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AQA AS Physics - June 2022 - Paper 2 Question 4.4 - struggling

Hi all, struggling with question 4.4 from AQA AS Physics June 2022 Paper 2. Not sure how to do it and don't understand the marks scheme.

Link to exam paper: https://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/AQA/AS-Paper-2/QP/June%202022%20QP.PDF

Any assistance would be great.
ms: https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2022/june/AQA-74072-MS-JUN22.PDF
Work out time of CD.
velocity upwards is unknown, but you know after max distance is travelled, the final v is 0.
Use s=vt-1/2at^2
This gets the displacement travelled from the platform to the max distance from the jump. To get distance just take absolute value.
The total distance is double this distance as it is symmetrical for a speed/time graph (distance is area of the graph)
(edited 1 month ago)
Actually it should be a straight line because constant acceleration but the main idea is still there
Original post by Thomas_02
Hi all, struggling with question 4.4 from AQA AS Physics June 2022 Paper 2. Not sure how to do it and don't understand the marks scheme.

Link to exam paper: https://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/AQA/AS-Paper-2/QP/June%202022%20QP.PDF

Any assistance would be great.

I can understand your confusion.

After seeing the MS, I realised that Q4.4 seems to ask for the vertical height travelled by the athlete rather than the distance travelled by the athlete. :smile:

However, after going through the calculation in detail, Q4.4 is asking the distance travelled by the athlete along the rail indeed.

The actual calculation is
12(9.81sin(30))(0.5×0.62)2×2 \frac{1}{2}(9.81 \sin(30^{\circ}))(0.5\times 0.62)^2 \times 2

which can be simplified to

12(9.81)(0.5×0.62)2 \frac{1}{2}(9.81)(0.5\times 0.62)^2
can be easily overlooked to say that we are calculating the vertical height.
(edited 1 month ago)
Reply 4
Her acceleration was 5 ms-2 when between A and B when her chair was accelerating towards the platform. But between C and D she pushes herself back on the rail so the acceleration is 5 ms-2?
Many thanks for your reply.
Original post by Thomas_02
Her acceleration was 5 ms-2 when between A and B when her chair was accelerating towards the platform. But between C and D she pushes herself back on the rail so the acceleration is 5 ms-2?
Many thanks for your reply.


Yes, along the rail, the acceleration is approximately 5 m/s2.
Reply 6
Original post by Eimmanuel
Yes, along the rail, the acceleration is approximately 5 m/s2.
Thanks for your prompt reply but I am confused why acceleration between C and D is equal to acceleration between A and B , can you elaborate on the reason please.
Original post by Thomas_02
Thanks for your prompt reply but I am confused why acceleration between C and D is equal to acceleration between A and B , can you elaborate on the reason please.

Maybe you want to share your thoughts about what you are confused with.
OR Are you thinking that the acceleration between C and D is NOT equal to the acceleration between A and B? If yes, why?
Reply 8
Original post by Eimmanuel
Maybe you want to share your thoughts about what you are confused with.
OR Are you thinking that the acceleration between C and D is NOT equal to the acceleration between A and B? If yes, why?
First of all thanks a million for being patience with my confusion.
My understanding is- between A and B athlete is coming down on rails from rest position and her right foot has touched the platform and after that she has pushed herself up the rail between C and D. So how the acceleration of the downwards motion of the athelete between A and B is equal to the upwards motion on the rail which she achieved by pushing her self upward on the rail by her right leg on the platform between C and D.

Once again, can’t thank you enough for your replies. Much appreciated.
Original post by Thomas_02
First of all thanks a million for being patience with my confusion.
My understanding is- between A and B athlete is coming down on rails from rest position and her right foot has touched the platform and after that she has pushed herself up the rail between C and D. So how the acceleration of the downwards motion of the athelete between A and B is equal to the upwards motion on the rail which she achieved by pushing her self upward on the rail by her right leg on the platform between C and D.

Once again, can’t thank you enough for your replies. Much appreciated.


It is always good that you share your thinking so that we know what to explain and leave out the unnecessary to focus on what benefits you the most.

Imagine you throw a ball vertically upward with an initial velocity of 20 m/s.
On the way to the maximum height, what is the acceleration of the ball?
On the way down from the maximum height to your hand, what is the acceleration of the ball?

Now, you throw the same ball vertically upward with an initial velocity of 40 m/s. Again, what is the acceleration the ball on the way up to the maximum height and down from the maximum height to back to your hand?

Note that we are ignoring air resistance or friction.
Reply 10
Original post by Eimmanuel
It is always good that you share your thinking so that we know what to explain and leave out the unnecessary to focus on what benefits you the most.

Imagine you throw a ball vertically upward with an initial velocity of 20 m/s.
On the way to the maximum height, what is the acceleration of the ball?
On the way down from the maximum height to your hand, what is the acceleration of the ball?

Now, you throw the same ball vertically upward with an initial velocity of 40 m/s. Again, what is the acceleration the ball on the way up to the maximum height and down from the maximum height to back to your hand?

Note that we are ignoring air resistance or friction.
In all the cases acceleration will be 9.81 ms^-2 while coming down and deceleration while going up.
Original post by Thomas_02
In all the cases acceleration will be 9.81 ms^-2 while coming down and deceleration while going up.


Indeed, you are right.
To be more precise, I would say, the ball will have the same acceleration (−9.81 m s−2) regardless of whether the ball is moving up or coming down or has a different initial speed.
Reply 12
Original post by Eimmanuel
Indeed, you are right.
To be more precise, I would say, the ball will have the same acceleration (−9.81 m s−2) regardless of whether the ball is moving up or coming down or has a different initial speed.
Got it, many thanks
Reply 13
Original post by Thomas_02
Got it, many thanks
I am really struggling to understand the concept of physics and feel that in A level physics I don’t have a proper book which can built up the physics so that I can do the challenging calculations questions. I have some books given to us by school but they don’t give enough information to make me prepare for difficult questions that come in question paper. I am really worried for my exams. Can you please suggest a good physics book that can gradually build up A level physics concepts. Many thanks in advance.

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