Question

A missile launcher, mass M, at rest on a horizontal track fires a missile, mass m, horizontally.
a) If the launch speed of the missile is v o , what (effective) coefficient of (kinetic) friction is
required in order to stop the launcher within x meters? (b) Missile launcher mass is 4400 kg,
missile mass is 110 kg, missile launch speed is 450 m/s, the effective coefficient of friction on
the launcher is 0.75. Find x.

Answer 1

Based on the conservation of momentum, the velocity of launcher V=vo*m/M.
The kinetic energy of launcher is E=1/2*M*V^2=1/2*m/M*m*vo^2.
Suppose the coefficient of friction is C.
The friction F=C*M*g.
The launcher is stoped, then F*x=E,then C=1/2*m^2/M^2*vo^2/g/x.
For the second question, the lauch speed of the missile vo is needed.