Mechanics (brother of physics)

[AlishaK]

Salaam and Hey !

Well, the title says it all. I am preparing for Mechanics P4 so, I thought we can share some problems and solutions. Bring it on people.

Best of luck!

 

[AlishaK]

Can anyone help me with Q6 and Q7 ii C, Even though i did both the questions, I'm still confused. Ca anyone explain me a bit. http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_42.pdf

Cheers.
Jazak Allah khair.

 

[00tanveer]

question 6
i. work done by driving force - work done against resistance to motion = change in kinetic energy + change in gravitational potential energy (learn this by heart for this exam)
OK, so, work done by driving force = 1200000 J ; work done against resistance = 1240X (where X is the distance AB )
change in kinetic energy = (1/2 x 16000 x 12^2) - (1/2 x 16000 x 15^2) = -648000 J
change in GPE = + (mgh) (You should mind signs coz cause they matter in calculations of energy change)
= + (16000 x 10 x Xsin(theta) ) here, h= Xsin(theta) (You should be able to get this trigonometry that I did here)
= +160000Xsin(theta)
So, 1200000 - 1240X = -648000 + 160000Xsin(theta)
solve this equation to find X. The value of sin(theta) is quoted.

ii. change in GPE = 2400000 J
=> mgh = 2400000
=> 16000 x 10 x BDsin(theta) = 2400000 once again the value of sin(theta) is known. So find BD.

iii. You solve this question the same way I have just done in question i. The difference is that the resistance are different in BC and CD. My hint will be to take BC = Y and CD = BD - Y and solve for Y coz you already know the value of BD.

There you go. I hope you understand all of this. If you are confused at any point knock me in a conversation. Cheers!

 

[00tanveer]

Oh and question 7 ii.c

v1−v=0.00005(t−200)^2−1
so if you put t = 0, you get v1 - v = 1
and for t = 400, you get v1 - v = -1
remember v1 - v is a quadratic function, so in the range 0<= t <= 400, -1 is a minimum value and this is no other minimum in between, and +1 is a maximum value
of v1 - v because there is no other maximum in between.
therefore the range of v1 - v is the inequality that you had to prove.

 

[Manobilly]

41/0/N 2009 q 4

 

[AlishaK]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Q7

 

[AlishaK]

41/0/N 2009 q 4

lemme check it out.

 

[AlishaK]

41/0/N 2009 q 4

it is pretty simple...
i) You need to resolve each forces i.e; of 40 cm string and 30 cm string, and to do that u need to find the angle that it makes with the horizontal which is the same as the angle that each string makes with the rod. So, as all the sides are given to me, im using Sine rule. 50/sin90=30/sinx...solving this gives us...x=36.9 deg (1dp) now you can find the other angle by simply subtracting 90-36.9=53.1 (as it's a right-angle triangle)...Taking the tension on string A as T1 and on string B as T2...we know that the sum of the vertical components of the two forces will be equal to the weight of P i.e; 5N. therefore, forming an eqn: T1 sin53.1 + T2 sin36.9 = 5
0.800T1 + 0.600 T2 = 5------eqn 1
Now for the horizontal comps:
T1cos53.1 = T2 cos 36.9
T1 = T2 0.800/cos53.1
T1=1.33T2-----eqn 2

Subs eqn 2 in eqn 1..
0.800*1.33T2 + 0.600 T2 = 5
Solving this.....gives us T2=3 N.
There u go!

 

[AlishaK]

for ii) As the frictional force of S is in the direction of the horizontal component of string B....hence, Friction force= T2cos36.9= 3 cos 36.9 = 2.4 N.

iii) using the formula Coeff of fric= friction force/ reaction force,
0.75= 2.4/x, x=3.2 N
Equating, Reaction force = W + 3 sin 36.9 (the vertical comp of the string the ring is attached to)
3.2= W +1.80, W=1.4 N

Hope I was of any help! Best of luck...and do pray for me.

 

[Manobilly]

Y

for ii) As the frictional force of S is in the direction of the horizontal component of string B....hence, Friction force= T2cos36.9= 3 cos 36.9 = 2.4 N.

iii) using the formula Coeff of fric= friction force/ reaction force,
0.75= 2.4/x, x=3.2 N
Equating, Reaction force = W + 3 sin 36.9 (the vertical comp of the string the ring is attached to)
3.2= W +1.80, W=1.4 N

Hope I was of any help! Best of luck...and do pray for me.

yes JazakAllah Alhamdulillah got it InshAllah pray for me too

 

[Manobilly]

I have another question 41/oct 2010 q3

 

[Adorkableme]

Guys please help for Q5 (a) from Oct/Nov 2012 .... http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_42.pdf
I got to the part of calculating the acceleration, its the next part that has left me confused....not sure please help!!!

 

[AlishaK]

Guys please help for Q5 (a) from Oct/Nov 2012 .... http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_42.pdf
I got to the part of calculating the acceleration, its the next part that has left me confused....not sure please help!!!

