mirror is Required?

Ray diagrams can be used to determine where a person must sight along a mirror in order to see an image of him/herself. As such, ray diagrams can be used to determine what portion of a plane mirror must be used in order to view an image. The diagram below depicts a 6-foot tall man standing in front of a plane mirror. To see the image of his feet, he must sight along a line towards his feet; and to see the image of the top of his head, he must sight along a line towards the top of his head. The ray diagram depicts these lines of sight and the complete path of light from his extremities to the mirror and to the eye. In order to view his image, the man must look as low as point Y (to see his feet) and as high as point X (to see the tip of his head). The man only needs the portion of mirror extending between points X and Y in order to view his entire image. All other portions of the mirror are useless to the task of this man viewing his own image.

PICTURE: http://www.physicsclassroom.com/class/refln/u13l2d1.gif

The diagram depicts some important information about plane mirrors. Using a cm-ruler, measure the height of the man (the vertical arrow) on the computer screen and measure the distance between points X and Y. What do you notice? The man is twice as tall as the distance between points X and Y. In other words, to view an image of yourself in a plane mirror, you will need an amount of mirror equal to one-half of your height. A 6-foot tall man needs 3-feet of mirror (positioned properly) in order to view his entire image.

But what if the man stood a different distance from the mirror? Wouldn't that cause the man to need a different amount of mirror to view his image? Maybe less mirror would be required in such an instance? These questions can be explored with the help of another ray diagram. The diagram below depicts a man standing different distances from a plane mirror. Ray diagrams for each situation (standing close and standing far away) are drawn. To assist in distinguishing between the two ray diagrams, they have been color coded. Red and blue light rays have been used for the situation in which the man is standing far away. Green and purple light rays have been used for the situation in which the man is standing close to the mirror.

PICTURE: http://www.physicsclassroom.com/class/refln/u13l2d2.gif

The two ray diagrams above demonstrate that the distance which a person stands from the mirror will not affect the amount of mirror which the person needs to see their image. Indeed in the diagram, the man's line of sight crosses the mirror at the same locations. A 6-foot tall man needs 3-feet of mirror to view his whole image regardless of where he is standing. In fact, the man needs the exact same 3-feet of mirror.

A common Physics lab involves using a tall plane mirror to explore the relationship between object height and the portion of mirror needed to view an image. A student stands a few meters from a planer mirror and views her image. With the student standing upright and still and staring at her feet, the lab partner moves a marker up and down the mirror until the sight location on the mirror is identified. The partner then marks this location on the mirror with an erasable marker. The process is repeated for the student staring at the tip of her head. Of course, being a lab, the procedure is subject to a variety of procedural and measurement error which may yield less than ideal results. The mirrors are occasionally mounted on a wall which is not perfectly vertical. Or a student will lean forward a slight amount, thus reducing his/her effective height. Or the mirror warps over the years leading to one which concave or convex rather than planar. Despite these potential complications, the 1:2 ratio between portion of mirror required to view the image and the height of the object is often observed.

Reference

• MARIS STELLA HIGH SCHOOL, Secondary 4 Preliminary Examination 2, PAPER 1 MULTIPLE CHOICE, NO. 22, 18th September 2008
• http://faculty.etsu.edu/gardnerr/einstein/e-reflection.jpg
• https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9Auue76yBpsaqU6-kEIqCYK5FKpdvgJFrj1rkwHdV-17G7tB4GQHUr43MrC0oOY-m9DduhAhHS-acG3xgtdxROlR3UIUYqpYk6qwhjz6BXsV1sz0vX8kivfJiP4uwC1s67D9gN6ZKtcc/s320/reflection.bmp
• http://www.physicsclassroom.com/class/refln/u13l2d.cfm
• http://www.physicsclassroom.com/class/refln/u13l2d1.gif
• http://www.physicsclassroom.com/class/refln/u13l2d2.gif

Secondary 4 Preliminary Examination Physics (18th September 2008)

From: MARIS STELLA HIGH SCHOOL, Secondary 4 Preliminary Examination 2, PAPER 1 MULTIPLE CHOICE, NO. 3, 18th September 2008

The acceleration of free fall is determined by timing the fall of a steel ball photo-electrically. The ball passes X and Y at times t_x and t_y, after released from P. Which of the following expression gives the correct computation of acceleration?

A) 2h/( t_x - t_y)
B) h/( t_y² – t_x²)
C) h²/( t_x - t_y)
D) 2h/( t_y² – t_x²)


Solution:








To obtain the displacement, we have to integrate the eq ④


With reference to the diagram above, to compute the acceleration, g, between time t_x and t_y, the time taken t = ( t_y - tx), displacement y = h, and initial velocity V_i = Vx

Substituting into eq ⑤





Supporting Explanations
Source: http://www.school-for-champions.com/science/gravity_equations_derivation.htm

Overview of distance and time relationships

The displacement a falling object travels in a given time is found by knowing that velocity is the change in displacement with respect to time:

v = dy/dt

Substituting for v in the equation v = gt + v_i and integrating, we get:

y = gt2/2 + vit

Note: Sincethe convention is that down is a positive direction, the downward distance y from the starting point is also positive.

