M1 pulleys

Post a link to the full question or a picture of the question.

When B hits the ground there is no tension and the only acceleration on A is gravity (acting against the motion)
At max height
0 = u^2 -2x9.8xs
u^2 = 3g/5 (calculated correctly earlier)
0 = 3g/5 -2gs
s = 3/10 = 0.3
The string becomes taut when it comes back down again another 0.3, so the total distance travelled is 0.6m

It is good to try to visualise what is happening and have faith that what you have been taught about Newton's laws and suvat will get you to the right answer

2 particles A and B have masses 2m and 3m respectively. The particles are connected by a light inextensible string which passes over a light smooth fixed pulley. The system is held at rest with the string taut. The hanging parts of the string are vertical and A and B are above a horizontal plane. The system is released from rest.

Show the tension in the string immediately after the particles are released is $\dfrac{12}{5} mg$
^done

After descending 1.5m, B strikes the plane and is immediately brought to rest. In the subsequent motion, A doesn't reach the pulley.

Find the distance travelled by A between the instant when B strikes the plane and the instant when the string next becomes taut.
^^ so far i've tried this
(tried to work out the extra little bit the particle travels when the string becomes loose)
tried to calculate the point at which the string becomes loose. The speed

u=0 v=v s=1.5 $a=\dfrac{g}{5}$

$v^2 = u^2 +2as$

$v^2 =0+3 \times \dfrac{g}{5}$

$v^2 =\dfrac{3g}{5}$

$v=\sqrt \dfrac{3g}{5}$

then tired to recalculate acceleration since when the string is loose it changes

resolving vertically up f=ma
no tension anymore so 2mg=2ma
a=g

then i tried to calculate that little extra bit of distance on top of the 1.5m

$u=\sqrt \dfrac{3g}{5}$ s=s v=0 a=g

$v^2 = u^2 +2as$

$0=\dfrac{3g}{5} +19.6s$

which gives -0.3m which i'm sure isn't right help pls

@Zacken

Okay.



Agree.



Good plan.



Disagree.

Resolve vertically up. Gravity acts downwards.

So a = -g.



Fix this. You should get +0.3 m.

Then double it.

When B hits the ground there is no tension and the only acceleration on A is gravity (acting against the motion)
At max height
0 = u^2 -2x9.8xs
u^2 = 3g/5 (calculated correctly earlier)
0 = 3g/5 -2gs
s = 3/10 = 0.3
The string becomes taut when it comes back down again another 0.3, so the total distance travelled is 0.6m

It is good to try to visualise what is happening and have faith that what you have been taught about Newton's laws and suvat will get you to the right answer

Ah thanks

Do i double it because it moves up a little when the string goes loose then travels the same distance when it moves back down a little?

It moves up when the string is loose. Then moves back down the same amount before it reaches the point at which the string is taut again. Hence the total distance travelled is 0.6 m whilst the displacement is 0 metres.

I got that too! Are u sitting it as well on the 9th of June???

No I'm a maths teacher with some spare time this week.

No I'm a maths teacher with some spare time this week.

The C1 paper was not really weird. Some of the questions required students to rapidly manipulate fractions but nearly everything was bog standard C1 chapter by chapter.
C2 was similar.
M1 - my advice would be to look at and understand people in lift problems. These are the hardest ones to get your head around.

Always draw a good forces diagram, even if they are not explicitly asked for - it becomes easier to resolve parallel and perpendicular to the plane. If you are uncomfortable with sin and cos in your calculation, just calculate them and substitute into the equations (use 4 d.p) before solving simultaneously.
Two connected particles around a pulley separate into two equations for each particle.
Work on the things you cant do well.

x

for part C impulse=mv-mu

where m=0.5 and $v=\sqrt \dfrac{3g}{5}$ u=0

is that right?

Nope. impulse = 3m(v-u).

oh so i just m wrong then? everything else is right?

Eh, I guess.