Posted by **Chad** on Tuesday, August 7, 2012 at 10:36pm.

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.

- Physics -
**bbwake**, Thursday, August 9, 2012 at 7:51am
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.

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