A 0.004-kg bullet travelinghorizontally with a speed of 1000 m/s enters an21.3-k

Question

A 0.004-kg bullet travelinghorizontally with a speed of 1000 m/s enters an21.3-kg door, imbedding itself10.1 cm from the side opposite the hingesas in the figure below. The 1.00-m-wide door is free to swing onits hinges. At what angular speed does the door swing open immediatelyafter the collision? (The door has the same moment of inertia as arod with axis at one end.) (rad/s) ? Calculate the energy of the door-bullet system (J) ? A 0.004-kg bullet travelinghorizontally with a speed of 1000 m/s enters an21.3-kg door, imbedding itself10.1 cm from the side opposite the hingesas in the figure below. The 1.00-m-wide door is free to swing onits hinges. At what angular speed does the door swing open immediatelyafter the collision? (The door has the same moment of inertia as arod with axis at one end.) (rad/s) ? Calculate the energy of the door-bullet system (J) ?

Explanation / Answer

the second part is easier to calculate. energy = initial KE of the bullet = (1/2)mv2 =.5(.004)106 = 2000 J after the collision, the bullet & door have rotational KE. (1/2)I2 = 2000 I = (1/3)ML2 = (1/3)(21.3)(12) =7.1 2 = 2000 * 2/I = 563.38 = 23.7 rad/s

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