Forces and tensions between masses check and help

[foonis]

Homework Statement

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

Homework Equations

∑Fb = ma = Fw - T
∑Fa = ma = T

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 = volt + .5at2 and Y - Yo = volt + .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 I am just not 100% positive.
Thanks!

 

[collinsmark]

Hello Foonis

Welcome to Physics Forums!

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

Sounds reasonable

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

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.

e) Determine the distance between the landing points of the two blocks.
- I used the kinematics equations X - Xo = volt + .5at2 and Y - Yo = volt + .5gt2
and found that the time from Block A leaving the table to hitting the floor was 2h/√(g)

Once again, how did h get out from under the square root?

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

I came up with a somewhat different answer.

 

[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?

 

[PhanthomJay]

Homework Statement

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

Homework Equations

∑Fb = ma = Fw - T
∑Fa = ma = T

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

good.

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

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?

e) Determine the distance between the landing points of the two blocks.
- I used the kinematics equations X - Xo = volt + .5at2 and Y - Yo = volt + .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

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.

 

[collinsmark]

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.

Your equation for the time agrees with mine.

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)

so far so good.

and by plugging that in my final answer for the distance between the blocks is now √h. How does all of this look now?

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?

 

[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+volt+.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 :)

 

[foonis]

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

 

[foonis]

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

Am I any closer?

 

[collinsmark]

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

Yes, that's the one.

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

'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.

 

[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?

 

[collinsmark]

'Looks right to me.

 

[foonis]

Thanks you helped a ton :D