Mechanics M1: Post your doubt here

[elbeyon]

Hii that's the attitude, I really appreciate.

yEah..tOtally agree brO..but dnt know if the paper will be as tough as the paper 1.:/
Inshallah everythings will be fine.
i've only 1 problem..zat is Timing..i usually gets out of time..can you just give me some advices brO
Jazakhallah!!

Thank you mates . TeKnOzOr timing is usually a problem for everyone. Try to be fast during your calculations but at the same time accurate too, otherwise its useless. Try to complete all the workings on time and make sure to maintain them neat and clean but if you just have few minutes left and a lot of questions to workout with then forget the above tips and just try to complete the works within time limit but make your that the one who is going to check your papers is at-least going to understand it. Something is better than nothing, isn't it ?

 

[GorgeousEyes]

Please explain Ju012 varient : 43 , no.7 with details , no.5 ii , no.4 ii
Thanks in advance

 

[Beca1206]

hey folks
when is ur mechanics exam??

 

[shikha95]

Why is the tension in M/J 2009 Q1 40N and not 50N?
And how do you use the formula R+T=W?

 

[Amy Bloom]

Nov11 p42 no6

WOW finally done it
try to understand in this mess lol

Yeah i understand them. except, i don't understand the formula u used in the 2nd image, first line, can u make it clearer to me? Thanks loads mate.

 

[Amy Bloom]

i have figured out how to do them do u still need help ?

Yeah I got the solution thanks iKhaled
Thanks to AhShun too!

 

[Amy Bloom]

Hi everyone, here are my doubts. I know the image appears scary, just i compiled all the doubts in one thing; only a few parts i require see the pink writings. I've included the answers for you to go faster. The paper was June 2011 P41

 

[Varuna0911]

Help me with ths pls..nov p41 Q7 all parts

 

[Varuna0911]

Help me with ths pls..nov p41 Q7 all parts

nov 2011 p41 Q7

 

[bamteck]

It is the force on pulley,i.e there is a force acting on the fixed pulley due to the tension in the string passing around the pulley....here, there is a downward force of 2T acting on the fixed pulley.due to the string and the attached loads.....T=4.55.....MAGNITUDE OF THE RESULTANT FORCE=2*4.55=9.1...HOPE IT HELPS....

Thanks

 

[AhShun]

nov 2011 p41 Q7

Nov11 p41 no7

umm... part(i) already done yesterday

 

[AhShun]

Hi everyone, here are my doubts. I know the image appears scary, just i compiled all the doubts in one thing; only a few parts i require see the pink writings. I've included the answers for you to go faster. The paper was June 2011 P41
View attachment 16662

Jun11 p41 no4

 

[AhShun]

Hi everyone, here are my doubts. I know the image appears scary, just i compiled all the doubts in one thing; only a few parts i require see the pink writings. I've included the answers for you to go faster. The paper was June 2011 P41
View attachment 16662

Jun11 p41 no5(ii),(iii)

 

[AhShun]

Hi everyone, here are my doubts. I know the image appears scary, just i compiled all the doubts in one thing; only a few parts i require see the pink writings. I've included the answers for you to go faster. The paper was June 2011 P41
View attachment 16662

Jun11 p41 no7(iii)

 

[Amy Bloom]

Jun11 p41 no7(iii)

Clear this for me please. I always get confused with this:
When B slacked, does the final velocity of B equal to the initial velocity of A? and why is 's' zero here?
thanks loads buddy, for everything. u're a boss!

 

[AhShun]

Clear this for me please. I always get confused with this:
When B slacked, does the final velocity of B equal to the initial velocity of A? and why is 's' zero here?
thanks loads buddy, for everything. u're a boss!

Consider particle B,
It's initial velocity was zero,u=0,
time to reach the ground=1.6
acceleration=2.5 <------- not free fall
final velocity when reaching ground= ???
v = u +at
v =4

When it is slacked, no tension in string anymore(it becomes a simple projectile motion)
Particle B should be going with same speed at any moment when still connected to A

umm.... s=0 because I wanted to calculate the time when displacement=0 . that is when particle returns to 'initial' position(see my small diagram)


"u're a boss! ," you are exagerrating a little LOL

 

[mehdi1028]

hey folks
when is ur mechanics exam??

Tomorrow ... 19th Oct

 

[Amy Bloom]

AhShun
Consider particle B,
It's initial velocity was zero,u=0,
time to reach the ground=1.6
acceleration=2.5 <------- not free fall
final velocity when reaching ground= ???
v = u +at
v =4 Ok fine.

When it is slacked, no tension in string anymore(it becomes a simple projectile motion) === okay so A will be going against gravity, -10ms-2, right?
Particle B should be going with same speed at any moment when still connected to A ==== so you mean to say that the initial velocity when B slacked (u) will be 4ms-1, right?

umm.... s=0 because I wanted to calculate the time when displacement=0 . that is when particle returns to 'initial' position(see my small diagram) ==== ok got it.


"u're a boss! ," you are exagerrating a little LOL ===humm no, I mean it, you always solve my doubts and give me clear explanations.

 

[AhShun]

Yes,you understand rapidly
if you take downward direction as positive, then acceleration can be positive,however the speed of A would be negative then

"so you mean to say that the initial velocity when B slacked (u) will be 4ms-1, right? " the final velocity of B would be the initial velocity of A

 

[Albert Einstein]

Hi could u help me wid the last part the answer is 0.5m