A bullet of mass m=27g and speed v=252 m/s perpendicularly hits an open rectangu

Question

A bullet of mass m=27g and speed v=252 m/s perpendicularly hits an open rectangular door at rest, stays encased in it, and puts it in motion. The door has a width L=1.26 m and a mass M=21.6 Kg. The bullet impact is at a distance d= 0.6 m from the door hinges. Neglect any type of friction (air resistance or hinges friction). The moment of inertia of the door is given by the formula: I= (1/3)*M*d^2.

- Calculate the initial (before impact) angular momentums, Lbi and Ldi, of the bullet and the door, respectively (in Kg m2/s).
- Explain whether this collision is elastic or inelastic, and explain its consequence on the door’s after-collision speed.
- Calculate the angular speed of the door just after the collision.

Explanation / Answer

initial (before impact) angular momentum of bullet = 0.027 * 252 * 0.6

                                                                                = 4.082 kg*m2/s

initial (before impact) angular momentum of door = 0    kg*m2/s

Moment of inertia of door = 11.43 kg*m2

angular speed of the door just after the collision = 4.082/(11.43 + 0.00972)

                                                                              = 0.3568 rad/sec

This collision is inelastic as initial rotational kinetic energy is not equal to final rotational kinetic energy .

Door after collision speed = 1.26 * 0.3568

                                          = 0.4495 m/sec

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