## Thursday, August 30, 2012

### The views from Fort Canning

The atmosphere in Fort Canning Park is so serene in the evenings. Worth to spend your time walking or jogging. Such lush greenery to relax your eyes. So serene you can hear your soul connecting to you.
The trees are so majestic and careless of the hustle just a block away.
The tress outlived those who planted them. They aged so gracefully. The parasites seems to keep the host company.
The flowers choose to spring out from the trunks.
I wonder the lighthouse can guide shoppers to the right direction.
What is a time ball?
Why won't anyone walk through this gate to the past.
Whenever one is in search for direction into the future.

## Sunday, May 13, 2012

### Circular movement

Humanly possible to exert a force sideways when one is moving forward.

## Tuesday, March 6, 2012

### Kinematics_Help

Tutorial 2
Question 1
ai) average acc is taking the velocity at t=4.7s minus the velocity at t=0 and dividing by 4.7s.
ii) The greatest instantaneous acc can be found by finding the point where the gradient is the steepest. Using a ruler and tilting it until you touch one point at the steepest gradient. Draw that line until you cover the whole graph for greater accuracy. Hence the gradient is taking the triangle where opp is 100 kmh-1 which is 23.6 ms-1 divide by the time taken the x axis ie. the line covers the whole graph.
iii) Estimate the area using trapezium formula. Area enclosed from t=2s to t=3s.
b) Construct a line from the peak to the x axis in a time of 3.7s . Since it is uniform deceleration the distance is area of a triangle. Calc the area enclosed.

Question 2
Draw the velocity-time graph of the driver. The area under the graph gives the distance the driver takes to stop hence the safe distance .Note during reaction time of 0.25s the car continue to move at constant speed.

Question 3
a) The s-t graph is a tedious process. You need to sum the area at every 2 s to obtain displacement . Key points to remember is that the area is negative below the time axis and positive above the time axis. At v = o at t=4s is the turning point of the s-t graph. The sequence of s at t =2,4,6,8,10 and 12s. The displacement points are -40, -50 , -40 ,0 , -50 and -70m. Note :1. when velocity is negative , the gradient is negative.
2. The velocity is constant at t = 6-8s, and t= 10-12s, the gradient is constant on the s-t graph.
b) The a-t graph is just the gradient of the v-t graph. Take note of positive and negative gradient which is fall above and below the x axis of the a-t graph. Graph looks like steps...

Question 4.
You need equation of motion to solve this very important question.
One involving s and time will be s=ut + 0.5at2. This equation is simplified if the motion is considered from the top where u = 0. Hence s = 0.5at2.
Consider first time t1 when the bottom edge reach the marker, s = 1.8 m .
The second time t2 is when the top edge reach the marker, s = (1.8 + L)m
Writing the two equations : 1.8 =0.5at'2 and 1.8+L =0.5at"2 but t"= t'+0.172s.

Question 5
First decide on the master arrow , down positive.
a) s = + 2.0 m , a= +9.81, and u = 0. Use v= u + at
b)repeat a) but find v using v2 = u2 + 2as.

c) The second leg involves upward motion , hence change master arrow to , up positive.
s = +1.5m, v = 0 ms-1, and a = -9.81ms-2 using v2= u2 + 2as find u
d) repeat the data in c) but use v= u + at, the u = +value
e) Drawing the v-t graph for the whole motion require one master arrow , down positive.
Velocity is down (+) at the start and up (-) after rebound. At the time of contact with the ground the direction changes from a large positive value to a negative value in a short time , hence the line is vertical. The gradient of this graph is constant as the acceleration due to gravity is g = + 9.81ms-2 throughout.
The area enclosed in the triangle is larger for the positive displacement s=2.0m and smaller for the negative displacement s = -1.5 m(up).

