Solving Formula Manipulation Problem: Block B Strikes Floor at t1

[Lucretius]

I am trying to manipulate a formula to get a known answer — only I am not getting it:

The necessary information: Two small blocks, each of mass m, are connected by a string of constant length 4h, with negligible mass. Block A is placed on a smooth tabletop (no friction), 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 T=0. Express all algebraic answers in terms of h, m, and g.

On to the problem I am stuck on: Block B strikes the floor and does not bounce. Determine the time t1 at which block B strikes the floor.

Now, the known answer is $$t_1=2\sqrt\frac{h}{g}$$

My attempted work goes as follows: I begin with the equation $d=\frac{1}{2}gt^2$ (falling body), getting $t^2$ by itself, I end up with the equation:

$$t^2=\frac{h}{.5g}$$

taking the square root of both sides then, I get

$$t=\sqrt\frac{h}{.5g}$$

If I got the steps right so far, I am completely at a loss as to what to do now.

 

[hotvette]

You are forgetting that the falling mass is pulling the mass on the table, thus your equation of motion of the falling mass is missing an upward force.

 

[Lucretius]

What would this upwards force be? Tension? Block B's falling acceleration is $\frac{g}{2}$

Instead of $d=\frac{1}{2}gt^2$, the formula would then become $d=\frac{1}{4}gt^2$ and later $t=\frac{h}{.5g}$right? I still don't know how the 2 ends up outside of the square root.

 

[hotvette]

I'm baffled. I came up with different answer than either of the ones you mentioned. Two questions:

1. How do you know the answer is $t_1=2\sqrt{\frac{h}{g}}$? I got $t_1=2\sqrt{\frac{2h}{g}}$. Could there be a typo in the answer?

2. Since this is a high school section, what concepts are you working with?
- Free body diagrams?
- Equations of motion? (e.g. F=ma)
- Energy equations? (e.g. sum of potential energy + kinetic energy, etc.)

Just want to know what level of dialog is appropriate.

 

[VietDao29]

hotvette, you are wrong. I think you should re-check your calculation. I think there's nothing wrong with the answer.
Lucretius, yes, block B's acceleration is g / 2. But there's a slight error in your arrangement:
$$\frac{1}{4}gt ^ 2 = h$$
$$\Leftrightarrow t ^ 2 = \frac{4h}{g}$$
$$\Leftrightarrow t = \pm \sqrt{\frac{4h}{g}} = \pm 2\sqrt{\frac{h}{g}}$$
Since t must be greater than 0. So you just choose:
$$t = 2\sqrt{\frac{h}{g}}$$
Viet Dao,

 

[hotvette]

Don't think so. I also got the acceleration = g/2. The answer is correct if the distance is h, which it isn't. The distance is 2h.

$$t_2 = \frac{4x}{g} = \frac{4*2h}{g} = \frac{8h}{g}$$

 

[VietDao29]

Don't think so. I also got the acceleration = g/2. The answer is correct if the distance is h, which it isn't. The distance is 2h.

Why can't the distance be h? You should re-read the problem.

... Block B is then released from rest at a distance h above the floor at T=0. Express all algebraic answers in terms of h, m, and g.

So is the answer the book provided right, then?
Viet Dao,

 

[hotvette]

Absolutely right, I missed it! Too late at night for me I guess. The right answer is indeed:

$$t = 2\sqrt{\frac{h}{g}}$$

So, what are the lessons here? I can think of 3:

1. Double check your math
2. Read the problem carefully
3. If you can, avoid working problems when you are tired