Secondary 4 Rate of Flow (Hey Math)

From: Hey Math! assignment (E-Maths), July 2010, Question #5, Revision Paper

A rectangular shaped bath tub can be filled with water flowing from two taps in 1 hour and 20 minutes. If the cold water tap alone takes 2 hours less than the hot water tap to fill the bath tub, find the time taken in hours by the hot water tap to fill the bath tub.

Select the correct answer from the following choices i, ii, iii or iv
i) 1⅓ hours
ii) 2 hours
iii) 4 hours
iv) 6 hours


Solution:
Let the volume of the bath tub be v m³
Let the rate of flow of the cold water tap be x m³/ h
Let the rate of flow of the hot water tap be y m³/ h

Let the time taken by the cold water tap to fill the tub be z hours


Using the cold water tap alone,
Volume, v = xz m³



Using the hot water tap alone,
Volume, v = y(z + 2) m³



Using both water taps at the same time,
Volume, v = (x + y)( 1⅓) m³ since 1 hour 20 minutes is 1⅓ hours

Substituting x and y into the equation,
v=(x + y)( 1⅓)

4(2z+2)=3z(z+2)
8z+8=3z²+6z
3z²-2z-8=0
(3z+4)(z-2)=0
Solving for z,
z = 2 or



Therefore, time taken by the cold water tap to fill the tub is 2 hours
And time taken by hot water tap is 2 + 2 = 4 hours

The answer is (iii).


Reference

• Hey Math! assignment (E-Maths), July 2010, Question #5, Revision Paper

Secondary 4 Circle and trigonometry (Hey Math)

From: Hey Math! assignment (E-Maths), July 2010, Question #7, Circle and trigonometry

An arc which subtends and angle of 60° at O is shown below. Find the ratio of r : R.



Select the correct answer from the following choices.
i) 1 : 2
ii) 1 : √2
iii) 1 : √3
iv) 1 : 3


Solutions:

With reference to the following diagram;



The radius ABO passed through the center of the circle ARC and bisects the angle EOD.
Triangle BCO is a right angled triangle with angles 30°, 60° and 90°.

By trigonometry,
if BC = r, then BO = 2r.

Therefore radius of the arc DAE = AB + BO = 3r
Given that the radius of the arc DAE = R,
That is, R = 3r
Therefore the ratio of r : R is 1 : 3.

The answer is (iv).


Reference

• Hey Math! assignment (E-Maths), July 2010, Question #7, Circle and trigonometry

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/AVvXsEiiqujIIX70JcbSTzFz8xJeQqiRx20Mcnn-UDYKYglDTx_5xPjr9D1IAPkyCEa90WY1ooErTZ-JKSzelp9Nu1yt-Wigo7ErTXF8FuYELGtiRKJRtLf0CRpPEFG_-Q2L4xkes3rBUEhuZk4/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/AVvXsEiiqujIIX70JcbSTzFz8xJeQqiRx20Mcnn-UDYKYglDTx_5xPjr9D1IAPkyCEa90WY1ooErTZ-JKSzelp9Nu1yt-Wigo7ErTXF8FuYELGtiRKJRtLf0CRpPEFG_-Q2L4xkes3rBUEhuZk4/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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From: TheSundayTimes, July 4, 2010, Page 7

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