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# The Physics PHYA2 thread! 5th June 2013 Watch

• View Poll Results: What mark do you think you got out of 70?
0-20
6
3.00%
21-40
12
6.00%
41-50
29
14.50%
51-60
79
39.50%
61-70
74
37.00%

1. (Original post by NedStark)
Possibly but I think that's specifically for that mystery liquid. Ball A didn't have an deceleration with oil, so I don't think B should either.
That's because A started in oil b started in air
2. (Original post by ClarkyC...)
Was it not decelerating?
Isn't decelerating when its velocity decreases?
3. (Original post by Bookaky)
For the graph about the ball and the oil I think it is debatable. Surely the ball would continue to accelerate in the oil for only a short period of time and then level off for terminal velocity?
This is correct if we're talking about the motion of ball B.

- Straight (linear) (+ve) acceleration up to hitting oil
- Vertical drop in velocity as ball hits oil
- Slight curve upwards as ball accelerates in the oil
- Levels out as new terminal velocity reached
- Vertical drop in velocity as ball hits bottom
4. (Original post by Davelittle)
Isn't decelerating when its velocity decreases?
It's velocity is decreasing
5. (Original post by ClarkyC...)
It's velocity is decreasing
Are you talking about the 6 marker or the graph? :P

On the 6 marker it's velocity didn't decrease at all
6. (Original post by Davelittle)
Isn't decelerating when its velocity decreases?
yep, how many marks you think I'll get if I basically did the other way around and said the velocity is largest at start and it decelerates? (Yeah I completely messed up)
7. (Original post by NabRoh)
This is correct if we're talking about the motion of ball B.

- Straight (linear) (+ve) acceleration up to hitting oil
- Vertical drop in velocity as ball hits oil
- Slight curve upwards as ball accelerates in the oil
- Levels out as new terminal velocity reached
- Vertical drop in velocity as ball hits bottom
Wait..... How can it be a straight line? If the ball is dropped from the rest, it should accelerate until it hits the oil surface. So, the graph is curve showing that velocity is increasing more with less time.
8. (Original post by Davelittle)
Are you talking about the 6 marker or the graph? :P

On the 6 marker it's velocity didn't decrease at all
Sorry no the velocity didnt decrease on the 6 marker but I'm not sure whether its decelerating or not. I'm going off another past paper as to why I think it's deceleration
9. (Original post by StalkeR47)
Wait..... How can it be a straight line? If the ball is dropped from the rest, it should accelerate until it hits the oil surface. So, the graph is curve showing that velocity is increasing more with less time.
that is what i thought too, but maybe they are considering acceleration to be due to gravity hence you get a straight line?
10. For the very last question, I put that:

There would be a spectrum of colours rather than just red
The fringe spacing will be smaller because the maxima are wider and the minima are smaller (I think this is wrong)
The central fringe would be white rather than red

How many marks would this be?
11. Right the diameter was 2.2x10-3 of the sphere and the density was 8100kgm-3. I can't get to 0.44N what am I doing wrong
12. (Original post by Boop.)
Thats what I put I think I remember it from an older past paper, I said that the ball didnt decelerate, but that it still accelerated in the fluid, the acceleration was decreasing with time, not sure if im right though..
Small balls of this type (hehe).. reach terminal velocity very fast, so when B was released it would have accelerated with constant acceleration and almost reached terminal velocity.. when it hit the oil the drag would increase by ALOT... so it's speed decreases as it moves through the oil, we know this acceleration is not constant/is non-uniform as the curve for A shows this.

A accelerated with non-uniform acceleration, so when B hits the oil it will decelerate with non-uniform deceleration. So the graph for B, I think, should be a straight line up to the point where it hits the oil then a curve slopping downwards and leveling off at terminal velocity.
13. (Original post by StalkeR47)
Wait..... How can it be a straight line? If the ball is dropped from the rest, it should accelerate until it hits the oil surface. So, the graph is curve showing that velocity is increasing more with less time.
a is constant so it would be a straight line (think about it, if gradient=a and a=g=9.81m/s^2 then the gradient has to be constant, what else is accelerating it?)
14. (Original post by ClarkyC...)
Right the diameter was 2.2x10-3 of the sphere and the density was 8100kgm-3. I can't get to 0.44N what am I doing wrong
Find r-> find V-> find m-> W=mg
15. (Original post by ClarkyC...)
Right the diameter was 2.2x10-3 of the sphere and the density was 8100kgm-3. I can't get to 0.44N what am I doing wrong
The diameter was 2.2x10-2

So W = (4/3) * pi * (2.2x10-2/2)^3 * 8100 * 9.81
16. how many squares did people get under the graph?
17. (Original post by ClarkyC...)
Sorry no the velocity didnt decrease on the 6 marker but I'm not sure whether its decelerating or not. I'm going off another past paper as to why I think it's deceleration
Its acceleration decreases, I don't think that is the same as deceleration?

If someone decelerates from 10m/s to 0m/s its velocity goes down, not up.
18. (Original post by MeerkatM)
how many squares did people get under the graph?
42 i think?
19. (Original post by FLLF)
For the very last question, I put that:

There would be a spectrum of colours rather than just red
The fringe spacing will be smaller because the maxima are wider and the minima are smaller (I think this is wrong)
The central fringe would be white rather than red

How many marks would this be?
Full marks the reasons not right for the second one but nothing is "inncorect" the reasons becuase the average wavelength of white light is smaller than red light
20. (Original post by kingmango)
I counted 43 squares for the area under the graph, would this be wrong?
Anybody know? for the energy stored in the spring, i counted 43 squares, so i did 43 x 0.025 = 1.075J. Would this be right???

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