Inelastic Launch Homework: Maximum Height & Velocity

[benf.stokes]

Homework Statement

A mass M attached to an end of a very long chain of mass per unit length $$\lambda$$
, is thrown vertically up with velocity $$v_{0}$$.
Show that the maximum height that M can reach is:

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

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

Homework Equations

$$F=\frac{dp}{dt}=\frac{dp}{dx}\cdot v$$

Conservation of energy cannot be used because inelastic collisions occur in bringing parts of the rope from zero velocity to v

The Attempt at a Solution

I start by setting up that the total mass at a position y is:
$$M_{total}=M+\lambda\cdot y$$ and thus the momentum at any position is given by:

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

$$F=\frac{dp}{dt}$$ I get an differential equation I can't solve.

Any help would be appreciated.

 

[anathema]

Thank you for your post. I understand your confusion and I will try to guide you through the solution step by step.

First, let's start with the equation F=dp/dt. This equation states that the net force acting on an object is equal to the rate of change of its momentum. In this problem, the net force acting on the mass M is gravity, which is given by F=mg, where g is the acceleration due to gravity. Therefore, we can rewrite the equation as:

mg = dp/dt

Next, we need to find an expression for the momentum of the mass M. We know that momentum is equal to mass times velocity, so we can write:

p = Mv

Substituting this into our equation, we get:

mg = d(Mv)/dt

Using the product rule for differentiation, we can rewrite this as:

mg = M(dv/dt) + v(dm/dt)

Since the mass M is constant, dm/dt = 0, so we can simplify the equation to:

mg = M(dv/dt)

Now, we need to find an expression for dv/dt. We can use the equation v=u+at, where u is the initial velocity, a is the acceleration, and t is time. In this case, we are looking for the velocity at a specific time, when the mass M reaches its maximum height. Therefore, we can write:

v = u + at = v0 - gt

Substituting this into our equation, we get:

mg = M(-g)

Solving for g, we get:

g = -mg/M

This is the acceleration of the mass M at any position y. Now, we can use this to find an expression for v. We know that v=dx/dt, where x is the displacement of the mass M at any position y. Therefore, we can write:

v = dx/dt = dy/dt = v0 - gt

Integrating both sides with respect to time, we get:

y = v0t - 1/2gt^2

Now, we need to find an expression for t when the mass M reaches its maximum height. At this point, the velocity of the mass M is 0, so we can write:

0 = v0 - gt

Solving for t, we get:

t = v0/g

Substituting this into our equation