Mechanics M1: Post your doubt here

[Asheeta]

Thanks Aheeta

Anytime Amigo

 

[Asheeta]

Nov08 7 , ii please

so let time at A be 0.
v = integration a dt
V= 10t - 0.15 (t^2) +c
at t=0 v=5 m/s
so V = 10t - 0.15 (t^2) +5
s = integration V dt
so S = 5(t^2) - 0.05(t^3) + 5t + c
at t=0 s=0
so S = 5(t^2) - 0.05(t^3) + 5t
for 3 s means that it has already travelled 1.25 m in 0.5 s
so t=2.5 s
at this time s=5(2.5^2) - 0.05(2.5^3) + 5(2.5)
so s= 43.0 m
so total distance fallen = 43.0 + 1.25 = 44.25 m

 

[GorgeousEyes]

so let time at A be 0.
v = integration a dt
V= 10t - 0.15 (t^2) +c
at t=0 v=5 m/s
so V = 10t - 0.15 (t^2) +5
s = integration V dt
so S = 5(t^2) - 0.05(t^3) + 5t + c
at t=0 s=0
so S = 5(t^2) - 0.05(t^3) + 5t
for 3 s means that it has already travelled 1.25 m in 0.5 s
so t=2.5 s
at this time s=5(2.5^2) - 0.05(2.5^3) + 5(2.5)
so s= 43.0 m
so total distance fallen = 43.0 + 1.25 = 44.25 m

thanks a loooot

 

[TeKnOzOr]

Aslkm...!!! what is the toughest paper m1 that exist.!?!
im stuck a bit in paper m1..mAybe bkoz im ruined

 

[Asheeta]

Aslkm...!!! what is the toughest paper m1 that exist.!?!
im stuck a bit in paper m1..mAybe bkoz im ruined

Doing the toughest paper now will not help... just try to be very good in what you are atleast you are sure to gain marks!
keep a good attitude and do not lose hope

 

[Amy Bloom]

Hello there. I need help for this question:

 

[Asheeta]

Hello there. I need help for this question:
View attachment 16613

initial velocity of A = final velocity of B = 2.6 m/s
When string is slack a= 10m/s^2
at max height v = 0 m/s
(v^2) = (u^2) + 2as
0 = 6.76 + (2*-10*s)
S = 0.338 m
v = u + at
0 = 2.6 -10t
t = 0.26 s (time to reach max height)
therefore time to reach max height and return to equilibrium position = 0.26*2 = 0.52 s
distance travelled by B = 0.65 m
time taken to travel this distance = 0.5 s as calculated above in (ii)
therefore T = 0.5 + 0.52 = 1.02 s

So distance travelled by A in 0.5 s = 0.65 m
dist travelled by A in reaching the max height and returning back to the equilibrium position = (0.338*2) = 0.676 m
so total distance travelled in T s = 0.65 + 0.676 = 1.326 m

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_42.pdf

Need help for no. 5(ii)

 

[Amy Bloom]

initial velocity of A = final velocity of B = 2.6 m/s
When string is slack a= 10m/s^2
at max height v = 0 m/s
(v^2) = (u^2) + 2as
0 = 6.76 + (2*-10*s)
S = 0.338 m
v = u + at
0 = 2.6 -10t
t = 0.26 s (time to reach max height)
therefore time to reach max height and return to equilibrium position = 0.26*2 = 0.52 s
distance travelled by B = 0.65 m
time taken to travel this distance = 0.5 s as calculated above in (ii)
therefore T = 0.5 + 0.52 = 1.02 s

So distance travelled by A in 0.5 s = 0.65 m
dist travelled by A in reaching the max height and returning back to the equilibrium position = (0.338*2) = 0.676 m
so total distance travelled in T s = 0.65 + 0.676 = 1.326 m

Thanks dear. But i don't get one thing. Look at this diagram:

Why is the distance here not (0.65 x 2), because i consider that both particles were initially at 0.65 on the ground, so when B slacks, A goes up 0.65 metres more, and is 1.3?

One more thing, can you explain me the part where it says "The string remains slack until A is at a height of 1.3 m above the ground for a second time"?
Does this mean, 2x (x + x) shown in the diagram above?

