A bullet of mass mb is fired horizontally with speed vi at a wooden block of mas

Question

A bullet of mass mb is fired horizontally with speed vi at a wooden block of mass mw resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest. Part A Which of the following best describes this collision? perfectly elastic partially inelastic perfectly inelastic Part B Which of the following quantities, if any, are conserved during this collision? kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy. Part C:

What is the speed of the block/bullet system after the collision?

Express your answer in terms of vi , mw , and mb .    vf = ?

Explanation / Answer

perfectly inelastic

only momentum is conserved and kinetic energy is not conserved

MbVi = (Mb + Mw)Vf

Vf = MbVi / (Mb + Mw)

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