Paper 42 guyss !!!!!!!

[MazenAlosali]

How did you guys do in paper 42 especially question 6 ??

 

[12rider21]

q6:how many marks was it?

 

[Asad rehman]

q6:how many marks was it?

3+3+3

 

[Asad rehman]

How did you guys do in paper 42 especially question 6 ??

only its 'a' part others didn't part get in time.

 

[12rider21]

what about the last question?

 

[HamzehAA]

which paper are you talking about

 

[Youssef Tawil]

which paper are you talking about

OCTOBER/NOVEMBER 2014 mechanics paper 42

 

[MazenAlosali]

Do anyone have answers ??

 

[Abdel Moniem]

Question 1i) you can use two methods for particle P u=11m/s a=-10m/s v=0m/s at maximum height. use v-u/t=a at=v-u t=v-u/a=-11/-10=1.1 seconds*2 to reach the ground. 2nd method set displacement to zero and use s=ut+1/2at^2=11t-5t^2=t(5-11t) t=11/5 seconds ii) Particle Q givens t=2.2 seconds a=10m/s^2 u=0m/s. use s=ut+1/2at^2=0(2.2)+1/2(10)(2.2)=24.2m. use s=(v+u/2)*t=v/2*2.2=24.2 v=22m/s

Q2ai) X components: +25*0.96-30*0.8=0
Y component*0.28+30*0.6-20=5
R=√(ΣFx)^2+(ΣFy)^2)=√0+25=5Newtons
If the force of 20N is removed the resultant would be 25 N. If the resultant is still the same as 5 N but in the negative y direction then the force replaced by 20 N is 30N

Q3ai Power at B=1.2*P at A
Driving force at B=0.96 *D at A
Initial Velocity=28m/s
Power=Driving Force*Velocity
P/P*1.2=D/D*0.96*28/V
1/1.2=1/0.96*28/V
1/1.2=28/0.96V
V=35m/s
3aii) Work done by train=Gain in K.E+Work done against Resistance.
=1/2(200,000)(35^2-28^2)+2.3*10^6

 

[Abdel Moniem]

4i) Forceof 40 N makes an angle of 30 with the vertical.
Block B mass=15kg
Resolve Vertically=R+40cos30+Xsin30=150
Resolve horizontally: 40sin30=Xcos30
20=√3 x/2
40/√3=x
ii) R=-40cos30-40/√3(sin30+150
R=103.8119
Ff=μR
10=103.8119μ
μ=0.096

5ai R=0.3g=3N
Ff=μR=0.7*3=2.1N
Work done=Frictional Force*Distance moved=2.1*0.9=1.89 J
5bithe potential energy of mass B doesn't change, so we can ignore it the change in potential energy of mass A will depend on it's change in height. its 0.9
Loss in P.E=mgh(final)-mgh(inital)=mg(hfinal-hinitial)=0.3*10*(0-0.9)=-2.7 J
5ciUse gain in K.E=Loss in P.E-Work done against Resistance.
=2.7-1.89=0.81
you just calculated the kinetic energy of the two mass system, and they are of equal mass moving at the same speed, so each has half of the kinetic energy.
0.405=1/2(0.3)(v)^2
v=√2.7
As the string breaks the particle is now under the effect of gravity a=10m/s^2 s=0.54 u=√2.7 use v^2=u^2+2as
v=√2.7+2(10*0.54)=3.67 m/s to reach the ground

 

[Abdel Moniem]

6ai) From motion A to B
R=mgcosα=9.6m mgsinα=ma 10*0.28=a a=2.8m/s^2
From motion B to C, the particle experiences some friction.
R=mgcosα mgsinα-Ff=ma
Ff=μR
10*0.28m-0.5(9.6m)=ma
-2m=ma
a=-2m/s^2
To find the distance AB lets add our givens a=2.8m/s^2 u=0m/s
use v^2=u^2+2as
v^2=2(2.8*s) v^2=5.6s
now we know that the final speed when reaching to point C is 2 m/s.
use v^2=u^2+2as
4=5.6s+2(-2*s) 4=1.6s s=2.5
From part A to B s=2.5 m u=0m/s a=2.8 m/s^2 use s=ut+1/2at^2 2.5=1/2(2.8)(t)^2, 2.5=1.4t^2 t=1.336
From part B to C s=2.5 m v=2 m/s a=-2 m/s^2 use s=vt-1/2at^2 2.5=2t-(1/2*-2*t^2) 2.5=2t+t^2 t^2+2t-2.5 t=0.8708
Add both times to get the time taken for P to move from A to C.

