Mathematics: Post your doubts here!

[Lyfroker]

can any one pls give me the formulas for finding workdone against resistnce of friction, with respect to kinetic, potential energy??
plss!! :'(

for partical moving down the plane : work done by driving force = 0
so 0 = gain in KE - loss in PE + work done against friction

partical moving up the plane : work done by the driving force = gain in PE + gain in KE + work done against resistance

 

[Lyfroker]

can anyone solve Q :6 (ii) P42 M/J/10

 

[Nibz]

cant solve Q:1, 3 ,4 P42 oct/nov/09 Q:3 , 5 and 7 Oct/Nov/2010 P42 plzzzzzzzzzz help . . .. . . . .plzzzzzzzzzz ​

Oct 2009.Q1. (i) Vertical component of weight = 12 sin 40
P acts opposite to it. So P = 12sin40P = 7.71 N

(ii) The horizontal component of P acts opposite to the vertical component of weight.
So Pcos40 = 12sin40
P = 12 sin40 / cos 40 or 12 ( tan 40) = 10.1 N

Q.3
Power = 24000 W
Mass = 1250 kg
Resistive Force = 600 N
Driving Force = Power / Velocity = 24000/v
When velocity is maximum, acceleration = 0
And when acceleration = 0 Driving Force = Resitive Force
So Driving Force = 600 N
600 = 24000/v
v= 40m/s

(ii)
When v= 15 m/s
Driving Force becomes 24000/15 = 1600N
Net Force = Driving Force - Resistive Force = 1600 - 600 = 1000 N
Net Force = mass x acceleration
1000 = 1250 x a
a = 0.8 m/s'2

Q4.
Normal = Mass x 10 cos theta
1.2 = 9.6m
m = 0.125 kg

(ii) Frictional component + 10 x mass x sin theta = mass (-acceleration (deceleration))
0.4 + (0.125) x (10) x (0.28) = 0.125 ( -a)
-a = 6
So decelration = 6 m/s'2

(iii) Frictonal component = 0.4
Weight's component = mg x sin theta = 0.35
since Frictional component is greater than weight's component. X remains at rest.

 

[Nibz]

can anyone solve Q :6 (ii) P42 M/J/10

For A:
Reaction = mass x 10
R = 2 N

T - uR = mass x acc.
T - (o.3 x 2) = 0.2a -> eq. 1

For B:
mass x 10 - T = mass x acc.
4.5 - T = 0.45 a -> eq. 2

Solve these simultaneously, to get a and T.
a= 6 m/s'2

For B:
v = ?
u = 0
a = 6 m/s'2
s = 1.3 m
v'2 - u'2 = 2as
v = 3.95 m/s

(ii)

When B hits the floor, there is no forward force acting on A. So the only force on it is the Frictional force.
Frictional Force = uR
uR = mass x deceleration
u (mass x 10 ) = mass x deceleration
both masses cancel out.
10 u = deceleration
10 x 0.3 = deceleration
3 m/s'2 = deceleration

so for A:
v= ?
a = -3
u = 3.95
s = 0.8 (2.8 - 2)

v'2 - u'2 = 2as
v = 3.29 m/s

 

[Lyfroker]

Oct 2009.Q1. (i) Vertical component of weight = 12 sin 40
P acts opposite to it. So P = 12sin40P = 7.71 N

(ii) The horizontal component of P acts opposite to the vertical component of weight.
So Pcos40 = 12sin40
P = 12 sin40 / cos 40 or 12 ( tan 40) = 10.1 N

Q.3
Power = 24000 W
Mass = 1250 kg
Resistive Force = 600 N
Driving Force = Power / Velocity = 24000/v
When velocity is maximum, acceleration = 0
And when acceleration = 0 Driving Force = Resitive Force
So Driving Force = 600 N
600 = 24000/v
v= 40m/s

(ii)
When v= 15 m/s
Driving Force becomes 24000/15 = 1600N
Net Force = Driving Force - Resistive Force = 1600 - 600 = 1000 N
Net Force = mass x acceleration
1000 = 1250 x a
a = 0.8 m/s'2

Q4.
Normal = Mass x 10 cos theta
1.2 = 9.6m
m = 0.125 kg

(ii) Frictional component + 10 x mass x sin theta = mass (-acceleration (deceleration))
0.4 + (0.125) x (10) x (0.28) = 0.125 ( -a)
-a = 6
So decelration = 6 m/s'2

(iii) Frictonal component = 0.4
Weight's component = mg x sin theta = 0.35
since Frictional component is greater than weight's component. X remains at rest.

thnx alooooooooooooooooooooooooooooooooooooottttttttttttttttttttttttt

 

[ousamah112]

plz help in may.june 06 q5 and q7!

 

[Nibz]

plz help in may.june 06 q5 and q7!

Paper? Link? Nothing?

 

[ousamah112]

Paper? Link? Nothing?

here http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s06_qp_4.pdf

 

[Lyfroker]

For A:
Reaction = mass x 10
R = 2 N

T - uR = mass x acc.
T - (o.3 x 2) = 0.2a -> eq. 1

For B:
mass x 10 - T = mass x acc.
4.5 - T = 0.45 a -> eq. 2

Solve these simultaneously, to get a and T.
a= 6 m/s'2

For B:
v = ?
u = 0
a = 6 m/s'2
s = 1.3 m
v'2 - u'2 = 2as
v = 3.95 m/s

(ii)

When B hits the floor, there is no forward force acting on A. So the only force on it is the Frictional force.
Frictional Force = uR
uR = mass x deceleration
u (mass x 10 ) = mass x deceleration
both masses cancel out.
10 u = deceleration
10 x 0.3 = deceleration
3 m/s'2 = deceleration

so for A:
v= ?
a = -3
u = 3.95
s = 0.8 (2.8 - 2)

v'2 - u'2 = 2as
v = 3.29 m/s

thnk u so muchhhhhh

 

[Nibz]

No problem.

