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machine gun and physics........help needed!!!?

A machine gun is used by a movie hero to open a heavy vault door of mass 20 tons. The bullets, 10 grams each, are sprayed at the rate of 50 per second always perpendicular to the door, and at its edge only, located 2 meters from the hinge. Each bullet simply falls to the floor after impact, right under the point of impact. If the door took 10 seconds to open from 0 degrees to 90 degrees, determine the speed of the bullets.

Public Comments

1. A mile an hour.
2. The equation: delta theta = 1/2 (alpha) t^2 (assuming the door started at rest). Solve for alpha, since time and the angle are given alpha (angular accel) = torque / (moment of inertia) you can get the moment of inertia of the door (rectangular lamina) from your book. It's a function of mass and the width, both of which are give. So you can get the torque The torque is F times the distance from the hinge to the edge where the bullets hit (since they come in perpendicular) So you can get the force F = change in momentum / change in time = (change in bullet velocity)(bullet mass) / (time between bullets) The bullets go from their initial velocity to zero velocity, so the change is just the velocity you need to find. Solve for it! Good luck!