Mechanics maths question help

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NewAccount

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Wasn’t sure where to post this sorry, but I don’t understand how you’re meant to calculate the second part0DDAD088-071F-4CBE-A56A-E2AD959EB795.jpeg
 
I don’t have any working for it, I think SUVAT is meant to be used but nothing I do works correctly
 
You "don't have any working" means that you haven't even tried! Surely you can write down the force on the lower particle? The pulley does not add any thing to the total force, just redirects it from vertical to horizontal. What does the word "smooth" in "smooth horizontal table" tell you?

Notice that the upper particle will still be moving after the first particle hits the ground and stops moving.
 
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NewAccount said:
I don’t have any working for it, I think SUVAT is meant to be used but nothing I do works correctly
Click to expand...
What you said is self-contradictory! Clearly you have done some work.

Believe it or not, showing us things you've done that didn't work can help us to help you. Please do so. At the very least, tell us what SUVAT means, and how you applied it.
 
S - displacement
U - initial velocity
V - final velocity
A - acceleration
T - time
There’s a set of equations using these letters for some mechanics questions
So to get the time taken for the mass m to hit the ground would be:
s - 0.5
u -
v - 0
a - -3.27
t -
s = vt - 1/2at^2
0.5 = - 1/2 x -3.27 x t^2
0.5 = 1.635t^2
0.5/1.635 = t^2
0.305.. = t^2
0.553 = t
t = 0.553
But then I don’t know how to get the rest of the time taken for mass 2m to hit the pulley after mass m hits the ground
Sorry I don’t know how to explain this well
 
NewAccount said:
But then I don’t know how to get the rest of the time taken for mass 2m to hit the pulley after mass m hits the ground
Sorry I don’t know how to explain this well
Click to expand...

Imagine pulling something on ice with a rope, even if you stop moving that thing will still moves toward you.... in other words the rope is not rigid and is not going to be in the way of moving object here... so here the rope is just gonna fell on top of the object”m”
So let’s go step by step
At the time 0, what are the forces acting on our objects (lets called them “2m” and “m”)?
What about when the “m” hits the ground? Do we have any forces acting on either of them?
 
At time zero I assume gravity and tension are acting on m, and tension is acting on 2m
When m hits the ground nothing is acting on 2m as the tension is zero though
 
NewAccount said:
Wasn’t sure where to post this sorry, but I don’t understand how you’re meant to calculate the second partView attachment 13408
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Did you draw the FBDs as I had suggested?

That is the first step (almost always) for most of the college mechanics problems.

Draw those FBDs and show us!
 
How do I draw an FBD for after ‘m’ hits the ground? There are no forces acting on 2m
 
NewAccount said:
At time zero I assume gravity and tension are acting on m, and tension is acting on 2m
When m hits the ground nothing is acting on 2m as the tension is zero though
Click to expand...
Perfect.... the rope is just transferring the gravity force on “m” to “2m”.... so at the beginning we only have gravity as force ... now you can calculate the speed of both m and 2m until m hits the ground
Then after m hits the ground there is no force.... Newton said something about not having any force on an object, remember that for solving the time that the 2m reaches the end
 
NewAccount said:
How do I draw an FBD for after ‘m’ hits the ground? There are no forces acting on 2m
Click to expand...
You draw FBDs prior to "hitting" the ground:

It will be used to calculate the acceleration and velocity of particle "2m".

Then you will use those numbers to finish rest of the problem.
 
I got that the acceleration is 3.27(rounded)
and I think the velocity is 1.808 but what then?
 
NewAccount said:
I got that the acceleration is 3.27(rounded)
and I think the velocity is 1.808 but what then?
Click to expand...
Now particle "2m" has to move (1-0.5 = ) 0.5 m before it hits the pulley - with an initial velocity of vi and acceleration "a=0".

So continue.....
 
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S - 0.5
U - 1.808
V - 0
A - ?
T - ?
S = 1/2(u + v) t
0.5 = 1/2(1.808)t
0.5 = 0.904t
0.5/0904 = t
0.553 = t
t = 0.553

0.553 + 0.553 = 1.106 Total time
But that doesn’t give the correct answer?
 
since there is no force on 2m anymore it will keep its velocity (no force—-> constant velocity) until it hits pulley
 
Oh so v = 1.808 not 0
I thought it’d be zero because it stops when it hits the pulley
 
Ok answer this question...when we let object m to fall, is there anything that is preventing m from free falling?
 
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