Qualification > Sciences
IGCSE PHYSICS DOUBTS HERE !!!!
~ Miss Relina ~:
--- Quote from: Vin on May 06, 2011, 08:42:01 pm ---Can you post a link, please?
--- End quote ---
http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0625%20-%20Physics/0625_w10_qp_32.pdf
here
Vin:
--- Quote from: Relina on May 06, 2011, 09:18:24 pm ---http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0625%20-%20Physics/0625_w10_qp_32.pdf
here
--- End quote ---
Q 2
a) Ep = mgh = 0.15 * 10 * 0.3
= 0.45 J
b i) Always remember, Kinetic energy of any body is maximum and Potential energy is minimum at the lowest point of its motion. (revise the line)
so the lowest point of the bob is 0.1 m (o.3 - 0.2 i.e height of the bob in fig. and length of the string)
here the kinetic energy is max and p e is min.
always Ep lost = Ek gain
so Ep at this point = 0.15 * 10 * 0.1
= 0.15 J
ii) Ek = 0.5 * m * v^2
0.15 = 0.5 * 0.15 * v^2
v = 1.4 ms-1
iii) 0.3, because the bob has only that much of energy to reach to the other side, energy is not lost or gained during the system. (extra info - Then again a question arises, at the left the length of the string is "more", if you see, at the release point the length of the string is small, so the velocity with which the bob will travel to left is greater, then will slow down, so ultimately both the energies during the motion add up to give the resultant energy of 0.45 J or energy at 0.3 to the right. If you know what I mean :P )
iv) drawing is easy. they haven't ask you for a precise scale drawing.
~ Miss Relina ~:
--- Quote from: Vin on May 06, 2011, 10:42:15 pm ---Q 2
a) Ep = mgh = 0.15 * 10 * 0.3
= 0.45 J
b i) Always remember, Kinetic energy of any body is maximum and Potential energy is minimum at the lowest point of its motion. (revise the line)
so the lowest point of the bob is 0.1 m (o.3 - 0.2 i.e height of the bob in fig. and length of the string)
here the kinetic energy is max and p e is min.
always Ep lost = Ek gain
so Ep at this point = 0.15 * 10 * 0.1
= 0.15 J
ii) Ek = 0.5 * m * v^2
0.15 = 0.5 * 0.15 * v^2
v = 1.4 ms-1
iii) 0.3, because the bob has only that much of energy to reach to the other side, energy is not lost or gained during the system. (extra info - Then again a question arises, at the left the length of the string is "more", if you see, at the release point the length of the string is small, so the velocity with which the bob will travel to left is greater, then will slow down, so ultimately both the energies during the motion add up to give the resultant energy of 0.45 J or energy at 0.3 to the right. If you know what I mean :P )
iv) drawing is easy. they haven't ask you for a precise scale drawing.
--- End quote ---
Thanks but in b revised the graph and i think it should be 0.2
Vin:
--- Quote from: Relina on May 07, 2011, 07:55:59 am ---Thanks but in b revised the graph and i think it should be 0.2
--- End quote ---
You're right :/
sorry, my bad.
Okay, so 0.1 is the CHANGE in height. It was 0.1 from the peg, now at perpendicular, it is 02. So delta h is 0.1
reason because when it is at the lowest position, there is still some potential energy, so we cannot take that as 0. so we have to consider the change in height, or change in potential energy.
~ Miss Relina ~:
--- Quote from: Vin on May 07, 2011, 11:53:56 am ---You're right :/
sorry, my bad.
Okay, so 0.1 is the CHANGE in height. It was 0.1 from the peg, now at perpendicular, it is 02. So delta h is 0.1
reason because when it is at the lowest position, there is still some potential energy, so we cannot take that as 0. so we have to consider the change in height, or change in potential energy.
--- End quote ---
plz we dont take it as 0.??? what
and i know that ur convined but my bad the march schme said ur previous answer is right???
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