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It ain't from CIE questions. It's from the Coursebook CD-ROM worksheet.
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Physics: Post your doubts here!
- Thread starter XPFMember
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death to the west xx
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I'll post the answers. You'll be surprised just as I was.
a. horizontal velocity= s/t = 2.5/0.6= 4.17m/s
b. Vertically final velocity is zero as ball passes over cross-bar.
v = u + at
0 = u - (9.81*0.6)
u = 5.89m/s (ans)
c. u = root over(4.17^2 + 5.89^2)
= 7.21m/s
d. angle = tan^-1 (5.89/4.17)
= 55degrees
The mark scheme totally trolled me!!! I don't get it. How can horizontal velocity be 2.5/0.6 ??
a. horizontal velocity= s/t = 2.5/0.6= 4.17m/s
b. Vertically final velocity is zero as ball passes over cross-bar.
v = u + at
0 = u - (9.81*0.6)
u = 5.89m/s (ans)
c. u = root over(4.17^2 + 5.89^2)
= 7.21m/s
d. angle = tan^-1 (5.89/4.17)
= 55degrees
The mark scheme totally trolled me!!! I don't get it. How can horizontal velocity be 2.5/0.6 ??
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An athlete bends down and then jumps straight up into the air. At the instant he starts his jump
three forces act on him – his weight, an upthrust from the air and the contact force with the
ground. What is the correct order in increasing magnitude of these three forces
A. contact force → weight → upthrust
B. upthrust → contact force → weight
C. upthrust → weight → contact force
D. weight → upthrust → contact force
Provide an explanation if you can answer. It's not from past papers.
three forces act on him – his weight, an upthrust from the air and the contact force with the
ground. What is the correct order in increasing magnitude of these three forces
A. contact force → weight → upthrust
B. upthrust → contact force → weight
C. upthrust → weight → contact force
D. weight → upthrust → contact force
Provide an explanation if you can answer. It's not from past papers.
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2.5m was the vertical distance. There's no way it can be used to calculate the horizontal distance. the two components are perpendicular.
I don't get how this is possible.
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okay i think this is simple but am kinda confused:
3bii.) - i don't get why to calculate the time only the vertical distance is considered
3biii.) - why do we use the horizontal velocity to calculate the speed.....and the time is the same as the journey of the ball before colliding with the wall??
Question paper: http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
Marking scheme:http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_21.pdf
please help urgently thanx so much.....May God bless you
3bii.) - i don't get why to calculate the time only the vertical distance is considered
3biii.) - why do we use the horizontal velocity to calculate the speed.....and the time is the same as the journey of the ball before colliding with the wall??
Question paper: http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
Marking scheme:http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_21.pdf
please help urgently
thanx so much.....May God bless you
- Messages
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reposting my question....please helppppp
okay i think this is simple but am kinda confused:
3bii.) - i don't get why to calculate the time only the vertical distance is considered
3biii.) - why do we use the horizontal velocity to calculate the speed.....and the time is the same as the journey of the ball before colliding with the wall??
Question paper: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Marking scheme:http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_21.pdf
please help urgently
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I know right!
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Aoa,
Can anybody help me with sketching this graph. Q2 part (b) iii
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
Can anybody help me with sketching this graph. Q2 part (b) iii
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_2.pdf
no 1cii
no 6
no 7
For 1cii, i dont know how to justify it..
For no 6, how do i know how many nodes are there? And why is 1/2lambda is 32.4 cm?
For no.7, i dont understand (a) (b) and (c)..how do i know which lamp is faulty? dont remember being taught this..
Please help. Waves in tube and electronics are my weakest topics
no 1cii
no 6
no 7
For 1cii, i dont know how to justify it..
For no 6, how do i know how many nodes are there? And why is 1/2lambda is 32.4 cm?
For no.7, i dont understand (a) (b) and (c)..how do i know which lamp is faulty? dont remember being taught this..
Please help. Waves in tube and electronics are my weakest topics
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Can someone please explain how they got to the velocity of the ball in question 3b (iii)
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
since the ball's initial velocity after collision is perfectly horizontal, take the horizontal distance moved and divide it by the time above(1.333)horizontal velocty stays constant throughout travel.
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3.b.i v^2=u^2 - 2gs
--> 0= 13^2 - (2*9.81 s)
--> s= 8.61 ~ 8.61m
3.b.ii v= u -gt Alternatively, you can also use horizontal velocity to calculate the time.
--> 0= 13 - 9.81*t time = distance/ horizontal velocity = 9.95/7.5 = 1.33 s
--> t= 1.325 s ~ 1.33 s
3. b.iii using vertical velocity, The horizontal velocity IS THE SPEED after rebounding from P. There is no vertical speed at P since in the beginning of the question
--> s= ut + .5 gt^2 it was said "The ball hits the wall at P with a velocity that is at right angles to the wall" This means there is no vertical velocity at P.
--> 8.61 = 4.905 t^2
--> t = 1.324 s
so horizontal velocity = 6.15/1.324 = 4.64 ~ 4.6 m/s
Hope that helps.
--> 0= 13^2 - (2*9.81 s)
--> s= 8.61 ~ 8.61m
3.b.ii v= u -gt Alternatively, you can also use horizontal velocity to calculate the time.
--> 0= 13 - 9.81*t time = distance/ horizontal velocity = 9.95/7.5 = 1.33 s
--> t= 1.325 s ~ 1.33 s
3. b.iii using vertical velocity, The horizontal velocity IS THE SPEED after rebounding from P. There is no vertical speed at P since in the beginning of the question
--> s= ut + .5 gt^2 it was said "The ball hits the wall at P with a velocity that is at right angles to the wall" This means there is no vertical velocity at P.
--> 8.61 = 4.905 t^2
--> t = 1.324 s
so horizontal velocity = 6.15/1.324 = 4.64 ~ 4.6 m/s
Hope that helps.
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tx soooooo much!!!! God bless3.b.i v^2=u^2 - 2gs
--> 0= 13^2 - (2*9.81 s)
--> s= 8.61 ~ 8.61m
3.b.ii v= u -gt Alternatively, you can also use horizontal velocity to calculate the time.
--> 0= 13 - 9.81*t time = distance/ horizontal velocity = 9.95/7.5 = 1.33 s
--> t= 1.325 s ~ 1.33 s
3. b.iii using vertical velocity, The horizontal velocity IS THE SPEED after rebounding from P. There is no vertical speed at P since in the beginning of the question
--> s= ut + .5 gt^2 it was said "The ball hits the wall at P with a velocity that is at right angles to the wall" This means there is no vertical velocity at P.
--> 8.61 = 4.905 t^2
--> t = 1.324 s
so horizontal velocity = 6.15/1.324 = 4.64 ~ 4.6 m/s
Hope that helps man.
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My pleasure.
- Messages
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An athlete bends down and then jumps straight up into the air. At the instant he starts his jump
three forces act on him – his weight, an upthrust from the air and the contact force with the
ground. What is the correct order in increasing magnitude of these three forces
A. contact force → weight → upthrust
B. upthrust → contact force → weight
C. upthrust → weight → contact force
D. weight → upthrust → contact force
Provide an explanation if you can answer. It's not from past papers.
three forces act on him – his weight, an upthrust from the air and the contact force with the
ground. What is the correct order in increasing magnitude of these three forces
A. contact force → weight → upthrust
B. upthrust → contact force → weight
C. upthrust → weight → contact force
D. weight → upthrust → contact force
Provide an explanation if you can answer. It's not from past papers.
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npkay...tx a lot
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can anyone please physics notes!!! its gonna help alot!! thank you