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OCR Physics A G481 2015 Unofficial Markscheme
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Graphs
Velocity-time
Force-length (not extension (doesn't start through the origin), not original length (length changes))
Similar scalar quantities
Stress and pressure, Pascal or N/m^2. Alternatively stress and Young's modulus, same units.
Material properties
Polymer, doesn't obey Hooke's law, large elastic region, no plastic deformation, returns to original length when tensile/compressive forces are removed...
Input power of a motor
Attach a mass to the hanger of the motor (after measuring it). Measure the initial height of the mass using a meter ruler. Turn the motor on full power and simultaneously start a timer. After some time has passed stop the timer and the motor. Measure the new height of the mass. Input power = (mass * g * (new height - initial height) / time) / 0.15
Prove units of k
k = drag/(area*v^2)
drag = force [N = kgms^-2]
k = kgm * s^-2 * m^-2 * s^2 * m^-2
k = kgm^-3
Resultant of A and B
Sum their vertical and horizontal components separately. Construct a triangle from the result of the new components and use Pythagoras to find the resultant force.
Is B experiencing terminal velocity
No. Drag must but equal and opposite to weight. In this instance it's perpendicular to weight.
Assumption when calculating extension
The object obeys Hooke's law and only undergoes elastic deformation during the experiment (saying elastic deformation on it's own isn't enough, look at polymer materials). Alternatively the cross-sectional area is constant.
Velocity-time
Force-length (not extension (doesn't start through the origin), not original length (length changes))
Similar scalar quantities
Stress and pressure, Pascal or N/m^2. Alternatively stress and Young's modulus, same units.
Material properties
Polymer, doesn't obey Hooke's law, large elastic region, no plastic deformation, returns to original length when tensile/compressive forces are removed...
Input power of a motor
Attach a mass to the hanger of the motor (after measuring it). Measure the initial height of the mass using a meter ruler. Turn the motor on full power and simultaneously start a timer. After some time has passed stop the timer and the motor. Measure the new height of the mass. Input power = (mass * g * (new height - initial height) / time) / 0.15
Prove units of k
k = drag/(area*v^2)
drag = force [N = kgms^-2]
k = kgm * s^-2 * m^-2 * s^2 * m^-2
k = kgm^-3
Resultant of A and B
Sum their vertical and horizontal components separately. Construct a triangle from the result of the new components and use Pythagoras to find the resultant force.
Is B experiencing terminal velocity
No. Drag must but equal and opposite to weight. In this instance it's perpendicular to weight.
Assumption when calculating extension
The object obeys Hooke's law and only undergoes elastic deformation during the experiment (saying elastic deformation on it's own isn't enough, look at polymer materials). Alternatively the cross-sectional area is constant.
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(Original post by Ngardnerpv)
sorry dude, honestly cant remember any of my numerical answers
sorry dude, honestly cant remember any of my numerical answers
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#4
Usually a teacher who has access to the paper after the exam will post an unofficial mark scheme on here.
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#5
(Original post by Throwaway6467)
Usually a teacher who has access to the paper after the exam will post an unofficial mark scheme on here.
Usually a teacher who has access to the paper after the exam will post an unofficial mark scheme on here.
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(Original post by j.greenwood98)
For scalar quantities for the same unit cant it be moment and torque?
For scalar quantities for the same unit cant it be moment and torque?
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#9
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#10
(Original post by Nuvertion)
Graphs
Velocity-time
Force-length (not extension (doesn't start through the origin), not original length (length changes))
Similar scalar quantities
Stress and pressure, Pascal or N/m^2. Alternatively stress and Young's modulus, same units.
Material properties
Polymer, doesn't obey Hooke's law, large elastic region, no plastic deformation, returns to original length when tensile/compressive forces are removed...
Input power of a motor
Attach a mass to the hanger of the motor (after measuring it). Measure the initial height of the mass using a meter ruler. Turn the motor on full power and simultaneously start a timer. After some time has passed stop the timer and the motor. Measure the new height of the mass. Input power = (mass * g * (new height - initial height) / time) / 0.15
Prove units of k
k = drag/(area*v^2)
drag = force [N = kgms^-2]
k = kgm * s^-2 * m^-2 * s^2 * m^-2
k = kgm^-3
Resultant of A and B
Sum their vertical and horizontal components separately. Construct a triangle from the result of the new components and use Pythagoras to find the resultant force.
Is B experiencing terminal velocity
No. Drag must but equal and opposite to weight. In this instance it's perpendicular to weight.
Assumption when calculating extension
The object obeys Hooke's law and only undergoes elastic deformation during the experiment (saying elastic deformation on it's own isn't enough, look at polymer materials).
Graphs
Velocity-time
Force-length (not extension (doesn't start through the origin), not original length (length changes))
Similar scalar quantities
Stress and pressure, Pascal or N/m^2. Alternatively stress and Young's modulus, same units.
Material properties
Polymer, doesn't obey Hooke's law, large elastic region, no plastic deformation, returns to original length when tensile/compressive forces are removed...
Input power of a motor
Attach a mass to the hanger of the motor (after measuring it). Measure the initial height of the mass using a meter ruler. Turn the motor on full power and simultaneously start a timer. After some time has passed stop the timer and the motor. Measure the new height of the mass. Input power = (mass * g * (new height - initial height) / time) / 0.15
Prove units of k
k = drag/(area*v^2)
drag = force [N = kgms^-2]
k = kgm * s^-2 * m^-2 * s^2 * m^-2
k = kgm^-3
Resultant of A and B
Sum their vertical and horizontal components separately. Construct a triangle from the result of the new components and use Pythagoras to find the resultant force.
