PAPER 1 MAY/JUNE 2013 PHYSICS DOUBTS HERE

[h4rriet]

Anybody doing the 2013 May/June Physics paper 1 can post their doubts here/discuss questions & answers here.

 

[bigboy]

Anybody doing the 2013 May/June Physics paper 1 can post their doubts here/discuss questions & answers here.

guys please, help with question 23 http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf

 

[h4rriet]

guys please, help with question 23 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

Pressure=rho x g x h. Rho=given, g=given, h=vertical height=5. Work it out and add to that the atmospheric pressure.

 

[bigboy]

tha

Pressure=rho x g x h. Rho=given, g=given, h=vertical height=5. Work it out and add to that the atmospheric pressure.

thanks a ton, how about this one?
A supermarket trolley, total mass 30 kg, is moving at 3.0 m s–1. A retarding force of 60 N is applied
to the trolley for 0.50 s in the opposite direction to the trolley’s initial velocity.
What is the trolley’s new velocity after the application of the force?

 

[h4rriet]

tha
thanks a ton, how about this one?
A supermarket trolley, total mass 30 kg, is moving at 3.0 m s–1. A retarding force of 60 N is applied
to the trolley for 0.50 s in the opposite direction to the trolley’s initial velocity.
What is the trolley’s new velocity after the application of the force?

Find the deceleration using a=F/m, and then choose a suvat equation to work out the final velocity.

 

[Oishee Asif]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

 

[hino]

tha
thanks a ton, how about this one?
A supermarket trolley, total mass 30 kg, is moving at 3.0 m s–1. A retarding force of 60 N is applied
to the trolley for 0.50 s in the opposite direction to the trolley’s initial velocity.
What is the trolley’s new velocity after the application of the force?

force=rate of change of momentum
F=mv-mu/t
F=-(mv-mu)/t
used minus outside the bracket since retarding force is in the opposite direction
u can use this formula to find the final velocity of the trolley

 

[hino]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

o/n 10 Q11
total momentum before=total momentum after
Mu=M1v1-M2v2
since the object is initially at rest
0=M1v1-M2v2
M1v1=M2v2
V1/V2=M2/M1

 

[hino]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

O/N 10 P11 q34 toatal resistance of the two wires=2*0.005*800
=8ohm
voltage lost in the wires =IR
=0.6*8
=4.8
total output emf of the power supply=4.8+16=20.8V

 

[h4rriet]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

M/J/09:
9. I=m(v-u). Give the velocities signs.
10.
11. Pressure increases as depth increases. Since the area is the same (p=F/A), the force too increases like that.
13. 1.2F=0.2 x 900.
15.
18. p=densityxgxh. h=the difference in height between the two tubes.

 

[Oishee Asif]

o/n 10 Q11
total momentum before=total momentum after
Mu=M1v1-M2v2
since the object is initially at rest
0=M1v1-M2v2
M1v1=M2v2
V1/V2=M2/M1

Thank you!

 

[Oishee Asif]

O/N 10 P11 q34 toatal resistance of the two wires=2*0.005*800
=8ohm
voltage lost in the wires =IR
=0.6*8
=4.8
total output emf of the power supply=4.8+16=20.8V

I feel like an idiot now; thank you so much!

 

[Oishee Asif]

M/J/09:
9. I=m(v-u). Give the velocities signs.
10.
11. Pressure increases as depth increases. Since the area is the same (p=F/A), the force too increases like that.
13. 1.2F=0.2 x 900.
15.
18. p=densityxgxh. h=the difference in height between the two tubes.

Thank you! However, what does 'I' stand for in 9?

 

[hino]

I feel like an idiot now; thank you so much!

your welcome

 

[hino]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

m/j 09
Q10

Consider first the conservation of momentum.
Momentum before = momentum after
0 = mv[1] - 2mv[2]
v[1] = 2v[2]
KE[1] = (1/2)mv[1]^2 =1/2m(2v[2]^2)
KE[2] = (1/2)2mv[2]^2
KE[1]/KE[2] = 4v[2]^2/2v[2]^2 = 2/1

Q15 P.E lost by X=P.E gained by Y
since X LOST half of its mass as well as height,Y gained m/2 mass and h/2 height
=m/2*h/2*g
=mgh/4

 

[Oishee Asif]

m/j 09
Q10

Consider first the conservation of momentum.
Momentum before = momentum after
0 = mv[1] - 2mv[2]
v[1] = 2v[2]
KE[1] = (1/2)mv[1]^2 =1/2m(2v[2]^2)
KE[2] = (1/2)2mv[2]^2
KE[1]/KE[2] = 4v[2]^2/2v[2]^2 = 2/1

Q15 P.E lost by X=P.E gained by Y
since X LOST half of its mass as well as height,Y gained m/2 mass and h/2 height
=m/2*h/2*g
=mgh/4

Amazing. Thank you again.

 

[hino]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

o/n 09 p11 Q1o look up the page for the answer

Q 14 Velocity at launch
E=1/2 mv^2
v=√2E/m
horizontal component of velocity at launch (u)
cos45=u/√2E/m
u=0.7√2E/m

at the highest point, the object has no vertical velocity, but has the same horizontal velocity it had at launch (if we ignore air resistance, horizontal velocity does not vary through the flight since there are no horizontal forces acting on the projectile)
so
vertical component=0 horizontal component=0.7√2E/m
now since we have both the components of velocity at the highest point we ll calculate the resultant velocity using pythagoras theorom which is going to be the same as the horizontal component of velocity
using this resultant velocity we ll calculate the K.E at the highest point
K.E=1/2 mv^2
K.E=1/2 m(0.7√2E/m)^2
K.E=0.5E (answer)

 

[Champ101]

hey guys add me in.

 

[hino]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

o/n 09 p11 Q 15
here the energy stored in the spring is equal to the K.Es of both the trollies
we ll use conservation of momentum to find the velocity of the trolley
mu=m1v1 -m1v2 (momentum before is zero)
0=2*2 -1*v2
v2=4ms^-1
since we have the final velocities of both the trollies (v1=2 v2=4)we ll calculate their kinetic energies
K.E1=1/2mv^2=1/2*2*2^2=4
K.E2=1/2mv^2=1/2*1*4^2=8
total energy stored =4+8=12 J

 

[hino]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

o/n 09 p11 Q22
area=1.96*10^-7m^2
young modulus=stress/strain
find strain with the help of data provided
u ll get 5.1*10^-4
since they are asking for the percentage strain simply multiply the strain with 100,u ll get the percentage