Mathematics: Post your doubts here!

[Lyfroker]

need help
i cnt solve Q:2(iii) and Q:6 and Q:7 in P4 may/june/2010

 

[hm12]

need help
i cnt solve Q:2(iii) and Q:6 and Q:7 in P4 may/june/2010

which variant?

 

[Lyfroker]

variant 2

 

[Suraj]

Is there any collections for tough questions???

 

[hm12]

need help
i cnt solve Q:2(iii) and Q:6 and Q:7 in P4 may/june/2010

question 2 part(iii) distance moved forward=distance moved to travel to original position so 1.08=1/2*(11-8)*v so v=0.72m/s

 

[Most_UniQue]

Can someone solve this? I got problem with these kind of questions

 

[hm12]

which variant?

question 6
i) mew=0.3 so F=0.3*2=0.6
4.5-T=0.45a
T-0.6=0.2a solving them simultaneously you get a=3.9m/s^2
Height of B above ground =2-(2.8-2.1)=1.3m
v^2-u^2=2as
v^2-0=2*1.3*6 so v=3.95m/s
ii)s=ut+1/2 at^2
t^2=1.3*2/6
t=0.658
v=u+at
v=0+6(0.658)=3.948
-F=ma
-0.6=0.2a so a=-3
v^2-u^2=2as
v^2-3.948^2=2*-3*(2.1-1.3)
v=3.29m/s

 

[Lyfroker]

thnks alooot

 

[iKhaled]

math P4 (Mechanics) may june 2007 question 5 ii) someone explain pls

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s07_qp_4.pdf

 

[Lyfroker]

question 6
i) mew=0.3 so F=0.3*2=0.6
4.5-T=0.45a
T-0.6=0.2a solving them simultaneously you get a=3.9m/s^2
Height of B above ground =2-(2.8-2.1)=1.3m
v^2-u^2=2as
v^2-0=2*1.3*6 so v=3.95m/s
ii)s=ut+1/2 at^2
t^2=1.3*2/6
t=0.658
v=u+at
v=0+6(0.658)=3.948
-F=ma
-0.6=0.2a so a=-3
v^2-u^2=2as
v^2-3.948^2=2*-3*(2.1-1.3)
v=3.29m/s

bt y F=0.3*2??????

 

[Lyfroker]

math P4 (Mechanics) may june 2007 question 5 ii) someone explain pls

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_4.pdf

(i) total work done = increase in KE + R
WD= 1/2m(v^2-u^2) + 5000
WD = 1/2*12500*(25^2-17^2) + 5000
WD = 2100000J+5000
WD = 2100KJ+5000
WD = 7100KJ

 

[Most_UniQue]

Can someone solve this? I got problem with these kind of questions

Can someone try to explain mine too

 

[shanky631]

I hope u got it!!

hey, yes dude thnx for the help.

 

[Most_UniQue]

(i) total work done = increase in KE + R
WD= 1/2m(v^2-u^2) + 5000
WD = 1/2*12500*(25^2-17^2) + 5000
WD = 2100000J+5000
WD = 2100KJ+5000
WD = 7100KJ

Hey why do they sometimes subtract R and sometimes its added. Im really confused

 

[hm12]

bt y F=0.3*2??????

F=mew* R where R= normal contact force which in this case is equal to 10m= 10*0.2=2N

 

[Lyfroker]

F=mew* R where R= normal contact force which in this case is equal to 10m= 10*0.2=2N

oh k thnks

 

[Lyfroker]

Hey why do they sometimes subtract R and sometimes its added. Im really confused

well i m not sure bt i thnk it depends if the particle is moving up or down

 

[Most_UniQue]

well i m not sure bt i thnk it depends if the particle is moving up or down

And if its moving up?

 

[Lyfroker]

And if its moving up?

as i said b4 m nt rele sure go to this website http://www.examsolutions.co.uk/maths-revision/index.php#Mechanics it might b helpful

 

[iKhaled]

(i) total work done = increase in KE + R
WD= 1/2m(v^2-u^2) + 5000
WD = 1/2*12500*(25^2-17^2) + 5000
WD = 2100000J+5000
WD = 2100KJ+5000
WD = 7100KJ

u misunderstood me i meant the second part, where u have to find the height :/