A 1.10-kg wooden block rests on a table over a large hole as in the figure below

Question

A 1.10-kg wooden block rests on a table over a large hole as in the figure below. A 5.60-g bullet with an initial velocity vi is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of 18.0 cm. (a) Describe how you would find the initial velocity of the bullet using ideas you have learned in this chapter. You can find the center of mass of the system based on how far the bullet stops inside the wooden block. Using the the work-energy theorem you can relate this the change in potential energy of the center of mass to the maximum height, Ug = MgyCM. Then, you can use the conservation of energy to relate this potential energy to the initial kinetic energy of the bullet. . Using the conservation of momentum, you can relate the speed of the block and bullet right after the collision to the initial speed of the bullet. Then, you can use the conservation of mechanical energy for the bullet-block-Earth system to relate the speed after the collision to the maximum height. Using the conservation of chemical energy you can relate the increased temperature of the block after the collision to the initial speed of the bullet. Then, you can use the conservation of mechanical energy for the bullet-block- Earth system to relate the speed after the collision to the maximum height. (b) Calculate the initial velocity of the bullet from the information provided. (Let up be the positive direction.)

Explanation / Answer

we can do this by working backwards...we know that the bullet block system has a final PE of

(M+m) gh and an initial KE of 1/2 (M+m)V^2 so that

V=speed of bullet/block = Sqrt[2 g h] = Sqrt[2x9.8m/s/s x 0.18m] = 1.88m/s

now we use conservation of momentum to find vi:

m vi = (M+m)V

vi=(1.10kg x 1.88m/s)/0.0056kg = 369.28m/s

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