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# more homework - past paper questions

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### #1claribell

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Posted 09 January 2007 - 07:52 PM

aah i really can't do physics and ive got my prelim soon

can anyone help me ive done part of this question but im unsure what to do next

the question is:

the fairway on a golf course is in two horizontal parts seperated by a steep bank.
A golf ball is given an initial velocity of 41.7m/s at an angle of 36 degrees to the horizontal
The ball reaches a maximum vertical height of 19.6m above the upper fairway - the ball hits the ground at point Q (sorry cant upload the diagram) The effect of air friction on the ball may be neglected.

(a) Calculate:
(i)the horizontal component of the initial velocity of the ball;
(ii)the vertical component of the initial velocity of the ball.

so far my working is:

U=41.7m/s
V=?
A=9.8m/s˛
S=19.6m
T=?

v˛=u˛ + 2as
=41.7˛ + (2*9.8*19.6)
=1738.89 + 384.16
=2123.05
v = 2123.05
=46.07
=46m/s

where do i go from here?

b)show that the time taken for the ball to travel from start to finish is 4.5s but i got a different answer? can someone please help me im confused

v=u + at
t=v-u/a
= 46 - 41.7/9.8
= 4.3/9.8
= 0.438
= 0.44 seconds

c)calculate the horizontal distance travelled by the ball.

another question is:

a student performs an experiment to study the motion of the school lift as it moves upwards
the student stands on the bathroom scales during the lift's journey upwards
the student records the reading on the scales at different parts of the lift's journey as follows:

part of journey - reading on the scales

At the start(while the lift is accelerating) - 678N
In the middle (while the lift is moving at a steady speed) - 588N
At the end (while the lift is decelerating) - 498N

(a)show that the mass of the student is 60kg
(b)calculate the initial acceleration of the lift
©calculate the deceleration of the lift
(d) during the journey, the lift accelerates for 1.0s, moves at a steady speed for 3.0s and decelerates for a further 1.0s before coming to rest. sketch the acceleration - time graph fot this journey

i think i am able to do part (b)© and (d) by myself

I would be very grateful if anyone could give me any help.

### #2John

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Posted 09 January 2007 - 09:52 PM

Split all the values into Horizontal and Vertical forces, and try going from there, anymore help just ask and i'll fully work through it

### #3claribell

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Posted 09 January 2007 - 09:58 PM

i done that so i was able to get the vertical velocity but i dont no how to get the horizontal would i use Vh=vcos angle ? so that i can have Vh and T to work out s will that work? for (ii)

### #4John

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Posted 09 January 2007 - 10:29 PM

What you are working with is Vectors(these should become piss easy by May)

So use Pythagoras, the sine rule or the cos rule to find the Vertical and Horizontal components of the Vector.

so lets treat it like a triangle

You know the hypotenuse is 41.7 m/s and the angle is 36 degrees

So work from there, hope that helps =)

### #5dfx

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Posted 09 January 2007 - 11:35 PM

QUOTE(claribell @ Jan 9 2007, 07:52 PM)

the question is:

the fairway on a golf course is in two horizontal parts seperated by a steep bank.
A golf ball is given an initial velocity of 41.7m/s at an angle of 36 degrees to the horizontal
The ball reaches a maximum vertical height of 19.6m above the upper fairway - the ball hits the ground at point Q (sorry cant upload the diagram) The effect of air friction on the ball may be neglected.

(a) Calculate:
(i)the horizontal component of the initial velocity of the ball;
(ii)the vertical component of the initial velocity of the ball.

so far my working is:

U=41.7m/s
V=?
A=9.8m/s˛
S=19.6m
T=?

v˛=u˛ + 2as
=41.7˛ + (2*9.8*19.6)
=1738.89 + 384.16
=2123.05
v = 2123.05
=46.07
=46m/s

where do i go from here?

Okay for part (a) all it is asking you is to resolve the velocity into its hrl and vrtl components. For that you use and

QUOTE(claribell @ Jan 9 2007, 07:52 PM)

b)show that the time taken for the ball to travel from start to finish is 4.5s but i got a different answer? can someone please help me im confused

v=u + at
t=v-u/a
= 46 - 41.7/9.8
= 4.3/9.8
= 0.438
= 0.44 seconds

c)calculate the horizontal distance travelled by the ball.

Your working worries me, it shows a lack of understanding of the basics. For parabolic motion (where you shoot off something at an angle), you always consider the vertical and horizontal bits seperately. Horizontally the acceleration is zero. Vertically the acceleration is g. Note though that although they are seperate "components", you share values between them. e.g. if you know the time it takes to get from P to Q, which you may have worked out considering the vertical motion, you can use that time for finding the horizontal distance, cause logically it is the same time. Confusing? More about this later, it should become clear with the working. Read on.

Make sure you have a really good read through of your notes and understand the whole horizontal/vertical difference cause its fundamental here!

For the time taken, consider the vertical motion. The ball travels upwards to its maximum peak height, and then is instantaneously at rest, and then starts coming back down under free fall to point Q. So we can split up the vertical motion into 2 bits, one for the upward journey, one for the downward journey:

For the first bit:
You launched the golf ball at U = 41.7Sin36 (thats the vertical initial velocity), the acceleration is 9.8, final velocity is 0, you want the time.

So t = v-u / a gives you 2.5 seconds.