 

[Nibz]

here http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_4.pdf

Q.6 (i)
K.E at A= 1/2 x 50 x 7'2 = 1225
K.E at B = 1/2 x 50 x 3'2 = 225
Loss = 1225 - 225 = 1000J

(ii)
Gain in P.E = m x g x h = 50 x 10 x 15 = 7500

(iii)
Mechanical Energy at A = K.E + P.E = 1225 + 0 = 1225
M.E at B = 225 + 7500 = 7725
Change in M.E = 6500

Change in ME = Net Work Done
Net work done = Wd by Driving Force - Wd by against Resistance
6500 = x - 1500
x = 8000
W.d by pulling force = 8000 J

(iv)
Pulling force = 45N
Horizontal component = 45cos theta
Wd by pulling force = 45 cos theta x 200
from (iii) Wd by pulling F = 8000

45 cos theta = 8000
theta = 27.3

 

[ousamah112]

Q.6 (i)
K.E at A= 1/2 x 50 x 7'2 = 1225
K.E at B = 1/2 x 50 x 3'2 = 225
Loss = 1225 - 225 = 1000J

(ii)
Gain in P.E = m x g x h = 50 x 10 x 15 = 7500

(iii)
Mechanical Energy at A = K.E + P.E = 1225 + 0 = 1225
M.E at B = 225 + 7500 = 7725
Change in M.E = 6500

Change in ME = Net Work Done
Net work done = Wd by Driving Force - Wd by against Resistance
6500 = x - 1500
x = 8000
W.d by pulling force = 8000 J

(iv)
Pulling force = 45N
Horizontal component = 45cos theta
Wd by pulling force = 45 cos theta x 200
from (iii) Wd by pulling F = 8000

45 cos theta = 8000
theta = 27.3

thankx nd q5 n q7??

 

[Lyfroker]

Q : 1 nd 2 plz hlp

 

[Nibz]

thankx nd q5 n q7??

Q.7
v of P = 1.3
Distance covered = x

V of Q = 1.3
Distance covered = y

10x mass x sin theta = mass x acc.
a = 10 sin theta

s = ut + 1/2 at^2
For P
x = 1.3t + 1/2 (10 sin theta) t^2
x = 1.3t + 5t^2 sin theta

Q moves in opp. direction so it's a = -10 sin theta

y = 1.3t + 1/2 ( -10 sin theta ) t^2
y = 1.3t - 5t^2 sin theta

Add both and you will get d = 2.6t

(ii)
When t = 2.5
d = 2.6 x 2.5 = 6.5

sin theta = 1.6 / d = 1.6 / 6.5 = 0.246

since a = 10 sin theta
a = 10 x 0.246 = 2.46 m/s'2

(iii)
Q:
u= 0
at max height v = 0
a = -2.46
v = u + at
t = 0.53 s

now s = ut + 1/2 at^2
s = 1.3 (0.53) + 1/2 (2.46) (0.53)^2
s = 1.03 m

 

[Nibz]

thankx nd q5 n q7??

For Q.5 you won't be able to get it without a diagram.

I will write down my solution only.

4√2 = 2Tcos45
T = 4N

10 x mass = 4
mass = 4/10 = 0.4 kg

(ii)
T = uR
and R = mass x 10
so T = u x mass x 10
4 = 0.8 x 10 x mass
Mass = 0.5

(iii)

for Q 10 x mass - T = mass x acc.
5 - T = 0.5 a


P :
R = mass x 10
R = 5 N

T- uR = mass x acc
T - 4 = 0.5 a

Solve simultaneously , you will get T = 4.5 N

 

[muneexa]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_41.pdf​

​

Can anyone please solve Q.3? I don't understand what directions to take for the x and y components.​

 

[ousamah112]

For Q.5 you won't be able to get it without a diagram.

I will write down my solution only.

4√2 = 2Tcos45
T = 4N

10 x mass = 4
mass = 4/10 = 0.4 kg

(ii)
T = uR
and R = mass x 10
so T = u x mass x 10
4 = 0.8 x 10 x mass
Mass = 0.5

(iii)

for Q 10 x mass - T = mass x acc.
5 - T = 0.5 a


P :
R = mass x 10
R = 5 N

T- uR = mass x acc
T - 4 = 0.5 a

Solve simultaneously , you will get T = 4.5 N

didnt get that Tcos45 part...:S
btw thankx alot

 

[Nibz]

didnt get that Tcos45 part...:S
btw thankx alot

T cos 45 is the horizontal component of the tension acting on Q. The same horizontal component is also for P.
so 2 T cos 45 = Force exerted by the pulley.

 

[ousamah112]

T cos 45 is the horizontal component of the tension acting on Q. The same horizontal component is also for P.
so 2 T cos 45 = Force exerted by the pulley.

how??:S diagram ??

 

[Nibz]

Sorry. Don't have time.