# Motion of two constrained masses

1. Feb 20, 2015

### PhMichael

1. The problem statement, all variables and given/known data
The attached images shows everything. (${\bf{e}}_1$ denotes the direction of X and ${\bf{e}}_2$ denotes the direction of Y).

Initially, the spring is force free when $X_{0}=0.4 m$ (which yields $Y_{0}=0.3 m$). Also, at this instant, the velocity of B is ${\bf{V}}_{B0}=-v_{0}{\bf{e}}_{2}$ such that the corresponding velocity of A is ${\bf{V}}_{A0}=v_{0}(Y/X){\bf{e}}_{1}=0.75v_0{\bf{e}}_{1}$.

I'm asked to find the minimum value of $v_0$ for which B hits the floor.

2. Relevant equations
Energy conservation yields
$$(\frac{1}{2}m_{B}V_B^{2}+\frac{1}{2}m_{A}V_A^{2})-(\frac{1}{2}m_{B}V_{B0}^{2}+\frac{1}{2}m_{A}V_{A0}^{2})+\frac{1}{2}k(X_{0}-L)^2=0$$
However, the kinematic constrains (L is constant) yields
$$XV_{A}+YV_{B}=0$$
So, when B hits the floor, $Y=0$ so that, at this instant, $V_{A}=0$. Hence,
$$V_B^{2}=(V_{B0}^{2}+\frac{m_{A}}{m_{B}}V_{A0}^{2})-\frac{k}{m_B}(X_{0}-L)^2=v_{0}^{2}(1+\frac{2}{3}(0.75)^{2})-\frac{100}{3}(0.4-0.5)^2$$
Now, my requirement is that $V_B=0$ when it hits the floor, so that the last equation yields
$$v_{0}\approx0.49 m/s$$
$$v_{0}\approx0.37 m/s$$
Am I doing/assuming something wrong here?

3. The attempt at a solution
See the previous item.

#### Attached Files:

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2. Feb 20, 2015

### Staff: Mentor

While it would help if you show more steps, I agree with your result. 0.37 m/s would be the speed of A.