# Forces and tensions between masses check and help

1. Sep 15, 2010

### foonis

1. The problem statement, all variables and given/known data
Two small blocks, each of mass m, are connected by a string of constant length (4h) and negligable mass. Block A is placed on a smooth tabletop as shown, and Block B hangs over the edge of the table. The tabletop is a distance (2h) above the floor. Block B is then released from rest at a distance (h) above the floor at time t=0. Express your algerbraic answers in terms of h, m, and g

2. Relevant equations
∑Fb = ma = Fw - T
∑Fa = ma = T

3. The attempt at a solution

a) Determine the acceleration of Block B as it descends.
- I came up with acceleration = gravity/2 because after finding the sum of the system of equations above I ended up with a(ma + mb) = mg(Fw) which simplifies to a=g/2

b) Block B strikes the floor and does not bounce. Determine the time it takes for block B to hit the floor.
- I used the equation 2h/g = t² to come up with 4h/(g/2) = t² which simplifies to
2h/√g = t

e) Determine the distance between the landing points of the two blocks.
- I used the kinematics equations X - Xo = Vot + .5at2 and Y - Yo = Vot + .5gt2
and found that the time from Block A leaving the table to hitting the floor was 2h/√(g) and by using that in the equation for movement along the x-axis I found that the distance between the two blocks after landing = h

How does all of this look to you? I feel fairly confident about it im just not 100% positive.
Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Sep 15, 2010

### collinsmark

Hello Foonis

Welcome to Physics Forums!
Sounds reasonable
Something is not right with the above, but I'm not quite sure what. How did the h get out from under the square root? Also, don't forget, Block B is not dropped from a height of 2h, it is dropped from a height of h.
Once again, how did h get out from under the square root?
I came up with a somewhat different answer.

3. Sep 15, 2010

### foonis

Oh, you're right, I didnt notice that and ran with it. So I plugged it all in again and I got b) 2h/g=t^2 as the base equation to get 2h/(g/2)=t^2 which should simplify to
2√(h/g)=t.
e) I fixed it to end up with the time between the block(a) leaving the table and hitting the floor as 2√(h/g) and by plugging that in my final answer for the distance between the blocks is now √h. How does all of this look now?

4. Sep 15, 2010

### PhanthomJay

good.
you have a math error in this equation, but the equation itself is not correct. Why did you use g when you already showed that a=g/2?
This is the wrong approach. When the first block hits the floor, the tension force becomes 0 (the string goes slack), but the block on the table continues to move, and then this becomes a parabolic motion problem as the 2nd block leaves the table and lands some distance away from the 1st block.

5. Sep 15, 2010

### collinsmark

Your equation for the time agrees with mine.
so far so good.
Oooh. Something doesn't seem right to me. You'll need to show your work for this one. How did you find the x-component of Block A's velocity?

6. Sep 15, 2010

### foonis

Wow this is a tough one haha hmm..

Well I used a kinematics equation to find the velocity as the block slides along the table after the other block hits the ground, which I set up like:
X=Xo+VoT+.5at^2, which i substituded the values to get:
X=0+(0)(2√(h/g))+.5(g/2)(2√(h/g))^2
so
X=.5(g/2)(2√(h/g))^2
so
X=.5(g/2)(4h/g)
so
X=(g/2)(2h/g)
so
X=2h/2
so
X=h?

Im extremely confused about this last one..

Thanks for all the help though, it is really saving me :)

7. Sep 15, 2010

### foonis

wait could I use the equation Vf = Vi + at ?
im going to try that and see what i get

8. Sep 15, 2010

### foonis

with that equation I got the velocity of Block A to be (g√h)/(√g)

Am I any closer?

9. Sep 15, 2010

### collinsmark

Yes, that's the one.
'Looks good
Now you know the velocity in the x-direction (which is constant when the block flies off the table) and the amount of time it's in the air. Solve for distance.

10. Sep 15, 2010

### foonis

so then i could just use distance=velocity x time
so
x=((g√h)/(√g))(2√(h/g))
so
x=2gh/g
so
x=2h

So the distance between the blocks after they land is 2h?

11. Sep 15, 2010

### collinsmark

'Looks right to me.

12. Sep 16, 2010

### foonis

Thanks you helped a ton :D

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