- #1

benf.stokes

- 71

- 0

Conservation of energy cannot be used because inelastic collisions occur in bringing parts of the rope from zero velocity to vA mass M attached to an end of a very long chain of mass per unit length [tex]\lambda[/tex]

, is thrown vertically up with velocity [tex]v_{0}[/tex].

Show that the maximum height that M can reach is:

[tex]h=\frac{M}{\lambda}\cdot \left [ \sqrt[3]{1+\frac{3\cdot \lambda\cdot v_{o}^{2}}{2\cdot M\cdot g}}-1 \right ][/tex]

and that the velocity of M when it returns to the ground is [tex]v=\sqrt{2\cdot g\cdot h}[/tex]

I start by setting up that the total mass at a position y is:

[tex]M_{total}=M+\lambda\cdot y[/tex] and thus the momentum at any position is given by:

[tex]p=(M+\lambda\cdot y)\cdot v[/tex] but I can't figure out an expression for v and using

[tex]F=\frac{dp}{dt}[/tex] I get an differential equation I can't solve.

Any help would be appreciated.