Mechanics M1: Post your doubt here

[GorgeousEyes]

can u post the answers to this question ?

Sure , 656 watt .

 

[Arjun Dhanak]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_4.pdf
Need help in question 6.
Thanks in advance.

 

[Arjun Dhanak]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_4.pdf
Can someone please help me in question 7.
Thanks in advance!

 

[leadingguy]

Emm , can't understand it :S can you repeat it in another way ?
also , your answer is different from what i have, So can you recheck your answer if you can ?
thanks .

my ans is t = 25 sec. so is urs



consider the distance travelled aftr 10s to be "S"


uptil 10s the car has traelled 12000m and police car has travelled 7500m. means now the police car is 4500m behind.

so the police car has to travel 4500 metres and the distance S . and we have to find the time when police cars completes 4500m and the distance "s" which the normal car has travelled aftr 10s. right????


the distance a normal car will travel aftr 10s will be S = t(1200) ...........................t= the time for covering distance "S"
now comming back to police car it also needs to cover that S distance to overcome the car plus 4500m !! remember??


so for police car distance will be (S+ 4500) = t (1500)................................................."t" is the time and it remains the same for both as we have to find the time when both are aside each other.

now substitute the vLUE OF "S" is the second eq. and find T. It will be 15sec.


10seconds ar already covered... the new time consumed is 15sec. so total time for travelling is 25seconds..


also see the graph above it will help u a lot

 

[iKhaled]

hey would u please help me with this question : http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_41.pdf number 4

dude this stupid question took me a long time to figure it out and at the end it just involves a lot of algebra thats it !

see the easiest way to do this question is to build up equations then substitute. Draw 2 triangles. the first one for the distance AB and the second one for the distance BC

lets first talk about the first stage ( distance AB):

u= u ms^-1
v = B ms^-1
a = a
t = 0.8 s
s1= 1.76m

s1 = ut + 1/2at^2
1.76 = 0.8u + 1/(0.8)^2a
1.76 = 0.8u + 0.32a

now lets tallk about the second stage ( distance BC ):

the initial velocity at B is the final velocity at B in stage one, so lets call this velocity 'B' and as u can see up i called the final velocty B but in the second stage it will be the initial velocity,

s2 = 2.16
t= 1.4 - 0.8 = 0.6
u = B
v= we dont care about it
a = a

s2 = ut+1/2at^2
s2= 0.6B + 1/2(0.6)^2 a
2.16 = 0.6B + 0.18a

now we have a problem, we have three unknowns ( u B and a ) in the 2 equations we made so we need to get rid of that b

v = u + at
B = u + 0.8a

and s1 + s2 = 3.92

0.8u + 0.32a +0.6B + 0.18a = 3.92
0.8u + 0.32a +0.6(u + 0.8a)+ 0.18a = 3.92

do some calcs and u will end up with an equation like this:

1.4u + 0.989a = 3.92

make another equation
s1 = 0.8u + 0.32a
1.76 = 0.8u + 0.32a

you have 2 equations now
first one : 1.76 = 0.8u + 0.32a
second one: 1.4u + 0.989a = 3.92

use elimination or simultaneous equation and find the value of u and a

any more questions feel free to ask and i will do my best to help u, god be with u in the coming exam

 

[leadingguy]

Sure , 656 watt .

the question is really too messy :S

 

[iKhaled]

the question is really too messy :S

gave a lot of time on it, i knew how to get the weight of the water though but i feel there is somethin really wrong with this question :S

 

[leadingguy]

gave a lot of time on it, i knew how to get the weight of the water though but i feel there is somethin really wrong with this question :S

lol

 

[iKhaled]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_4.pdf
Can someone please help me in question 7.
Thanks in advance!

(i) R = mgcosθ
R = 10m cos 21
R = 9.336m

while the particle is going upward u have 2 forces acting opposite to its direction of motion...first the component of the weight and F
F = ma

lets find the acceleration of the particle
v = u + at
0 = 10-2a
a= -5

-mgsinθ-F = -5m
-10msin 21 - F = -5m
F = 1.416m

(ii) F = μR
μ = 1.416/9.336
μ = 0.152

(iii) while the particle is traveling downward only F is acting against the direction of motion
mgsinθ - F = ma
10msin21 - 1.416m = ma
a = 2.17 ms^-2

lets find the distance it moved ( note the distance moved from P to the highest point is the same distance it moved while it is goin back to P ) so,

s = 1/2(v + u)t
s = 1/2(10)(2)
s= 10m

v^2 = u^2 + 2as
v= √(0 + 2(2.17)(10))
v = 6.58 ms^-1

 

[GorgeousEyes]

my ans is t = 25 sec. so is urs



consider the distance travelled aftr 10s to be "S"


uptil 10s the car has traelled 12000m and police car has travelled 7500m. means now the police car is 4500m behind.

so the police car has to travel 4500 metres and the distance S . and we have to find the time when police cars completes 4500m and the distance "s" which the normal car has travelled aftr 10s. right????


the distance a normal car will travel aftr 10s will be S = t(1200) ...........................t= the time for covering distance "S"
now comming back to police car it also needs to cover that S distance to overcome the car plus 4500m !! remember??


so for police car distance will be (S+ 4500) = t (1500)................................................."t" is the time and it remains the same for both as we have to find the time when both are aside each other.

now substitute the vLUE OF "S" is the second eq. and find T. It will be 15sec.


