A Bullet Is Fired into a Wooden Block
Part A = perfectly inelastic
Part B = momentum only
Part C = vf = mb ˙ vi / (mb + mw)
Solution Below:
A bullet of mass m is fired horizontally with speed 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 within the block, the block, with the bullet in it, is traveling at speed vf.
Part A
Which of the following best describes this collision?
- perfectly elastic
- partially inelastic
- perfectly inelastic
This type of collision (where the two objects stick together) is perfectly 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
In a perfectly inelastic collision, only momentum is conserved.
momentum only
Part C
What is the speed of the block/bullet system after the collision?
Express your answer in terms of vi, mw, and mb.
Express your answer in terms of vi, mw, and mb.
vf = mb ˙ vi / (mb + mw)
vf = mb ˙ vi / (mb + mw)
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