IAL Edexcel physics Jan 2014 Unit 1 Question 17

Hello everyone, I would really appreciate if i could get help on this question.

https://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/WPH01_01_que_20140115.pdf

This is my first time making a post, I am not sure how I can share the question apart from providing a link to the question. If there are better ways to share questions, let me know.

The first issue is that i am unable to understand the graph which makes answering the questions impossible.

I have tried using the mark scheme and examiner's report to help me understand but they've just left me with more questions than answers.

Why is the graph shaped like that?

Why does the graph have a semi horizontal section if this is a graph on vertical displacement only ?

If the whole graph is for vertical displacement, doesn't that mean i can get the velocity by working out the gradient at any section of the graph? If yes, why does the examiner's report say that we had to calculate the gradient of the semi horizontal section to find the velocity?

Reply 1

mqb2766

Original post by Daada

The marble is falling vertically under the action of gravity. So the vertical displacement will be a quadratic as a function of time. This is most easily seen in the first vertical drop phase as it starts off with zero displacement and velocity (gradient), then the displacement downwards increases quadratically. In the middle phase when it hits the channel base, the vertical velocity is initially reduced to near zero (small gradient) and the vertical acceleration is also amost zero as the channel's floor reaction is similar to the gravitational force so there is ~zero net force on the mass. So the displacment again increases quadratically, but the curvature is very much smaller and almost negligible. Then the final vertical drop phase, its similar to the first phase.

The instantaneous velocity is the gradient of the tangent to the graph at a point. In the middle phase, it says the speed (vertical velocity) remains constant (its pretty much a straight line) so the tangent corresponds to that line so the vertical velocity (speed) corresponds to the gradient of that line