Mechanics help!!

[mousa]

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

[astarmathsandphysics]

 When I get home

[astarmathsandphysics]

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

[astarmathsandphysics]

2nd question

[astarmathsandphysics]

Resolve paralell to plane Pcos 30-mgsin30=0
P=mgcot30=18/tan30

[mousa]

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

[astarmathsandphysics]

Wait

[astarmathsandphysics]

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

[mousa]

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??

[astarmathsandphysics]

when falling freely under gravity the acceleration is always -9.8

[mousa]

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?

[mousa]

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 

[astarmathsandphysics]

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

[mousa]

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!!