# Mechanics Question

## Homework Statement

A mass m with velocity v0 makes an inelastic collision with second mass M that is suspended by a string of length L. The velocity v0 is perpendicular to the vertical string. After the collision the combined masses, m+M, rotate in a vertical plane around the point of suspension of the string. Find the minimum value v0 such that the string remains always under tension for a complete rotation.

∑Fr=mv2/r
T=1/2mv2
U=mgh

## The Attempt at a Solution

The condition for the an inflexible string to remain under tension is that it never goes slack, i.e. length=L for the entire rotation.

Since the collision is inelastic, momentum is conserved while kinetic energy is not. So,

mv0=(m+M)v' where v' is the magnitude of the x-directed velocity of the combined masses (m+M) after the collision. so v'=m/(m+M)v0.

At the bottom of the vertical circle (In Polar Coordinates),
letting m'=(m+M)

∑Fr= Tb-m'g=m'v'2/L (*)

At the top,

∑Fr=Tt+m'g=m'v''2/L (**)

Using Energy,

1/2m'v'2=1/2m'v''2+m'g(2L) (***)
v'2=v''2+4gL

v''2=v'2-4gL

I am going to pause here to see if there is anything I am missing.