 

[Abdel Moniem]

7) v=-0.2(3)^2+4(3)-15=-4.8m/s
Use the following points: (0,0) (3,-4.8) a=Change in Y/Change in X=-4.8/3=-1.6m/s^2
at maximum velocity acceleration=0 m/s^2
Integrate with respect to t and set it to 0
-0.4t+4=0
t=10 seconds plug t=10 seconds into v=-0.2(10)^2+4(10)-15=5 m/s
Average Velocity=Total Distance Travelled/Total time taken
Use the area under graph to find the distance from t=0 to t=3 and then integrate v to find the distance and set your limits from 5 to 3
s=-0.2t^3/3+4t^2/2-15t
-0.2(5)^3/3+4(5)^2/2-15(5)--0.2(3)^3/3+4(3)^2/2-15(3)=-4.53333 now add it to the area from A to B 1/2*-4.8*3=-7.2
-7.2-4.53333=-11.733333
-11.73333/5=-2.34
for the last part integrate and set your limits from 15 to 5
(-0.2(15)^3/3) +4(15)^2/2-15(15)+0.2(5)^3/3-4(5)^2/2+15(5)=33.333 add it to the distance from time t=0 to t=5 which is 11.733
11.7333+33.333/15=3.00m/s

 

[salma saher]

q6:how many marks was it?

9 marks 3 for each i ii and iii

7) v=-0.2(3)^2+4(3)-15=-4.8m/s
Use the following points: (0,0) (3,-4.8) a=Change in Y/Change in X=-4.8/3=-1.6m/s^2
at maximum velocity acceleration=0 m/s^2
Integrate with respect to t and set it to 0
-0.4t+4=0
t=10 seconds plug t=10 seconds into v=-0.2(10)^2+4(10)-15=5 m/s
Average Velocity=Total Distance Travelled/Total time taken
Use the area under graph to find the distance from t=0 to t=3 and then integrate v to find the distance and set your limits from 5 to 3
s=-0.2t^3/3+4t^2/2-15t
-0.2(5)^3/3+4(5)^2/2-15(5)--0.2(3)^3/3+4(3)^2/2-15(3)=-4.53333 now add it to the area from A to B 1/2*-4.8*3=-7.2
-7.2-4.53333=-11.733333
-11.73333/5=-2.34
for the last part integrate and set your limits from 15 to 5
(-0.2(15)^3/3) +4(15)^2/2-15(15)+0.2(5)^3/3-4(5)^2/2+15(5)=33.333 add it to the distance from time t=0 to t=5 which is 11.733
11.7333+33.333/15=3.00m/s

ma3lesh enta gebt el 2s2ella men feen wla enta lsssa fakerhom ??

 

[Youssef Tawil]

9 marks 3 for each i ii and iii

ma3lesh enta gebt el 2s2ella men feen wla enta lsssa fakerhom ??

justpastpapers.com

 

[abdulkhan]

4i) Forceof 40 N makes an angle of 30 with the vertical.
Block B mass=15kg
Resolve Vertically=R+40cos30+Xsin30=150
Resolve horizontally: 40sin30=Xcos30
20=√3 x/2
40/√3=x
ii) R=-40cos30-40/√3(sin30+150
R=103.8119
Ff=μR
10=103.8119μ
μ=0.096

5ai R=0.3g=3N
Ff=μR=0.7*3=2.1N
Work done=Frictional Force*Distance moved=2.1*0.9=1.89 J
5bithe potential energy of mass B doesn't change, so we can ignore it the change in potential energy of mass A will depend on it's change in height. its 0.9
Loss in P.E=mgh(final)-mgh(inital)=mg(hfinal-hinitial)=0.3*10*(0-0.9)=-2.7 J
5ciUse gain in K.E=Loss in P.E-Work done against Resistance.
=2.7-1.89=0.81
you just calculated the kinetic energy of the two mass system, and they are of equal mass moving at the same speed, so each has half of the kinetic energy.
0.405=1/2(0.3)(v)^2
v=√2.7
As the string breaks the particle is now under the effect of gravity a=10m/s^2 s=0.54 u=√2.7 use v^2=u^2+2as
v=√2.7+2(10*0.54)=3.67 m/s to reach the ground

4ii U =0.107