Question 6
Graph work required with some patience and perserverance.
a) The area under a-t graph is the v value .
Take the time markings to be 1s, 2s, 3s and 4s for easy reference.
Graph A, the area is increasing and the value v is negative. The increase in area is greater at t = 0 to 1 s and slower from t= 1 to 2s . Similarly t= 2-3 and 3-4s. Draw a curve which is steeper at both ends. If you calc the area and plot the graph will be more accurately drawn but not necessary.
Graph B requires the same method. The a = 0 at two occasions t=1 and 3s. Hence there is a turning point on the v-t graph at those two times. At t= 2 s the positive area and negative area cancels out, hence the v = 0 .The area is + and then - to cancel out and hence v= 0 .Ignore the axes of graph at the top of page, the negative sine graph is the v-t graph of graph B.

Question 7
Equation 1 for stone 1: 200 = 0.5 at2
Equation 2 for stone 2 : 200 = 20t' + 0.5at'2.
t=6.39s and t'= 4.66s.
Comparing t - 6.39s and 4.66+1s = 5.66s, the second stone will hit the ground first.
Hence the first stone is still above ground.
s= 0.5 (9.81)(5.66)2 = 157 m . Hence it's 200-157 = 43 m above the ground.

Question 8
a) Follow A , positive to the right
b) i) graph B crosses the axis , v = 0 ms-1
ii) A and B intersect, the v is the same and hence they are closest
iii) numerical value gets smaller , ignore sign . Graph B , first quarter and last quarter section .
iv) positive acc is positive gradient of graph , at 1st , 2nd and 3rd quarter section.
The graph will be used for discussion as it is quite complex .

## Friday, March 2, 2012

### Congratulations !

Here is a special flower for all who are celebrating their achievements in their A level examination results.I see fairness in the Physics examination results.

Best Wishes

## Monday, February 27, 2012

### Measurement_TYS

24. percentage error in n = 25.5 %
Hence 25.5% of value is 0.3 x 10-3
Hence the written as (1.1 + 0.3) x10-3 kgm-1s-1

25. The average of several diameter reading is taken : 1. To reduce the random error
2. The wire may not be perfectly circular.

26. See definition in lecture notes
28. aii) Measure zero error and take zeror error into account to remove systematic error.
Measure the diameter several times by rotating the sphere. To remove random error .
b) Measure the volume of the mercury inside the capillary tube. The formula , density - mass/volume can be used to find the volume of mercury. The volume of the tube can thus be used to find the diameter if the length is known. The volume of a cylinder is used.
The uncertainty due to diameter measurement is (0.2 / 25 )x100 = 0.8%. This percentage is close to the diameter variation of 1%. Hence it is not possible to detect the changes in diameter. See pic of travelling microscope.
29. Parallax error exist as the specimen (metal cylinder) is a distance away from the ruler. To reduce this error , use a set square to point at the ruler (touching the ruler scale reading ).
See how a traveling microscope works.
Look into microscope and read the reading R1 (vernier scale) . Move the microscope by turning the knob which moves the microscope to read R2. The length moved by the microscope is R1-R2. Hence the diameter of the capillary tube can be measured.

## Friday, February 17, 2012

### Lesson_102

Question 3:
To find period T , divide time by N
A student 1 took N=9 , hence period is T= t/9, the correct measurement.
Instead he counted as N=10 as he started with 1 instead of 0. Hence his period T' is t/10.This in smaller than the actual , hence a drop in perod.To calc percentage , take the reading - actual and divide by actual. Hence the percentage drop is indicated with a negative sign and (T-To)/To. Which is (T/To)- 1. Hence, (10/9 -1)100%. The answer is B

## Thursday, November 17, 2011

### Query 1

Intensity is the same all over the surface if the r is the same . Hence it must be perpendicular to the direction of travel of wave.

next: If ratio of frequency means ratio of wavelength also. I dont think there is any special thing you need to know concerning this ratio.
Overlap means some colours of higher order are seen at smaller angles before the lower order colours ended. Eg. third order blue is seen than second order red. ( 3rd than 2nd order)hence overlap is said to have occured at higher orders. It may happen at higher orders depending on the d (diffraction grating)
See my post label waves, july 16 , overlapping spectrum ,got some pic.

## Tuesday, June 21, 2011

### Consultation on 24th June

Need me to piece together what is still puzzling you. Consultation available on 24 th June. Friday .
12pm to 4pm .
On the hour.