Is This an Elastic or Inelastic Collision?

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The scenario involves a bullet passing through a block of wood, prompting a discussion on whether the collision is elastic or inelastic. The consensus is that it is an inelastic collision since kinetic energy is not conserved, despite the objects separating afterward. The momentum conservation formula was applied, leading to a calculated final velocity of the block at 2.21 m/s. The distinction between inelastic and perfectly inelastic collisions is clarified, noting that most collisions result in separation. Understanding these principles is crucial for accurately solving similar physics problems.
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Homework Statement

A bullet with a mass of 6.00 g is fired through a 1.25 kg block of wood on a frictionless surface. The initial speed of the bullet is 896 m/s, and the speed of the bullet after it exits the block is 435 m/s. At what speed does the block move after the bullet passes through it?


The attempt at a solution

Should I treat this as an inelastic collision even though the two items separate at the end?

If so, using the (m1)(v1)i + (m2)(v2)i = (m1)(v1)f + (m2)(v2)f formula, I got v2f (final vel. of block) = 2.21 m/s

Do you agree?
 
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ariana0923 said:
Should I treat this as an inelastic collision even though the two items separate at the end?

If so, using the (m1)(v1)i + (m2)(v2)i = (m1)(v1)f + (m2)(v2)f formula, I got v2f (final vel. of block) = 2.21 m/s
Looks good.

The collision is inelastic, because KE is not conserved. (Calculate it and see for yourself.) It's not a completely inelastic collision, because they do separate. In any case, all you need to solve it is momentum conservation.
 
ariana0923 said:
Should I treat this as an inelastic collision even though the two items separate at the end?

Most collisions result in the two bodies separating in the end; the case whereby the two bodies coalesce together is a special case of inelastic collisions - a perfectly inelastic collision.
 
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