 

[Asheeta]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_42.pdf

Need help for no. 5(ii)

Using s = ut + (1/2)a(t^2)
sP = 12t - 5(t^2)
sQ = 7t - 5(t^2)
3 sP = 8 sQ
3{12t - 5(t^2)} = 8 {7t - 5(t^2)}
36t - 15t^2 = 56t - 40t^2
5t( 5t - 4) = 0
so t = 4/5 = 0.8 s

V=u+at
vP = 12 +(-10*0.8) = 4 m/s
vQ = 7 + (-10*0.8) = -1 m/s (travelling in the opposite direction)

 

[elbeyon]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_42.pdf

Need help for no. 5(ii)

Use the equation s =ut + 0.5at^2 for both the particles and replace the values of "u" as given by the question and "a" as -10 ms^-2.

For p:
Hp = 12t - 5t^2

For q:
Hq = 7t - 5t^2

Then use the equation given by question (i.e 3hp = 8hq ) and you will obtain the value of t. Then use the equation v = u + at to find out the velocities of each particles.

Hope this helps.

 

[Asheeta]

Thanks dear. But i don't get one thing. Look at this diagram:
View attachment 16620
Why is the distance here not (0.65 x 2), because i consider that both particles were initially at 0.65 on the ground, so when B slacks, A goes up 0.65 metres more, and is 1.3?

One more thing, can you explain me the part where it says "The string remains slack until A is at a height of 1.3 m above the ground for a second time"?
Does this mean, 2x (x + x) shown in the diagram above?

so the first part:
this is because A was already 0.65 m above the ground. It only moves an additional 0.65 m, its height is 1.3 m but its displacement remains 0.65 m

second:
this implies the time when A is at first at 1.3 m above the ground i.e when B hits the ground
second time implies when A goes to a max height and returns back to its original position i.e at 1.3m , again above the ground

Hope it is ok now.

 

[Amy Bloom]

so the first part:
this is because A was already 0.65 m above the ground. It only moves an additional 0.65 m, its height is 1.3 m but its displacement remains 0.65 m

second:
this implies the time when A is at first at 1.3 m above the ground i.e when B hits the ground
second time implies when A goes to a max height and returns back to its original position i.e at 1.3m , again above the ground

Hope it is ok now.

Okay got it. Thanks!

 

[elbeyon]

People out here please develop a habit of hitting the "Like" button when someone helps you. It will encourage them to further help you when ever you need them. Will you loose anything by just hitting the like button when someone helps you ? So please keep you ego aside and hit the "Like" button when someone helps you. Please try to appreciate the time they spent on helping you.

 

[bamteck]

Using s = ut + (1/2)a(t^2)
sP = 12t - 5(t^2)
sQ = 7t - 5(t^2)
3 sP = 8 sQ
3{12t - 5(t^2)} = 8 {7t - 5(t^2)}
36t - 15t^2 = 56t - 40t^2
5t( 5t - 4) = 0
so t = 4/5 = 0.8 s

V=u+at
vP = 12 +(-10*0.8) = 4 m/s
vQ = 7 + (-10*0.8) = -1 m/s (travelling in the opposite direction)

Thanks mate

 

[bamteck]

Use the equation s =ut + 0.5at^2 for both the particles and replace the values of "u" as given by the question and "a" as -10 ms^-2.

For p:
Hp = 12t - 5t^2

For q:
Hq = 7t - 5t^2

Then use the equation given by question (i.e 3hp = 8hq ) and you will obtain the value of t. Then use the equation v = u + at to find out the velocities of each particles.

Hope this helps.

Thanks mate

 

[Amy Bloom]

People out here please develop a habit of hitting the "Like" button when someone helps you. It will encourage them to further help you when ever you need them. Will you loose anything by just hitting the like button when someone helps you ? So please keep you ego aside and hit the "Like" button when someone helps you. Please try to appreciate the time they spent on helping you.

Rightly said! and i would add something, most importantly, thank them.

 

[elbeyon]

Rightly said! and i would add something, most importantly, thank them.

Yeah ! That's the main idea that I am trying to convey.

 

[Varuna0911]

Could you help me with 5(ii) plz?

 

[iKhaled]

hey i need help in may june 2012 p42 the last question the last part 7(ii)