## Help with #57 and #72

rrr
Posts: 5
Joined: Mon Jan 22, 2007 11:28 am

### Help with #57 and #72

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

rjharris
Posts: 101
Joined: Wed May 10, 2006 6:48 pm
for those w/o the test in hand, could you post the questions? thanks.

rrr
Posts: 5
Joined: Mon Jan 22, 2007 11:28 am
57. A stream of water of mass density, p (rho), cross-sectional area A, and speed, v, strikes a wall that is perpendicular to the direction of the stream, as shown in the firgue. The water then flows sideways across the walls. The force exerted by the stream on the wall is

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?

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

rjharris
Posts: 101
Joined: Wed May 10, 2006 6:48 pm
on 57, (a) and (c) are the only ones with the correct units. and (c) contains an h, which is entirely irrelevant. so (a) must be the answer.

/ i dont feel like thinking about how to actually do the problem at the moment.

CPT
Posts: 77
Joined: Mon Jan 15, 2007 2:05 am
Hey rrr,

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,

rrr
Posts: 5
Joined: Mon Jan 22, 2007 11:28 am

### Almost get it...

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

CPT
Posts: 77
Joined: Mon Jan 15, 2007 2:05 am
Consider the situation like this:

Code: Select all

``````Before:

_______________________.____.
/                      /    /|
/<--------  x  ------->/ dx / |
/                      /    /  |
/______________________/____/   |
|                      |    | A |
|                      |    |   /
|                      |    |  /
|                      |    | /
|______________________|____|/

After

_______________________.
/                      /|
/<--------  x  ------->/ |
/                      /  |
/______________________/   |
|                      | A |
|                      |   /
|                      |  /
|                      | /
|______________________|/
``````
The ammount of water that has splashed past the wall is a volume of length dx = v.dt {v is the velocity and dt is the infinitesimal time}, and the cross section area is 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?

rrr
Posts: 5
Joined: Mon Jan 22, 2007 11:28 am

### Thanks for all the help!!

I finally get it.

CPT
Posts: 77
Joined: Mon Jan 15, 2007 2:05 am
Hey, glad to be able to help. Just keep at it and things'll get familiar soon.

Good Luck!