Stuck with Mechanics dynamics question help!

What do you mean by 5am? Are you talking about the time? Why r u awake at 5am then?

I did the a) part and the answer was 4.66ms^-1. I don't know how to do the b) part so anyone please help.

Using suvat equations.. and the acc u found in part a find the velocity of B when it reaches the ground(in other words its final velocity)
That final velocity will be the same as the speed of A initially then use suvat again to find v as A reaches the pully with the same acceleration

Do you want any links for good tutorials on this? its about 1 hour (maybe) but its worth watching

Ok that would be useful. Thanks for your help.

The key thing to understand is that the speed of the object on the right side of the pully as it hits the ground will be the same as the speed of a at that instant

I did that, but the answer is different from the book's answer.
My working:
B:u=0ms^-1
s= 0.6m
a= 4.66ms^-2
v=?

v²=u²+2as
v²=5.592
v= 2.36 (to 3s.f)

A:u=2.36ms^-1
s=1m
a=4.66ms^-2
v=?

v²=u²+2as
v²=14.912
v= 3.86 (to 3s.f)

But the answer in the book is 1.57ms^-1.

Your working for particle B is correct.

The surface is rough. Once particle B reaches the ground, it stops moving, and the string stops being taught. As a result, A no longer experiences any tension force from the string, as it is no longer taught.

A will have moved 0.6m and be travelling at 2.36m/s at this point (same magnitude of motion as B). Then A will no longer experience tension force, but will still experience friction force opposing its motion, as the table is rough. And A will have a further 0.4m to move before reaching P.

Find the friction force acting on A, then find the acceleration of A, and then apply that over a distance of 0.4m.

With this method I got an answer of 1.56m/s which is 0.01m/s off the textbook answer, probably due to a rounding error earlier on.

can I apply the value of v (particle B) for u(particle A) as I did earlier?

Yes, as particle A will be moving at the same velocity as particle B just before B hits the ground.

But a and s for particle A are different to what you used earlier, as at this point A will accelerate due to friction (opposing motion), not tension, and will have already travelled 0.6m (same distance B travelled) so will only have a further 0.4m to travel.

Yes, as particle A will be moving at the same velocity as particle B just before B hits the ground.

But a and s for particle A are different to what you used earlier, as at this point A will accelerate due to friction (opposing motion), not tension, and will have already travelled 0.6m (same distance B travelled) so will only have a further 0.4m to travel.

Can you show me your working, please? I couldn't do it.

Ok, sorry, I was out.

FULL SOLUTION:

Ok, sorry, I was out.

FULL SOLUTION:

Using suvat equations.. and the acc u found in part a find the velocity of B when it reaches the ground(in other words its final velocity)
That final velocity will be the same as the speed of A initially then use suvat again to find v as A reaches the pully with the same acceleration

I did that, but the answer is different from the book's answer.
My working:
B:u=0ms^-1
s= 0.6m
a= 4.66ms^-2
v=?

v²=u²+2as
v²=5.592
v= 2.36 (to 3s.f)

A:u=2.36ms^-1
s=1m
a=4.66ms^-2
v=?

v²=u²+2as
v²=14.912
v= 3.86 (to 3s.f)

But the answer in the book is 1.57ms^-1.