IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: mousa on February 05, 2010, 02:58:57 pm

Title: Mechanics help!!
Post by: mousa on February 05, 2010, 02:58:57 pm
plz , i need your help doing those Questions.



June 2009, Q6 part(iv)
June 2006 ,Q5 part(i)
November 2004 , Q2 part (ii)

Thanks in advance
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 05, 2010, 03:41:25 pm
 When I get home
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 06, 2010, 09:25:40 am
6i)s=1/at^2 so a=2s/t^2=2m/s^2
ii)v=u+at=2*9.6=1.2m/s
s=
u=1.2
v=0
a=-9.8
t=
v^2=u^2+2as
0=1.44-19.6s so s=1.44/19.5
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 06, 2010, 09:28:31 am
2nd question
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 06, 2010, 09:31:02 am
Resolve paralell to plane Pcos 30-mgsin30=0
P=mgcot30=18/tan30
Title: Re: Mechanics help!!
Post by: mousa on February 06, 2010, 11:35:13 am
Quote from: astarmathsandphysics on February 06, 2010, 09:25:40 am
6i)s=1/at^2 so a=2s/t^2=2m/s^2
ii)v=u+at=2*9.6=1.2m/s
s=
u=1.2
v=0
a=-9.8
t=
v^2=u^2+2as
0=1.44-19.6s so s=1.44/19.5

mm, sir i am confused, why have you done it this way??, wont the maximum height= 0.36 m ( which it was initially at )+ the total distance partical A will cover while falling which is 0.36 m aswell ///

thanks for the help with the other questions :)
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 06, 2010, 12:18:15 pm
Wait
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 06, 2010, 12:25:09 pm
o.36+extra distance travelled by B after A hits the ground - during this time the acceleration chages from 2m/s^2 to -9.8m/s^2
Title: Re: Mechanics help!!
Post by: mousa on February 06, 2010, 01:41:00 pm
Quote from: astarmathsandphysics on February 06, 2010, 12:25:09 pm
o.36+extra distance travelled by B after A hits the ground - during this time the acceleration chages from 2m/s^2 to -9.8m/s^2

why??? ::) does the acceleration change??
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 06, 2010, 01:44:25 pm
when falling freely under gravity the acceleration is always -9.8
Title: Re: Mechanics help!!
Post by: mousa on February 06, 2010, 02:36:28 pm
Quote from: astarmathsandphysics on February 06, 2010, 01:44:25 pm
when falling freely under gravity the acceleration is always -9.8

so what exactly is the 2 m/s^2 that we found?? and about which particle are you talking about?
Title: Re: Mechanics help!!
Post by: mousa on February 06, 2010, 02:44:30 pm
Quote from: astarmathsandphysics on February 06, 2010, 09:25:40 am
6i)s=1/at^2 so a=2s/t^2=2m/s^2
ii)v=u+at=2*9.6=1.2m/s
s=
u=1.2
v=0
a=-9.8
t=
v^2=u^2+2as
0=1.44-19.6s so s=1.44/19.5


one thing else, why did you take acce as 2 m/s2 while finding v= 1.2 and then took it as -9.8?? i am confuseeeeeeeeeeeeeed  :P
Title: Re: Mechanics help!!
Post by: astarmathsandphysics on February 06, 2010, 03:05:52 pm
There are tow SEPARATE INSTANCES
before A hits the foor a=2 when a hits the floor v=1.2
After A hit the floor the final velocity of A becomes the initial velocity of B and a becomes -9.8

WHENEVER YOU HAVE A COLLISION YOU HAVE TO USE SUVAT ALL OVER AGAIN
Title: Re: Mechanics help!!
Post by: mousa on February 06, 2010, 04:48:28 pm
Quote from: astarmathsandphysics on February 06, 2010, 03:05:52 pm
There are tow SEPARATE INSTANCES
before A hits the foor a=2 when a hits the floor v=1.2
After A hit the floor the final velocity of A becomes the initial velocity of B and a becomes -9.8

WHENEVER YOU HAVE A COLLISION YOU HAVE TO USE SUVAT ALL OVER AGAIN

Thank you sir!!, i finally got it!!