Rearranging y = gt²/2 + v_it and solving the quadratic equation for t gives you:

t = [ −vi± √(vi2+ 2gy) ]/g

This equation can create some confusion because of the plus-or-minus sign. If the object is thrown downward, the plus (+) sign is used. If the object is thrown upward, the sign depends on the object's position with respect to the starting point.

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Reference

• MARIS STELLA HIGH SCHOOL, Secondary 4 Preliminary Examination 2, PAPER 1 MULTIPLE CHOICE, NO. 3, 18th September 2008
• http://www.school-for-champions.com/science/gravity_equations_derivation.htm

Secondary School Mid-Year Holiday is a Nightmare!

Holiday (noun) is known as:

• A period of time spent away from home for enjoyment and relaxation
• (Often plural) Chiefly Brit & NZ a period in which a break is taken from work or studies for rest or recreation
• A day on which work is suspended by law or custom, such as a bank holiday

From: http://www.thefreedictionary.com/Holiday
By Collins Essential English Dictionary 2nd Edition 2006 © HarperCollins Publishers 2004, 2006



From http://www.rd.ca/cms/images/image/extraburnout_230_y0u.jpg


School Assignment (noun) is defined as:

• A school task performed by a student to satisfy the teacher

From: http://dictionary.reference.com/browse/school+assignment?qsrc=2446
By WordNet® 3.0, © 2006 by Princeton University.


Nightmare (noun) is known as:

• A dream arousing feelings of intense fear, horror, and distress.
• An event or experience that is intensely distressing.
• A demon or spirit once thought to plague sleeping people.

From: http://www.answers.com/topic/nightmare

This coming Secondary School Mid-Year Holiday is from 29 May 2009 till 28 June 2009, almost the same as the Great Singapore Sales. It is supposed to be a period in which a break is taken from work or studies for the students. Students can take this opportunity to catch up with what they are weak in and have the rest of the time for rest or recreation activities.

Unfortunately this is not the case. The schools have been very enthusiastic in preparing and creating new lessons, worksheets, assignments, forums and quiz for almost every subject to be carried out by the students during the holiday period. Just to name a few subjects, English, Chemistry, Physics, Social Studies, E-Maths, A-Maths, Chinese, ..... For Chinese, they have to read two story books and complete the online assignments.


From http://genychina.com/wp-content/uploads/2009/02/technology-overload2.gif

During the school holiday, the students have to go back to take their English Oral Communication exam. Meanwhile they are told of a guide book, which cannot be found in any bookshop, to refer to. Even before the Oral exam has started, the schools have started posting articles, worksheets, assignments and lessons for the students to work on.

We are to suppose that the teachers would like to have their holidays and put the plight of their students to the back of their mind. The MOE has the slogan "Teach less, Learn more", which they intend to strictly adhere to. After all, a fully paid holiday is hard to come by, especially during this recession period. Brown harvests and green shoots have no value if there are iron bowls full of rice. Ah Kong's money seems to be easily planned and obtained.


Two headed tree ?

To have an idea of the amount of time required to carry out the tasks, they are tabulated for easy understanding and calculation.





















Estimated number of days available = 20
Time required per day = 224 ÷ 20 = 11.2 hours
This hour is 100% more than that of a normal 5.5hrs school day.

The student also has another 5 days of school related activities to attend.
It will be a very exhausting holiday.

When the school re-opens on 29 June 2009 there is a group of students selected to perform one week of official duties for the Asian Youth Game hosted in Singapore. During this one week, the rest of the students will attend classes normally whereas those on duty will be denied this opportunity. Instead they will have problems catching up with their studies. What the school is willing to do is just to keep the homework papers for them. Even that is also to be done by a school colleague living nearby. There is no plan on how to remedy the missed lessons. This is a very stressful situation for the student and this “crime” is allowed to be committed by the school which participated in the AYG.

Reference

• http://www.thefreedictionary.com/Holiday
• Collins Essential English Dictionary 2nd Edition 2006 © HarperCollins Publishers 2004, 2006 - Holiday definition
• http://dictionary.reference.com/browse/school+assignment?qsrc=2446
• WordNet® 3.0, © 2006 by Princeton University - school assignment definition.
• http://www.answers.com/topic/nightmare
• http://dictionary.reference.com/browse/nightmare
• http://genychina.com/wp-content/uploads/2009/02/technology-overload2.gif
• http://www.rd.ca/cms/images/image/extraburnout_230_y0u.jpg