OCR A2 Physics 2012 June Exam question 2 help please - Projectile Motion

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Reda2
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I'm doing this past paper, and I don't fully understand question 2 part (c).

Why is the GPE not equal to the KE? I know it says because it is still moving in the mark scheme, but why does that mean that the GPE is not the same?
Also if you look at the next part of the question (d) in the mark scheme it says accept mgh = 1/2mv^2 which does not make sense as they just said GPE is not equal to KE at the top.

Past Paper: http://www.ocr.org.uk/Images/131307-...-mechanics.pdf
Mark Scheme: http://www.ocr.org.uk/Images/135330-...anics-june.pdf

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Reda
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uberteknik
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(Original post by Reda2)
I'm doing this past paper, and I don't fully understand question 2 part (c).

Why is the GPE not equal to the KE? I know it says because it is still moving in the mark scheme, but why does that mean that the GPE is not the same?
Also if you look at the next part of the question (d) in the mark scheme it says accept mgh = 1/2mv^2 which does not make sense as they just said GPE is not equal to KE at the top.

Past Paper: http://www.ocr.org.uk/Images/131307-...-mechanics.pdf
Mark Scheme: http://www.ocr.org.uk/Images/135330-...anics-june.pdf

Thanks
Reda
At point A (the point of throwing the ball), the instantaneous velocity (and hence the unresolved KE) is 30 degrees to the horizontal. This is the initial total KE of the ball imparted to it by the thrower. Hence the initial \rm KE = \frac{1}{2}mv^2 uses v = 24ms-1.

From that point onwards, the ball experiences the gravitational force acting in a vertical direction only. In which case, the initial velocity at point A, must be resolved into individual horizontal and vertical velocity components. Use trigonometry to do this. i.e. Total KE is conserved and resolved into two initial velocity components - horizontal and vertical.

When the ball reaches point B, the vertical KE component has reached zero and converted to PE. However there is still the horizontal component remaining intact because we are told to ignore air resistance. Again energy conservation means that the vertical PE and horizontal KE must equal the initial KE.

Hence only some of the initial KE has converted to PE (the vertical component only) with the horizontal component of KE unaltered.

mgh = \frac{1}{2}mv^2 of part 2 d). refers to the vertical velocity component only and NOT the initial velocity of 24 ms-1.
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Reda2
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(Original post by uberteknik)
At point A (the point of throwing the ball), the instantaneous velocity (and hence the unresolved KE) is 30 degrees to the horizontal. This is the initial total KE of the ball imparted to it by the thrower. Hence the initial \rm KE = \frac{1}{2}mv^2 uses v = 24ms-1.

From that point onwards, the ball experiences the gravitational force acting in a vertical direction only. In which case, the initial velocity at point A, must be resolved into individual horizontal and vertical velocity components. Use trigonometry to do this. i.e. Total KE is conserved and resolved into two initial velocity components - horizontal and vertical.

When the ball reaches point B, the vertical KE component has reached zero and converted to PE. However there is still the horizontal component remaining intact because we are told to ignore air resistance. Again energy conservation means that the vertical PE and horizontal KE must equal the initial KE.

Hence only some of the initial KE has converted to PE (the vertical component only) with the horizontal component of KE unaltered.

mgh = \frac{1}{2}mv^2 of part 2 d). refers to the vertical velocity component only and NOT the initial velocity of 24 ms-1.
Sorry for the late reply. Thank you so much for your answer, it helped me understand it. Thank you.
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