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Quatum Mechanics Primer Isaac Physics

I have a slight doubt with this question: https://isaacphysics.org/questions/qmp_ch1_q3?board=qmp_ch1_1_6&stage=all

So to find the period I equated the EPE to KE? I am not sure how to tackle it,help would be appreciated. This is my working
Original post by GrizzlyBear24
I have a slight doubt with this question: https://isaacphysics.org/questions/qmp_ch1_q3?board=qmp_ch1_1_6&stage=all
So to find the period I equated the EPE to KE? I am not sure how to tackle it,help would be appreciated. This is my working

@mqb2766 :smile:
Reply 2
Original post by grizzlybear24
I have a slight doubt with this question: https://isaacphysics.org/questions/qmp_ch1_q3?board=qmp_ch1_1_6&stage=all
So to find the period I equated the EPE to KE? I am not sure how to tackle it,help would be appreciated. This is my working

Id have thought a sketch would be a v shape and youd analyse motion along the plane(s) so get the distance and resolved acceleration (force) and hence the time? However, isaac isnt giving, so there may be something Im misunderstanding.
(edited 4 weeks ago)
Original post by mqb2766
Id have thought a sketch would be a v shape and youd analyse motion along the plane(s) so get the distance and resolved acceleration (force) and hence the time? However, isaac isnt giving, so there may be something Im misunderstanding.

I may be completely out of order here, but should you integrate something, it is because in the QMP book it does resolve time in another exercise by integrating. However, I am not entirely sure why they do that. ( infact it is the previous exercise that they integrate in). I can't show the exercise where the integrate nor can I show my working as it does not allow me to upload an image but they essentially use energies (kE and potential )
(edited 4 weeks ago)
Original post by GrizzlyBear24
I may be completely out of order here, but should you integrate something, it is because in the QMP book it does resolve time in another exercise by integrating. However, I am not entirely sure why they do that. ( infact it is the previous exercise that they integrate in). I can't show the exercise where the integrate nor can I show my working as it does not allow me to upload an image but they essentially use energies (kE and potential )


This is purely kinematics using only suvat NO integration is required.
Find the time taken for the mass to slide down along the slope where you can express the slope distance with the horizontal displacement using trigo ratio.
Multiply the time taken by 4 to get the period.
Original post by Eimmanuel
This is purely kinematics using only suvat NO integration is required.
Find the time taken for the mass to slide down along the slope where you can express the slope distance with the horizontal displacement using trigo ratio.
Multiply the time taken by 4 to get the period.

I got the answer now, but would equating the potential energy to the kinetic energy when the ball is in the juntion, so you get that v=sqrt(2*g*x_0*tan(theta)) and then the time t to go down one slope is t=v/(g*sin(theta)), where v is the velocity calculated using energy and gsing(theta) is the acceleration of the particle? Then multiply by four. Would that also be a good approach or is it a bit nonsense? By the way, thanks for the help I had been with that problem for two days now.
Reply 6
Original post by grizzlybear24
I got the answer now, but would equating the potential energy to the kinetic energy when the ball is in the juntion, so you get that v=sqrt(2*g*x_0*tan(theta)) and then the time t to go down one slope is t=v/(g*sin(theta)), where v is the velocity calculated using energy and gsing(theta) is the acceleration of the particle? Then multiply by four. Would that also be a good approach or is it a bit nonsense? By the way, thanks for the help I had been with that problem for two days now.

Thats ok as well, though the suvat way (resolve along the plane) seems more elementary as its constant acceleration and you can pick the relevvant suvat equation you need, without having to combine two (admittedly simple). Youd have effectively used
v^2 - u^2 = 2as
and
v - u = at
rather than using
s = ut + 1/2 at^2
directly.
(edited 4 weeks ago)
Original post by mqb2766
Thats ok as well, though the suvat way (resolve along the plane) seems more elementary as its constant acceleration and you can pick the relevvant suvat equation you need, without having to combine two (admittedly simple). Youd have effectively used
v^2 - u^2 = 2as
and
v - u = at
rather than using
s = ut + 1/2 at^2
directly.

Yes, I have tried using energy and I do get the same answer, it is just that I did not simplify.

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