Pulley and Mass question, helpl

[Oatmeal11]

Homework Statement

Two masses are connected by a light string passing over a light, frictionless pulley as in the figure below. The m1 = 5.30 kg object is released from rest at a point 4.00 m above the floor, where the m2 = 2.85 kg object rests.

a) Determine the speed of each object when the two pass each other.
b) Determine the speed of each object the moment the 5.3 kg mass hits the floor.
c) How much higher does the 2.85 kg mass travel after the 5.3 kg mass hits the floor?

Homework Equations

KE=(1/2)mv^2
F=ma
y=volt+.5at^2

The Attempt at a Solution

I treated it as one system and made a free body diagram. I inserted it into the equations, sum of F=ma and added up force of tension on string +force of gravitiy=a(m1+m2) I solved for acceleration then inserted that into the v^2=vo^2+2ax. Unfortunately the answers are all wrong and I'm not even sure i started this problem rigiht. Helpl??

 

[PeterO]

Homework Statement

Two masses are connected by a light string passing over a light, frictionless pulley as in the figure below. The m1 = 5.30 kg object is released from rest at a point 4.00 m above the floor, where the m2 = 2.85 kg object rests.

a) Determine the speed of each object when the two pass each other.
b) Determine the speed of each object the moment the 5.3 kg mass hits the floor.
c) How much higher does the 2.85 kg mass travel after the 5.3 kg mass hits the floor?

Homework Equations

KE=(1/2)mv^2
F=ma
y=volt+.5at^2

The Attempt at a Solution

I treated it as one system and made a free body diagram. I inserted it into the equations, sum of F=ma and added up force of tension on string +force of gravitiy=a(m1+m2) I solved for acceleration then inserted that into the v^2=vo^2+2ax. Unfortunately the answers are all wrong and I'm not even sure i started this problem rigiht. Helpl??

If you treat the masses and string as one system - with the pulley merely providing a convenient direction change - then Tension is an internal force and can be ignored.

The external forces are the weight of each mass, in its appropriate direction. So net force is Mg - mg.

So Mg - mg = (M + m).a

You can work out which mass is M and which mass is m

Once you have acceleration it sounded like you were doing the right thing.

EDIT: you can work the first two parts using conservation of energy - Gravitational Potential Energy and Kinetic Energy of the system.

 

[Oatmeal11]

Thanks for explaining the thing about tension! I used the equation Mg-mg=(m+M)a and got 2.94 for the acceleration, sticking that into the velocity equation,v^2=Vo^2+2ax; Vo=0, a= 2.94, h=4 ;so v^2=2(2.94)(4) the speed it has when the 5.3kg mass hits the floor is 4.85. However, I am not sure how to solve part A( speed when two masses pass each other) because I don't know what height it will be at.
Also part c is confusing because why would the other mass keep traveling after the 5.3kg mass hits the floor?

 

[PeterO]

Thanks for explaining the thing about tension! I used the equation Mg-mg=(m+M)a and got 2.94 for the acceleration, sticking that into the velocity equation,v^2=Vo^2+2ax; Vo=0, a= 2.94, h=4 ;so v^2=2(2.94)(4) the speed it has when the 5.3kg mass hits the floor is 4.85. However, I am not sure how to solve part A( speed when two masses pass each other) because I don't know what height it will be at.
Also part c is confusing because why would the other mass keep traveling after the 5.3kg mass hits the floor?

Firstly: if one mass started 4 m above the other, they will have traveled 4 m when the upper mass has hit the floor.
How far will they travel so that they are at the same height? One travels up, the other travels down.

Secondly: When the the 5.3kg mass hits the floor, why should it stop? - Because it has hit the floor of course
When the 5.3 kg mass hits the floor, why should the 2.85 kg mass stop? - because it has hit the ... anyone? ... anyone? .. Beuller? Oh that's right it hasn't hit anything, so it won't stop! Well not immediately, but even a bullet fired straight up into the air eventually stops and this mass is going a lot slower than a bullet out of a gun.

[I hope you understood the reference to the movie "Ferris Beuller's Day Off"]

 

[Oatmeal11]

So the two masses will pass each other at a height of two! Thanks!
However, I am not sure I still understand why the 2.85 kg will keep moving; a bullet keeps moving but it is not attatched to a string like the mass in this problem is. Wouldn;t the length of the string limit how high the mass can move?
p.s. I didn't understand the reference...sorrry...but I did LOL a little.(:

 

[gneill]

Pushing a rope is not nearly as effective as pulling it.

 

[Rayquesto]

This is for personal understanding and quite possibly yours, if a system in motion is affected by the same acceleration, will everything else in that system of motion contain the same change in distance? This is how I see this problem. If you know that, hen you could find out how much time it takes to contain the same position.

 

[PeterO]

So the two masses will pass each other at a height of two! Thanks!
However, I am not sure I still understand why the 2.85 kg will keep moving; a bullet keeps moving but it is not attatched to a string like the mass in this problem is. Wouldn;t the length of the string limit how high the mass can move?
p.s. I didn't understand the reference...sorrry...but I did LOL a little.(:

By the look of the diagram, the 2.85 would have to travel a further 4+ metres before the string went tight, or it hit the pulley.
I think you will find it will continue a much smaller distance than that.

[Why do I expect that?
I know the system will accelerate at a rate smaller than g! Why, because if the 5.3 kg mass was NOT attached to the 2.85 kg mass, it would accelerate at g. Because it is tied to the other mass, I know the acceleration will be less.
The acceleration takes place over 4 m, reaching some final speed.
Then the 2.85 decelerates at an acceleration equal to g.
At the lower rate it took 4m to build up the speed - at acceleration g it will take less than 4m to stop.
That is the way I see the situation - I use the formulas to get accurate figures to match that thinking]

 

[Oatmeal11]

I looked at the answer key and the correct answer to that last part is 1.2m. I am still not sure how that was found but your reasoning makes sense I guess. Thanks for all the help!(:

 

[Rayquesto]

for the last part, I would calculate the velocity during the time it takes for the other block to reach the ground, then multiply that time times the acceleration of the blocks, then I would use a kinematic equation to find the distance it goes with gravity and that velocity you would have calculated.

 

[PeterO]

I looked at the answer key and the correct answer to that last part is 1.2m. I am still not sure how that was found but your reasoning makes sense I guess. Thanks for all the help!(:

You correctly calculated a speed of 4.85 m/s for the masses at the point where the 5.3 reaches the ground.
The 5.3 immediately stops [hits the floor]
the 2.85 now becomes a body moving up, but under the influence of gravity.
[It would be just like you tossing up a catching a ball, and managing to give it an initial speed of 4.85 m/s up].

At maximum height, its velocity will be zero - if it wasn't it would be going higher!

You can best use:

v_f² = v_i² + 2.a.d

Using the figures you have calculated, you get an answer of 1.2 m.

You could use v_f = v_i + a.t and d = v_i.t + (0.5).a.t² but that requires you to work out the time on your way through.