Mechanics M1: Post your doubt here

[Kirabo Takirambudde]

Please help with part (ii) i get part 1 but part (ii) makes no sense and neither does the marking scheme

 

[Gunner7]

Please help with part (ii) i get part 1 but part (ii) makes no sense and neither does the marking scheme

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)

 

[leadingguy]

can any one explain how the highlighted part is to be tackled down??? I am having problems in understanding it atm. thanku. I concerned the m,s bt it is totally differnt from my wa of attempting. here is the ms.

 

[iKhaled]

can any one explain how the highlighted part is to be tackled down??? I am having problems in understanding it atm. thanku. I concerned the m,s bt it is totally differnt from my wa of attempting. here is the ms.

this question is simple brother. i will explain two ways of solving it.

first way:

calculate the initial energy and the final energy and the difference of these two is the energy wasted on air resistance cuz uk energy is conserved, cant just disappear..

initial energy: mgh + 1/2mv^2
initial energy= 6.2(10)(6.2) + 1/2(0.6)(5.2)^2
initial energy = 45.31 J

final energy ( energy when it reached the ground) = mgh + 1/2mv^2
final energy= 0.6(10)(0) + 1/2(0.6)(12)^2
final energy = 43.2 J

see here u notice there is some energy missing so obviously thats the energy wasted on air resistance which is,
45.31- 43.2 = 2.11 J

second way:

calculate the work done by air resistance while the particle was moving upwards and the work done by air resistance while the particle was moving downward..
when the particle was moving upwards the air resistance was equal to:

F+R = ma
R = 0.6(10.4) -6
R = 0.24 N

WD while particle moving upwards,

WD = FD
WD = 0.24 X 1.3
WD = 0.312 J

when the particle was moving downwards the air resistance was equal to:

F - R = ma
6-R = (0.6)(9.6)
R = 0.24 N

WD while partile was moving downwards:

WD = FD
WD = 0.24 X 7.5
WD= 1.8 J

total work done,
WD = 1.8 + 0.312
WD = 2.11 J

 

[leadingguy]

this question is simple brother. i will explain two ways of solving it.

first way:

calculate the initial energy and the final energy and the difference of these two is the energy wasted on air resistance cuz uk energy is conserved, cant just disappear..

initial energy: mgh + 1/2mv^2
initial energy= 6.2(10)(6.2) + 1/2(0.6)(5.2)^2
initial energy = 45.31 J

final energy ( energy when it reached the ground) = mgh + 1/2mv^2
final energy= 0.6(10)(0) + 1/2(0.6)(12)^2
final energy = 43.2 J

see here u notice there is some energy missing so obviously thats the energy wasted on air resistance which is,
45.31- 43.2 = 2.11 J

second way:

calculate the work done by air resistance while the particle was moving upwards and the work done by air resistance while the particle was moving downward..
when the particle was moving upwards the air resistance was equal to:

F+R = ma
R = 0.6(10.4) -6
R = 0.24 N

WD while particle moving upwards,

WD = FD
WD = 0.24 X 1.3
WD = 0.312 J

when the particle was moving downwards the air resistance was equal to:

F - R = ma
6-R = (0.6)(9.6)
R = 0.24 N

WD while partile was moving downwards:

WD = FD
WD = 0.24 X 7.5
WD= 1.8 J

total work done,
WD = 1.8 + 0.312
WD = 2.11 J

I really appreciate ur involvement in the querry thanks a lot I was just making sum silly mistake so was nt ablr to come up with the correct ans. thnks

 

[iKhaled]

I really appreciate ur involvement in the querry thanks a lot I was just making sum silly mistake so was nt ablr to come up with the correct ans. thnks

no problem brother

 

[GorgeousEyes]

A car is travelling at a constant speed of 72 km/h and passes a stationary police car. The police car immediately gives chase, accelerating uniformly to reach a speed of 90 km/hr in 10 s and continues at this speed until he overtakes the other car. Find:
(a) the time taken by the police to catch up with the car,
(b) the distance travelled by the police car when this happens

 

[leadingguy]

A car is travelling at a constant speed of 72 km/h and passes a stationary police car. The police car immediately gives chase, accelerating uniformly to reach a speed of 90 km/hr in 10 s and continues at this speed until he overtakes the other car. Find:
(a) the time taken by the police to catch up with the car,
(b) the distance travelled by the police car when this happens

please post the ans as welll so then ur posts will b answered soon.

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

Please help me for no. 6 (ii) iKhaled

 

[GorgeousEyes]

please post the ans as welll so then ur posts will b answered soon.

a) t=25 s
b) 500 m

 

[doorman1995]

A car is travelling at a constant speed of 72 km/h and passes a stationary police car. The police car immediately gives chase, accelerating uniformly to reach a speed of 90 km/hr in 10 s and continues at this speed until he overtakes the other car. Find:
(a) the time taken by the police to catch up with the car,
(b) the distance travelled by the police car when this happens

well its 46s for past a and 1.025km for part b if u cnfrm answers we can proceed with explanation

 

[GorgeousEyes]

well its 46s for past a and 1.025km for part b if u cnfrm answers we can proceed with explanation

a) t=25 s
b) 500 m

 

[doorman1995]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

Please help me for no. 6 (ii) iKhaled

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

 

[iKhaled]

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

 

[leadingguy]

a) t=25 s
b) 500 m

this graph might help u....

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


now the speeds are 1500 and 1200 ms-1
uptil 10s the car has traelled 12000m and police car has travelled 7500m. means now the police car is 4500m behind.


the distance a normal car will travel will be S = t(1200) t= the time for covering distance "S"
now comming bac to police car it also needs to cover that S distance plus 4500m !!

so for police car (S+ 4500) = t (1500) t = time remains same as we are finding the time when both have covered same distance . take reference from graph.

now just find out T it will be 15 seconds.... means the total time is 15 = 10


FOR THE SECOND PART UR ANS IS WRONG

 

[GorgeousEyes]

this graph might help u....

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


now the speeds are 1500 and 1200 ms-1
uptil 10s the car has traelled 12000m and police car has travelled 7500m. means now the police car is 4500m behind.


the distance a normal car will travel will be S = t(1200) t= the time for covering distance "S"
now comming bac to police car it also needs to cover that S distance plus 4500m !!

so for police car (S+ 4500) = t (1500) t = time remains same as we are finding the time when both have covered same distance . take reference from graph.

now just find out T it will be 15 seconds.... means the total time is 15 = 10


FOR THE SECOND PART UR ANS IS WRONG

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 .

 

[GorgeousEyes]

An engine works at a steady rate to pump water , initially at rest , through a vertical height of 5 m and then discharges it at a speed of 8m/s through a pipe of cross section 10 cm square . at what rate is the engine working ? ( 1 cubic meter of water = 1000 kg )

 

[iKhaled]

An engine works at a steady rate to pump water , initially at rest , through a vertical height of 5 m and then discharges it at a speed of 8m/s through a pipe of cross section 10 cm square . at what rate is the engine working ? ( 1 cubic meter of water = 1000 kg )

can u post the answers to this question ?

 

[daviruss]

hey would you please help me to solve this question: http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_41.pdf number 4

 

[daviruss]

hey would u please help me with this question : http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_41.pdf number 4