Woot! I'm doing the same paper.
Hopefully, you got the acceleration as -1m/s^2. Now we gotta use newtons second law or simply F=ma. And to do that, we need to know the forces acting on the particle and as we're supposed to find coeff of fric, it's pretty obvio that we need to find the Friction force and the Normal Reaction force first. The weight of the particle would b 'mg'. now resolving it with one component parallel to the plane i.e; mgsin30 (theta from i), and the other component perpendicular to the inclined plane i.e; mgcos30. Therefore, the Normal Reaction force (N) is equal to the perpendicular component that is 'N=mgcos30',
Coeff of fric=Friction force(F)/N; now as the particle slides down, we know that the force (parallel component from weight) acting downwards must be more than the friction force acting upwards. so we can form an eqn of 'mgsin30-F=m*-1 (this nothing but F=ma) where F=coeff of fric*N, F=Coeff of fric* mgcos30, hence, subs F in the previous eqn: mgsin30-Kmgcos30=-m (K is coeff of friction here) solve it now. cancel all the m's. final step wd be: 5-8.66K=-1, 8.66K=6, K=0.693 (3 sfig).

Hopefully I was clear enough! Best of luck! Remember me in ur prayers.

 

[AlishaK]

I have another question 41/oct 2010 q3

if u havn't got it yet, please give me sometime, i'll get back at that one!
Cheerio!

 

[Mayedah]

Oh and question 7 ii.c

v1−v=0.00005(t−200)^2−1
so if you put t = 0, you get v1 - v = 1
and for t = 400, you get v1 - v = -1
remember v1 - v is a quadratic function, so in the range 0<= t <= 400, -1 is a minimum value and this is no other minimum in between, and +1 is a maximum value
of v1 - v because there is no other maximum in between.
therefore the range of v1 - v is the inequality that you had to prove.

Can u explain to me its first part ? where do i get -1 from ?

 

[AlishaK]

Can u explain to me its first part ? where do i get -1 from ?

Sure,
v=(0.04t-0.00005t^2), V1= u+at => V1=1+0.02t
Hence, V1-V=(1+0.02t)-(0.04t-0.00005t^2)...solving this.....when u reach, 0.00005t^2-0.02t+1 => Use complete the square method.(from ur AS maths)
0.00005(t^2 - 400t)+1 (I just divided the first two terms by 0.00005 excluding the '+1')
0.00005[(t^2-400) +200^2 - 200^2]+1
0.00005[(t-200)^2 - 40000]+1
0.00005(t-200)^2 - 0.00005*40000 +1
0.00005(t-200)^2-2+1
Therefore, 0.00005(t-200)^2 - 1.

Hopefully u understand. Make sure to revise Completing the square method from P1.
Best of luck!

 

[VinnCh]

Woot! I'm doing the same paper.
Hopefully, you got the acceleration as -1m/s^2. Now we gotta use newtons second law or simply F=ma. And to do that, we need to know the forces acting on the particle and as we're supposed to find coeff of fric, it's pretty obvio that we need to find the Friction force and the Normal Reaction force first. The weight of the particle would b 'mg'. now resolving it with one component parallel to the plane i.e; mgsin30 (theta from i), and the other component perpendicular to the inclined plane i.e; mgcos30. Therefore, the Normal Reaction force (N) is equal to the perpendicular component that is 'N=mgcos30',
Coeff of fric=Friction force(F)/N; now as the particle slides down, we know that the force (parallel component from weight) acting downwards must be more than the friction force acting upwards. so we can form an eqn of 'mgsin30-F=m*-1 (this nothing but F=ma) where F=coeff of fric*N, F=Coeff of fric* mgcos30, hence, subs F in the previous eqn: mgsin30-Kmgcos30=-m (K is coeff of friction here) solve it now. cancel all the m's. final step wd be: 5-8.66K=-1, 8.66K=6, K=0.693 (3 sfig).

Hopefully I was clear enough! Best of luck! Remember me in ur prayers.

Strangely I know how to solve the question but have no single idea on how to find the acceleration. Mind teaching me?

 

[AlishaK]

Strangely I know how to solve the question but have no single idea on how to find the acceleration. Mind teaching me?

They have given us 4.8 secs the new time. Now it took 0.8 already, so we subtract, that is 4.8-0.8=4 sec
Therefore a=0-4/4 => -1 m/s^2
Hope u got it now? tell me if you didn't.
Cheers mate!

 

[VinnCh]

They have given us 4.8 secs the new time. Now it took 0.8 already, so we subtract, that is 4.8-0.8=4 sec
Therefore a=0-4/4 => -1 m/s^2
Hope u got it now? tell me if you didn't.
Cheers mate!

Oh. I just had a moment of realisation. You mean a=0-4/4 is actually a=(v-u)/t ? My god. How can I even forget that. Thank you so so much!

 

[AlishaK]

Oh. I just had a moment of realisation. You mean a=0-4/4 is actually a=(v-u)/t ? My god. How can I even forget that. Thank you so so much!

Yes, that's the formula. It's okay..happens.