Need help with 2 prac questions (1 Viewer)

[davidbarnes]

With the 'Analysis' part, how would I do question 3?



With the 2nd prac above again its with the 'Analysis'. How woudl I do question 1 and 2? Has just got me confused that one.

A speedy reply would be greatly appreciated.

 

[twilight1412]

lol man ... havent been here since hsc..
since noone is answering your questions i'll do it

the first link
question 3
you are given a formula somewhere along the line
its telling you to get it into the form y=kx
which is the formula of a straight line passing through the origin
it then tells you that y = T²
and so whatever else you have must = kx
and if x = l
then whatever is left is equal to k
which is your gradient

second link
to answer these questions
YOU NEED TO DO THE EXPERIMENT
otherwise its just pointless
question 1 is basically telling you to pull out a stop watch and ruler
and measure how far the ball is down the slope every second
question 2
v = d/t
where d is distance (cm)
and t = time (secs)
for every second find the distance it travelled divided by the time it took to get between the 2 points WHICH SHOULD BE 1 SEC

and that my friend is the basic gist of the answer

 

[Forbidden.]

With the 'Analysis' part, how would I do question 3?



With the 2nd prac above again its with the 'Analysis'. How woudl I do question 1 and 2? Has just got me confused that one.

A speedy reply would be greatly appreciated.

1. Your graph should display a straight-line relationship. Draw a line of best fit and evaluate the gradient.

2. Rearrange the pendulum equation given earlier to the form; T² = kl where k is a combination of constants

3. Compare this formula with the general equation for a straight line: y = kx. This comparison shows that if T² forms the y-axis and length, l, forms the x-axis, the expression you derived for mass, m, in step 2 should correspond to the gradient of the graph you have drawn. Write down your expression: gradient = _____________ (complete)

Prepare for my thorough explanation. Once you have read it, you should know how to answer the questions well.
It seems Question 1 and 2 kind of go together.

Firstly, know that the symbols and values mean before approaching the question.
Hopefully you know that the constants don't change and variables do change.

T - dependent variable (depends on independent variable i.e. l) - period of the pendulum - seconds (s)
2π - constant
g - constant - acceleration due to gravity at the surface - metres per second squared (ms^-2)
l - independent variable (You can manipulate this i.e. change the length of the string) - length of the pendulum string - metres (m)

Now k is a constant as in 1, 2, 3 and so on .... draw (or imagine) y = x, y = 2x and y = 3x. You notice that three of these functions form a straight line.
So graphs in the form y = kx form a straight line !

(A part of Question 3 but should be known beforehand)

If T = 2π√(l/g), then squaring the entire formula gives T² = 4π²(l/g).
Why can't you use T = 2π√(l/g) ?
Because it is in the form y = √x which doesn't give a straight line but the T² does because all square roots are eliminated by squaring and that gives a straight line !

The k in T² = kl means shove all the constants into k and leave T² and l apart.
Refer to the image below

So the k is 4π²/g

(This should be Question 2)

You remember y = mx + b ?
Omit the b because if l = 0, there is no period T, there is no y-intercept (i.e. no cutting of the y-axis).
So there is y = mx, now the x which is the x-axis representing the length of the string (l) versus the y which is the y-axis representing the period squared (T²) of the pendulum.
Graph it.

(This should be Question 1)

Now m is the gradient, it is the constant in the pendulum formula, hence it is 4π²/g
The grandient is found by m = rise / run (or use the y = y₂ - y₁ / x₂ - x₁), find its value and use it to find g.
Make g the subject therefore g = 4π²/m

(This should be Question 3)

Note m isn't the mass, it is neglected in this experiment, it doesn't appear in the formula T = 2π√(l/g), and neither does amplitude.
(But you do need to keep the angle below 10⁰ to keep the experiment as a simple pendulum)
(This information was taken from a Physics teacher on the HSC Advice Line)

 

[davidbarnes]

second link
to answer these questions
YOU NEED TO DO THE EXPERIMENT
otherwise its just pointless
question 1 is basically telling you to pull out a stop watch and ruler
and measure how far the ball is down the slope every second
question 2
v = d/t
where d is distance (cm)
and t = time (secs)
for every second find the distance it travelled divided by the time it took to get between the 2 points WHICH SHOULD BE 1 SEC

and that my friend is the basic gist of the answer

We didn't use a stopwatch or time it in anway when we done this.

 

[twilight1412]

lol forbidden how long did that take you to write up?

 

[Forbidden.]

I wish I had the Jacaranda when I did the HSC.
Our projectile motion practical was different to the one in the Jacaranda textbook.
But,
The time of flight is independent from the initial velocity.
From the shape of the graph we can see the range is directly proportional to the initial horizontal velocity.

lol forbidden how long did that take you to write up?

Haha, don't remember !

We didn't use a stopwatch or time it in anway when we done this.

You should have, it is an essential equipment in finding the time of flight.

 

[twilight1412]

lol? projectile prac? omg i didnt even get one
i suggested bottle rockets over the back oval ... rejected straight away

 

[Forbidden.]

lol? projectile prac? omg i didnt even get one
i suggested bottle rockets over the back oval ... rejected straight away

Our one consists of a soccer ball being pushed off a table on a balcony, it travels in the path of half a parabola, very similar to the one in the HSC 2003 paper.
LOL at your suggestion, it could be rejected because the practical isn't as simple as the trajectory path would be a full parabola which introduces more variables such as the angle of trajectory.

 

[twilight1412]

but it will never be a true parabola unless its in a perfect vaccum and under the effects of only one gravitational field
which is techinically impossible lol
air resistance will see to it that it will never be parabolic