June 2010 Mechanics doubts

[arlery]

Q.4, 5 (ii), 7

http://www.xtremepapers.me/CIE/Internat ... _ms_43.pdf

 

[waleedsmz]

Question 4:

i) By resolution of the forces on the surface of the triangle on particle A, and by using the law D - R = ma
Driving force is the component of weight which is 0.36X 10 X sin60
Resisting force is the tension
mass is 0.36
acceleration is 0.25
0.36 X 10 X sin60 - T = 0.36 X 0.25
T= 3.03N

 

[waleedsmz]

Question 4:

ii) Again, by resolution of the forces on the surface of the triangle on particle B, and by using the law D - R = ma
Driving force would be the tension which is 3.03N
Resisting would be both the friction and the component of weight
mass = 0.24
a= 0.25

3.03 - ( F + 0.24 X 10 X sin60 ) = 0.24 X 0.25
F = 0.889

Another look on resolution on particle B:
Uprward force = downward force
Component of weight = Normal Reaction "R"
0.24 X 10 X cos60 = R
R= 1.2

Since F = COF X R
0.899 = COF X 1.2
COF= 0.74

 

[arlery]

Thanks, but what about the other questions?

 

[waleedsmz]

5) ii)
Alright since the ball hasn't left contact with the table and no new forces act on it, then it has the same deceleration

By using the rule V^2= U^2 + 2as on the distance BC
V = 0 ( comes to rest at C )
A = -0.25
S = 2
0 = U^2 + 2 X -0.25 X 2
U = 1

As for the time:
S = ( U + V ) / 2 X t
2 = ( 1 + 0 ) / 2 X t
t= 4

I'm sorry I'm quiet busy these days because I have mechanics and physics on the same day
I'll try to answer the third question as soon as possible.

Can you help me with my question though ? viewtopic.php?f=26&t=6780

 

[waleedsmz]

Question 7:

I really am not an expert in this question, but I'll do my best and hope that someone corrects me if I'm wrong.

i) Considering the two boxes one body, as the boxes starts to move when a force of 3150 N is supplied, then the resisting force is equal to 3150 N.
And because there could be only one resisting force which is friction, then friction is equal to 3150 N.
Normal reaction force would be equal to the weight of the two boxes.
COF X R = F
COF X 4500 = 3150
COF = 0.7

 

[waleedsmz]

ii) I don't really get that part so I can't answer it :/

iii) Since maximum acceleration is 2
D - R = ma
D - 3150 = 450 X 2
D = 4050

:/

Please help me with my question: viewtopic.php?f=26&t=6780

 

[arlery]

Yeah Im trying to solve that q fr u right nw. As soon as it's done, as I'll post a reply. Don't worry.

 

[arlery]

Here's how we solve part ii

Friction = 2000*0.2
400= ma
400/200=a
a= 2 m/s^2