So this is how long the ball takes to reach the top of its parabolic path, if you imagine it half way through its flight from P to Q, it's now right at the top. It's now going to plummet downwards to Q. Again we can ignore the horizontal bit and treat the vertical motion independently.

So we know it's got a U = 0 right at the top when it comes to "rest" in mid air i.e. when it stops travelling upwards and is about to turn to come back down. It's going to accelerate freely courtesy of gravity at a = 9.8 and you also know that the vertical height it travels is s = 19.6m. Now you could use s = ut + 0.5at^2 and solve the quadratic for t, but that's complicating things. Instead use v^2 = u^2 + 2as and solve for V the final vertical velocity as it's coming back down and about to hit point Q. If you solve this you end up with V = 19.6. Then you know V, U, a and you need the time t, which ends up nicely as 2 seconds.

So the total time is 2.5 + 2 = 4.5s

For the total distance travelled, we'll look at the horizontal component since the distance is horizontal too!

So we know U = 41.7Cos36

We also know it cannot accelerate horizontally (the reason we consider acceleration vertically is obviously cause g acts the whole time, but horizontally there's no force pushing it or stopping it, so will remain at a constant velocity.... and that's the bottom line cause Newton said so). Therefore, a = 0.

We also know the time it takes to get from P to Q, because we worked that out earlier. This was the confusing bit I was on about, but I hope it makes sense now. If you chuck something half way accross the room (on a parabolic path i.e. launch it at an angle), the time it takes to get to the top of its motion and back down is obviously the total time for the horizontal bit too!

So you know the time, the velocity and since a = 0, you can work out s = vt where v is the horizontal component 41.7Cos36.

Trajectories are abit tricky but once you get it all, it's very intuitive. Just read your stuff and ask any questions you may have.

QUOTE(claribell @ Jan 9 2007, 07:52 PM)

another question is:

a student performs an experiment to study the motion of the school lift as it moves upwards
the student stands on the bathroom scales during the lift's journey upwards
the student records the reading on the scales at different parts of the lift's journey as follows:

part of journey - reading on the scales

At the start(while the lift is accelerating) - 678N
In the middle (while the lift is moving at a steady speed) - 588N
At the end (while the lift is decelerating) - 498N

(a)show that the mass of the student is 60kg
(b)calculate the initial acceleration of the lift
©calculate the deceleration of the lift
(d) during the journey, the lift accelerates for 1.0s, moves at a steady speed for 3.0s and decelerates for a further 1.0s before coming to rest. sketch the acceleration - time graph fot this journey

i think i am able to do part (b)© and (d) by myself

I would be very grateful if anyone could give me any help.

For part (a), look at it this way. What are the forces acting on the bathroom scales? You have the weight of the student, and the pushing/pulling/upthrust force of the lift. So you are given 3 sets of forces, and all you can use is F = ma. But you don't know the accelerations associated with these forces. What you do know, however, is that when it is moving at a steady speed (i.e. constant velocity, no acceleration, so no extra force there to make it accelerate), the only force acting on the scales will be the students weight and the upthrust of the lift, which is equal to the weight cause there is no acceleration either upwards or downwards. So 588 = mg, solve for m to get 60Kg. Hope that helps.

### #6claribell

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Posted 10 January 2007 - 05:38 PM

thank you john

yeah i'm not very good at this V,Vh and Vv i got confused some how. I went to see the teacher about it today and i got even more confused so that didn't help but the way you took it step by step did thank you very much dfx

### #7dfx

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Posted 10 January 2007 - 05:58 PM

QUOTE(claribell @ Jan 10 2007, 05:38 PM)

and thank you dfx i understand it more now but why do you use u=71.7cos36 why wouldn't ouy just say u=41.7 ?

You're welcome. To the question:

71.7Cos36? Perhaps you mean 41.7Cos36, i.e. why did I use 41.7Cos36 instead of just 41.7? Well the reason is because of the way you deal with these questions: you split them into vertical components and horizontal components. Because velocity is a vector, you can resolve it into its vertical and horl components. Like John said, think about it as a right-angled triangle with 41.7 as the hypotenuse and 36 the angle between the hypotenuse and the adjacent sides. Then using ratios you can work out what either the adjacent or the opposite sides are, and that gives you the components. HSN has some really good notes that should help you.

This page goes over exactly what your question covers, and it explains it bit by bit with diagrams really well:
http://id.mind.net/~zona/mstm/physics/mech...alSolution.html

And the HSN notes:
http://www.hsn.uk.net/files/HSN61100.pdf for a basic introduction and particularly see COMPONENTS on page 6.

See also http://www.hsn.uk.net/files/HSN61200.pdf page 6 for an introduction to 2D motion, projectiles and some good problems, rather different from yours but same principles nonetheless, and quite common in the Highers.

Can I ask whether you do maths?

### #8claribell

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Posted 10 January 2007 - 06:24 PM

yes i realised why you had used 41.7 oops i didnt mean 71.7 lol

yes i took maths and im actually good at it lol i also took chemistry, english and modern studies

and suprisingly i got 100% in my NAB for the radiation and matter unit its just this part of the course the teacher didn't explain to well as i got confused and so did my friends.

once again thank you especially for the links

### #9dfx

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Posted 10 January 2007 - 10:37 PM

Ah excellent, nae bother. Yeah I know I didn't really get this myself till around easter time when it clicked during a 2 week revision course if I'm honest! If you can get a 100% on radiation and matter then you'll be fine.

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