P1 CIE physics doubts

[sweetie]

Q10, 11 and 32

thanx 4 any help!!!!

[~ A.F ~]

I'm on it

[~ A.F ~]

Q10, 11 and 32

thanx 4 any help!!!!

PS; I haven't got the answers..soc heck'em pls :

Q.10: v1-v2/u2+u1=1

therefore u1+u2=v2-v1------> D

Q.11:

You must write down the equations for both objects:

1) box-----> T-6= 2a----> T=28+6
2) 2 KG---> 2g-T=2a----> T=2g-2a

now we got 2 Ts equations:

8a+6=2g-2a

10a= 2g-6

a= 1.4---->A

Hope you got that

[nid404]

thanks + rep  

it goes up by 2

[~ A.F ~]

Thank you Nid   

[sweetie]

PS; I haven't got the answers..soc heck'em pls :

Q.10: v1-v2/u2+u1=1

therefore u1+u2=v2-v1------> D

Q.11:

You must write down the equations for both objects:

1) box-----> T-6= 2a----> T=28+6
2) 2 KG---> 2g-T=2a----> T=2g-2a

now we got 2 Ts equations:

8a+6=2g-2a

10a= 2g-6

a= 1.4---->A

Hope you got that

for Q10 , arent v1 and v2 moving in same direction , so their signs should be the same , nd the signs of u1 and u2 should be diff.?

[~ A.F ~]

lol no..

there's that thing called coefficient of restitution, e,=

e= v1+- v2/u1+-u2

sign is positive when directions are similar..e=1 when collision is PERFECTLY elastic

[astarmathsandphysics]

For restitution
If moving in same direction closing speed or parting speed = bigger speed - smaller spedd
If movin towards each other clsong spped =sum of speeds

[astarmathsandphysics]

And if they are moving apart separation speed = sum of speeds

[sweetie]

ok ppl,
cn sum1 xplain Q32 as well nd Q11 and 10 too  ( am not gettin them )
Thank You

[astarmathsandphysics]

year? I suppose p1

[sweetie]

its nov08 ( i've attached the file)

[astarmathsandphysics]

10 approach speed=u1+u2
separation speed=v2-v1
they are equal for elastic collisions  D
11.res for both particles using F=ma
for 2g particle 2g-T=2a (1)
for 6g particle T-6=6a (2)
(1) +(2) 2g-6=8a so a/(2*9.8-6)/8=1.7 B
32)1/R=1/100+1/10+1/10+1/10+1/10+1/10+1/10=61/100
R=100/61 B

[dodo]

hi would please help me in answering this question and plesa while answering explain how u progressed to this result
the paper is may/june CIE physics 2009
question 5 part b)

[nid404]

hi would please help me in answering this question and plesa while answering explain how u progressed to this result
the paper is may/june CIE physics 2009
question 5 part b)

here u go

there is a formula u need to know

phase difference=( 2pi/ lamda)  X path difference

phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi  X path difference.
path difference is S₂M - S₁M 
find S₂M using Pythagoras = root of 80² + 100²
                                                   =128
path difference= 128-100=28
wavelength= 2pi/pi X 28
               =56cm                 
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz   wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas