Mechanics Question

[Dania]

Okay, so I did this question and checked my answer close to a million times and still found my tension to be 44.4N whilst the mark scheme says 40N. Can someone please help me here?

Here is the question below, and the extract from the mark scheme.

[S.M.A.T]

Here

[Dania]

Oh wow, I just made myself look like an idiot. Thank you so much

[S.M.A.T]

lol .........sometime this happens to me

U welcome

[Dania]

Thanks for the rep, By the way.

[Dania]

Help me, Asif

[S.M.A.T]

Here Dani

[Dania]

Thank You, Asif. You help me so much.

I have another question. Take a look at the attached file, thanks

P.S Don't bother yourself answering (i) for me. I'm only interested in (ii), but you'll have to do the whole question to effectively help me out.

[nid404]

The frictional force provides forward acceleration since there is no more tension(once block B is on the ground)

[S.M.A.T]

The frictional force provides forward acceleration since there is no more tension(once block B is on the ground)

Garfield u are wrong  .....frictional force is an opposing force so it always try to decelerate an object

When particle B is about to hit the floor the speed of A is also 3.95ms^-2,this is because the acceleration of A&B are same before B hit the ground and they both travel the same distance(1.3m).
Once particle B is on the ground the tension in the string become zero and the only force acting on A is frictional force and it is decelerating particle A.
therefore

u=3.95ms^-1

frictional force=ma
-0.06g=0.2a
a=-0.3g
s=2.1-1.3=0.8m(1.3m is already travelled by A when B was about to hit the ground)

(v^2)=(u^2)+2as
(v^2)=(3.95^2)+(2*-0.3g*0.8 )
v=3.29

[thecandydoll]

I feel so upset screwing up my p1 math.
i think i land up with 54/75 or something


so  doing well in mechanics,is there a chance i can get a b?
I have a doubt.
Mechanics Paper 4.M/J 2003.
q2

[elemis]

First we draw a set of axis. Then we place the forces on them as shown in the first pciture.

As they are asking for the resultant force in the direction PQ we must resolve for forces in the y axis.

i.e. : 10 cos 30 + 10 cos 30 - 6 cos 60= 14.3 N

The direction 'perpendicular to PQ' is basically the x axis. So we resolve forces in the x axis.

10 cos 60 + 6 cos 30 - 10 cos 60 = 5.20 N

For the last part we know the magnitude of the horizontal and vertical forces. The resultant is the red line. (see 2nd pic)

Using Pythagoras Theorem : 5.2² + 14.3 ² = 231.53

Hence, = 15.2 N (3 s.f.)

[nid404]

Garfield u are wrong  .....frictional force is an opposing force so it always try to decelerate an object

negative acceleration...my bad -_-

[Deadly_king]

I feel so upset screwing up my p1 math.
i think i land up with 54/75 or something


so  doing well in mechanics,is there a chance i can get a b?
I have a doubt.
Mechanics Paper 4.M/J 2003.
q2

Hmm........I guess it is possible

You need to catch up in p4 though!

[thecandydoll]

Where does the 6cos60 come from
OMGGGG I DONT GETTT IT.

brain has shutdown.

First we draw a set of axis. Then we place the forces on them as shown in the first pciture.

As they are asking for the resultant force in the direction PQ we must resolve for forces in the y axis.

i.e. : 10 cos 30 + 10 cos 30 - 6 cos 60= 14.3 N

The direction 'perpendicular to PQ' is basically the x axis. So we resolve forces in the x axis.

10 cos 60 + 6 cos 30 - 10 cos 60 = 5.20 N

For the last part we know the magnitude of the horizontal and vertical forces. The resultant is the red line. (see 2nd pic)

Using Pythagoras Theorem : 5.2² + 14.3 ² = 231.53

Hence, = 15.2 N (3 s.f.)