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Physics: Post your doubts here!

Soldier313

Soldier313

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okay i think this is simple but am kinda confused::cry:

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??:confused:

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
 
Soldier313

Soldier313

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reposting my question....please helppppp

Soldier313 said:
okay i think this is simple but am kinda confused::cry:

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??:confused:

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:( thanx so much.....May God bless you
Click to expand...
 
M

Mattman

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Unicorn said:
Can someone please explain how they got to the velocity of the ball in question 3b (iii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_21.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Click to expand...
:) 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.
 
00tanveer

00tanveer

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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. :)
 
Soldier313

Soldier313

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00tanveer said:
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 man. :)
Click to expand...
tx soooooo much!!!! God bless:)
 
00tanveer

00tanveer

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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.
 
Unicorn

Unicorn

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Mattman said:
:) 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.
Click to expand...

00tanveer said:
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 man. :)
Click to expand...


ohh ok

thanks :D
 
Saad (سعد)

Saad (سعد)

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leosco1995 said:
www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2 d (ii)

The force calculated in (i) is 75 N and the resistance force is 23 N.. so shouldn't the power be the net force * velocity? In the marking scheme, the forces have been added instead. Why?
Click to expand...

They are added, because the question asks for the total output power supplied by the boy's legs. The boy's legs first provide a force equal to the resistive force (23 N) in order to overcome the resistance. Then a further 75 N is supplied by the boy's legs, which causes his acceleration of 1.12 m/s. So the total force supplied by the boy's legs is = 23 + 75 N = 98 N.

Power = Force x velocity
= 98 x 4.5
= 441 W.

Hope that helps insha' Allah!
 
L

leosco1995

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Saad (سعد) said:
They are added, because the question asks for the total output power supplied by the boy's legs. The boy's legs first provide a force equal to the resistive force (23 N) in order to overcome the resistance. Then a further 75 N is supplied by the boy's legs, which causes his acceleration of 1.12 m/s. So the total force supplied by the boy's legs is = 23 + 75 N = 98 N.

Power = Force x velocity
= 98 x 4.5
= 441 W.

Hope that helps insha' Allah!
Click to expand...
It did help. Thanks a lot. :)
 
hussamh10

hussamh10

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leosco1995 said:
www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2 d (ii)

The force calculated in (i) is 75 N and the resistance force is 23 N.. so shouldn't the power be the net force * velocity? In the marking scheme, the forces have been added instead. Why?
Click to expand...
Because the boy had to do work against the resistance as well as to maintain the speed till 4.5 so the both forces will add (the one which cause acceleration and against the friction).Hope i helped
 
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