Two blocks, a string, and a spring.

1. Jul 24, 2012

AbigailM

1. The problem statement, all variables and given/known data
Two blocks A and B with respective masses $m_{A}$ and $m_{B}$ with respective masses $m_{A}$ and $m_{B}$ are connected via a string. Block B is on a frictionless table, and block A is hanging at a vertical distance h from a spring with spring constant k that is at its equilibrium position. The blocks are initially at rest. Find the velocity of A and B when the spring is compressed by an amount $\delta y =m_{A}g/k$. Determine the maximum compression $\delta y_{max} of the spring in terms of [itex]m_{A}, m_{B}$, g and k. (Hint: what happens to the motion of the blocks when $\delta y-m_{A}g/k$?)

2. Relevant equations
$\delta y=m_{A}g/k$ (Eq 1)

$(m_{A}+m_{B})gh=\frac{1}{2}(m_{A}+m_{B})v^{2}$ (Eq2)

$\frac{1}{2}m_{A}v^{2}=m_{A}g\delta y - \frac{1}{2}k\delta y^{2}$ (Eq 3)

3. The attempt at a solution
From Eq2 $v_{B}=\sqrt{2gh}$

Solve Eq3 for v and substitute in Eq1. Then we can subtract our new equation from Eq2:
$v_{A}=\sqrt{2gh}-\sqrt{m_{A}/k}g$

To find $\delta y_{max}$ substitute $v=\sqrt{2gh}$ into Eq3:
$m_{A}gh=m_{A}g\delta y - \frac{1}{2}k\delta y^{2}$

Now solve for $\delta y$:
$\delta y=\frac{m_{A}g-\sqrt{m_{A}^{2}g^{2}-2km_{A}g}}{k}$

Does this look correct? Thanks for the help

2. Jul 25, 2012

PhanthomJay

You seem to be breaking up the motion into stages (you don't necessarily have to) and coming up with incorrect equations. You must consider the energies of both blocks when applying the conservation of energy equations. The block on the table still has PE at any stage of the motion, and it still has KE as block A hits the spring.