## Help with #57 and #72

### Help with #57 and #72

Could anyone please explain the answers to #57 and #75 on the GRE 0177?

The answer is pv^2A.

72. Two identical blocks are connected by a spring. The combination is suspended, at rest, from a string attached to the ceiling, as shown in the figure. The string breaks suddenly. Immeadiately after the string breaks, what is the downward acceleration of the upper block?

The answer is 2g.

There are diagrams with the problem, but I couldn't draw them. The link to the test is http://www.ets.org/Media/Tests/GRE/pdf/Physics.pdf

I hope you've figured out the questions by now. If not here's a hint:

The volume of water hitting the slab in time dt is:

*A*(

**v**.dt), it's velocity changes from v to 0 in the direction of flow, and mass is just ρ time

*V*, so that shouold do it for the first one.

The second question: Initially everything's at rest so, by writing the equations for the lower block you can see that the tension in the spring is : kx = mg. Now after the thread breaks, the upper block has a net force of: mg(gravity) + mg(spring) acting on it, so the net force is 2mg.

Hope that helps,

### Almost get it...

Thank you, but could you expand a little more on #57?

Code: Select all

```
Before:
_______________________.____.
/ / /|
/<-------- x ------->/ dx / |
/ / / |
/______________________/____/ |
| | | A |
| | | /
| | | /
| | | /
|______________________|____|/
After
_______________________.
/ /|
/<-------- x ------->/ |
/ / |
/______________________/ |
| | A |
| | /
| | /
| | /
|______________________|/
```

Therefore mass of water = ρ (A v dt)

This much water had velocity v initially so its momentum would have been:

ρ (A v dt) v = ρ A v^2 dt

Its final momentum is zero as it stops moving in that direction, so change in momentum is: ρA(v^2) dt

Force = d(momentum)/dt = ρ A v^2

Clearer?