Is B experiencing terminal velocity
No. Drag must but equal and opposite to weight. In this instance it's perpendicular to weight.
Assumption when calculating extension
The object obeys Hooke's law and only undergoes elastic deformation during the experiment (saying elastic deformation on it's own isn't enough, look at polymer materials).
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(Original post by tomzorz)
I assumed it was that is undergoes only elastic deformation/no plastic deformation/obeys Hooke's Law
I assumed it was that is undergoes only elastic deformation/no plastic deformation/obeys Hooke's Law
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(Original post by John Fluffy Bunny)
Don't you just draw a line from A to B to construct a vector triangle and then yay Pythagoras to work out its magnitude? Also I messed up the graph I've labelled it F against x overall I don't know if I'm even gonna get an a which is sad because when doing past papers I always used to get at least 90 percent
Don't you just draw a line from A to B to construct a vector triangle and then yay Pythagoras to work out its magnitude? Also I messed up the graph I've labelled it F against x overall I don't know if I'm even gonna get an a which is sad because when doing past papers I always used to get at least 90 percent
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#13
I stated that the assumption is that the cross-section area is constant throughout the metal strip.
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(Original post by 1z3)
I stated that the assumption is that the cross-section area is constant throughout the metal strip.
I stated that the assumption is that the cross-section area is constant throughout the metal strip.
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#15
(Original post by Nuvertion)
Pythagoras can only be used on right-angled triangles.
Pythagoras can only be used on right-angled triangles.
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(Original post by John Fluffy Bunny)
Yeah i did that I believe I got root 5 by using the grid
Yeah i did that I believe I got root 5 by using the grid
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(Original post by tman96)
For the scalar quantities is torque and moments an answer
For the scalar quantities is torque and moments an answer
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#19
(Original post by Nuvertion)
Graphs
Velocity-time
Force-length (not extension (doesn't start through the origin), not original length (length changes))
Similar scalar quantities
Stress and pressure, Pascal or N/m^2. Alternatively stress and Young's modulus, same units.
Material properties
Polymer, doesn't obey Hooke's law, large elastic region, no plastic deformation, returns to original length when tensile/compressive forces are removed...
Input power of a motor
Attach a mass to the hanger of the motor (after measuring it). Measure the initial height of the mass using a meter ruler. Turn the motor on full power and simultaneously start a timer. After some time has passed stop the timer and the motor. Measure the new height of the mass. Input power = (mass * g * (new height - initial height) / time) / 0.15
Prove units of k
k = drag/(area*v^2)
drag = force [N = kgms^-2]
k = kgm * s^-2 * m^-2 * s^2 * m^-2
k = kgm^-3
Resultant of A and B
Sum their vertical and horizontal components separately. Construct a triangle from the result of the new components and use Pythagoras to find the resultant force.
Is B experiencing terminal velocity
No. Drag must but equal and opposite to weight. In this instance it's perpendicular to weight.
Assumption when calculating extension
The object obeys Hooke's law and only undergoes elastic deformation during the experiment (saying elastic deformation on it's own isn't enough, look at polymer materials). Alternatively the cross-sectional area is constant.
Graphs
Velocity-time
Force-length (not extension (doesn't start through the origin), not original length (length changes))
Similar scalar quantities
Stress and pressure, Pascal or N/m^2. Alternatively stress and Young's modulus, same units.
Material properties
Polymer, doesn't obey Hooke's law, large elastic region, no plastic deformation, returns to original length when tensile/compressive forces are removed...
Input power of a motor
Attach a mass to the hanger of the motor (after measuring it). Measure the initial height of the mass using a meter ruler. Turn the motor on full power and simultaneously start a timer. After some time has passed stop the timer and the motor. Measure the new height of the mass. Input power = (mass * g * (new height - initial height) / time) / 0.15
Prove units of k
k = drag/(area*v^2)
drag = force [N = kgms^-2]
k = kgm * s^-2 * m^-2 * s^2 * m^-2
k = kgm^-3
Resultant of A and B
Sum their vertical and horizontal components separately. Construct a triangle from the result of the new components and use Pythagoras to find the resultant force.
Is B experiencing terminal velocity
No. Drag must but equal and opposite to weight. In this instance it's perpendicular to weight.
Assumption when calculating extension
The object obeys Hooke's law and only undergoes elastic deformation during the experiment (saying elastic deformation on it's own isn't enough, look at polymer materials). Alternatively the cross-sectional area is constant.
also whats this question about the assumption made in the experiement. i think i said something about constant area throughout the material somewhere in the paper, can someone remind me what that question was about again?
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#20
(Original post by nmjasdk)
I never found any angles to calculate compenents of A and B to find AB resultant. i just drew a line between the 2 points and measures its length, and used the scale 1cmm = 1km.
also whats this question about the assumption made in the experiement. i think i said something about constant area throughout the material somewhere in the paper, can someone remind me what that question was about again?
I never found any angles to calculate compenents of A and B to find AB resultant. i just drew a line between the 2 points and measures its length, and used the scale 1cmm = 1km.
also whats this question about the assumption made in the experiement. i think i said something about constant area throughout the material somewhere in the paper, can someone remind me what that question was about again?
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