A 0.005-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an

Question

A 0.005-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 18.7-kg door, imbedding itself 9.9 cm from the side opposite the hinges as in the figure below. The 1.00-m-wide door is free to swing on its hinges. (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? O yes O no Explain your answer. When hiting the door, the bullet has angular momentum relative to the axis rotation (b) Is mechanical energy conserved in this collision? Answer without doing a calculation. yes O no (c) At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.) rad/s (d) Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision. (Give the kinetic energy of the door-bullet system just after collision.) Need Help?

Explanation / Answer

Moment of inertia of door = I = mass of door*width of door^2 / 3 = 18.7*1^2 / 3 = 6.23kgm^2 and Moment of inertia of bullet after embedding = mass of bullet*distance from the hinge to the point where it is embedded^2 = 0.005*(1m-0.099m)^2 = 4.06*10^-3

Now, about the hinge, Initial angular momentum = Final angular momentum

So, 0.005*1.00*10^3*(1m-0.099m) = (6.23+ 4.06*10^-3)*angular speed

So, angular speed = 0.7226 rad/s

So, the angular speed at which the door swing open immediately after the collision is 0.7226 rad/s

Now, Kinetic Energy of the door bullet system just after collision = 0.5*(6.23+ 4.06*10^-3)*0.7226^2 = 1.63J

The kinetic energy of the bullet before the collision is 0.5*0.005*(1.00*10^3)^2 = 2500J

So, the kinetic energy of the door bullet system just after collision is less than the kinetic energy of the bullet before the collision

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