# alevel physicsWatch

Announcements
#1
I don't understand why this graph would decelerate after constant acceleration?
its question 3b ii

http://filestore.aqa.org.uk/subjects...2-QP-JUN13.PDF
0
2 years ago
#2
Steel ball B -

Constant acceleration when falling in air.
Then hits oil - suddenly there is a lot of drag because the ball is moving quite quickly and the oil is thick.
This means there is a resultant force upwards - so the ball will decelerate.
As it slows down the drag force also decreases until it balances the weight, there is then no resultant force so no acceleration.
When it hits the bottom there will be a sudden deceleration.
0
#3
(Original post by Fiziks)
Steel ball B -

Constant acceleration when falling in air.
Then hits oil - suddenly there is a lot of drag because the ball is moving quite quickly and the oil is thick.
This means there is a resultant force upwards - so the ball will decelerate.
As it slows down the drag force also decreases until it balances the weight, there is then no resultant force so no acceleration.
When it hits the bottom there will be a sudden deceleration.
I understand that but the problem is with the curve part of the graph i don't understand why it's that shape, thought it would've been like an e^-x graph.
0
2 years ago
#4
I'm not sure I understand, an e^-x curve would look like the one given in the mark scheme, wouldn't it? Perhaps you could add a sketch of what you were expecting so I can see what you mean
0
2 years ago
#5
(Original post by ace-master)
I understand that but the problem is with the curve part of the graph i don't understand why it's that shape, thought it would've been like an e^-x graph.
It doesn't matter what type of curved graph it is, as long as it curves and then becomes horizontal. I believe resistive forces are proportional to v^2; I could be wrong, but I don't think it matters anyway.
0
#6
(Original post by Fiziks)
I'm not sure I understand, an e^-x curve would look like the one given in the mark scheme, wouldn't it? Perhaps you could add a sketch of what you were expecting so I can see what you mean
sorry i had to do it this way, would've took a picture but my phone camera isnt good.
0
2 years ago
#7
(Original post by ace-master)
sorry i had to do it this way, would've took a picture but my phone camera isnt good.
I would give a brief answer to why graph B is incorrect.

When the ball is falling in the air, the ball does not reach terminal velocity.
0
2 years ago
#8
(Original post by ace-master)
I understand that but the problem is with the curve part of the graph i don't understand why it's that shape, thought it would've been like an e^-x graph.

I am not sure if I understand your question. The graph for 3b(ii) is in fact of the form e-x when the ball is moving in the oil. It has to be if the drag force inside the oil is proportional to velocity. If the drag force is proportional to the square of the velocity, the exponential function is “more complicated”.

You can set up the differential equation from Newton’s second law and solve for the velocity.
0
#9
Finally figured it out!!!
The ball is not decelerating, its acceleration is decreasing to 0 that's when it reaches terminal velocity.
that's why it curves like that.
Thanks for trying to help guys appreciate it
0
2 years ago
#10
(Original post by ace-master)
Finally figured it out!!!
The ball is not decelerating, its acceleration is decreasing to 0 that's when it reaches terminal velocity.
that's why it curves like that.
Thanks for trying to help guys appreciate it
The ball does decelerate when it hits the oil - we can see this as the velocity decreases over time (the line goes down). The difference between your graph and the correct one is how the gradient changes.

First, remember that the gradient of a velocity time graph is the acceleration (deceleration if it's sloping down). On your graph after it hits the oil it starts flat then goes steeply down. This means that there is a small deceleration at first then it starts to decelerate more and more.

The correct graph shows the opposite, there is a big deceleration to start with then it levels out. To understand this you need to think of the forces. The resistive drag force depends on the speed. When the ball hits the oil it is moving very quickly, it has to push lots of oil out of the way so there is a big drag. This gives us the big (steep) deceleration. As it slows down the drag force is less and so it decelerates more gently, the line levels off.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cardiff University
Sat, 26 Oct '19
• Brunel University London
Sat, 26 Oct '19
• University of Lincoln
Sat, 26 Oct '19

### Poll

Join the discussion

Yes (102)
24%
No (323)
76%