10seconds ar already covered... the new time consumed is 15sec. so total time for travelling is 25seconds..




also see the graph above it will help u a lot

Got it .Thank you a lot

 

[GorgeousEyes]

gave a lot of time on it, i knew how to get the weight of the water though but i feel there is somethin really wrong with this question :S

lol

It's okay , thank you for your effort

 

[bamteck]

well the formula and idea involved is s=ut +1/2at^2 now first calculate distance travelled by particle p and then by particle q since they are 4.9m apart there difference should be equal to 4.9m.
firstly note that the ques says q is released when p pases point A hence we will take time for p as 2+t since p has been traveeling 2s before u will obtain an expression for s then for q using the same formula but value of time should be t subract Sp-Sq=4.9

Thanks !
Im confused

 

[bamteck]

bamteck here is the graph to the question u asked me yesterday.. i just woke up >.<
the y- axis is the velocity and the x-axis is the time

Thank you

 

[daviruss]

dude this stupid question took me a long time to figure it out and at the end it just involves a lot of algebra thats it !

see the easiest way to do this question is to build up equations then substitute. Draw 2 triangles. the first one for the distance AB and the second one for the distance BC

lets first talk about the first stage ( distance AB):

u= u ms^-1
v = B ms^-1
a = a
t = 0.8 s
s1= 1.76m

s1 = ut + 1/2at^2
1.76 = 0.8u + 1/(0.8)^2a
1.76 = 0.8u + 0.32a

now lets tallk about the second stage ( distance BC ):

the initial velocity at B is the final velocity at B in stage one, so lets call this velocity 'B' and as u can see up i called the final velocty B but in the second stage it will be the initial velocity,

s2 = 2.16
t= 1.4 - 0.8 = 0.6
u = B
v= we dont care about it
a = a

s2 = ut+1/2at^2
s2= 0.6B + 1/2(0.6)^2 a
2.16 = 0.6B + 0.18a

now we have a problem, we have three unknowns ( u B and a ) in the 2 equations we made so we need to get rid of that b

v = u + at
B = u + 0.8a

and s1 + s2 = 3.92

0.8u + 0.32a +0.6B + 0.18a = 3.92
0.8u + 0.32a +0.6(u + 0.8a)+ 0.18a = 3.92

do some calcs and u will end up with an equation like this:

1.4u + 0.989a = 3.92

make another equation
s1 = 0.8u + 0.32a
1.76 = 0.8u + 0.32a

you have 2 equations now
first one : 1.76 = 0.8u + 0.32a
second one: 1.4u + 0.989a = 3.92

use elimination or simultaneous equation and find the value of u and a

any more questions feel free to ask and i will do my best to help u, god be with u in the coming exam

haha thx man/sir may allah reward you for your work! best thing is to eliminate easiest way thx you. and inshallah i need to get an A in this exam cuz i kinda messed up in my p1 paper. btw if i get in m1 A inshallah and got C in p1 lasama7allah. well i be able to get the B in total ?

 

[iKhaled]

haha thx man/sir may allah reward you for your work! best thing is to eliminate easiest way thx you. and inshallah i need to get an A in this exam cuz i kinda messed up in my p1 paper. btw if i get in m1 A inshallah and got C in p1 lasama7allah. well i be able to get the B in total ?

how many questions did u leave in this p1 and what do u think ur score will be ?

 

[daviruss]

how many questions did u leave in this p1 and what do u think ur score will be ?

told more than C

 

[iKhaled]

told more than C

yeah i know it depends actually if its a high C or a low C got me ? so approx how many marks u think u have lost ?

 

[Kirabo Takirambudde]

Let's assume Car's Driving force is 'X', at the Bottom F =X and and at the top,F = 3 x (X) = 3X

Power = F x V ,At bottom Power = 6 x (X) = 6X , At top Power = 6X x 5 = 30X

Power = F x V, 30X = 3X x V , V at top = 30X / 3X = 10 , and then u solve it normally like (i)

THANKS!!!!!

 

[daviruss]

yeah i know it depends actually if its a high C or a low C got me ? so approx how many marks u think u have lost ?

 

[daviruss]

yeah i know it depends actually if its a high C or a low C got me ? so approx how many marks u think u have lost ?

like 20 marks i may not get the B even if i get an A in m1 ?