Inelastic/elastic collision?

  • Thread starter ariana0923
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  • #1
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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?
 

Answers and Replies

  • #2
Doc Al
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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.
 
  • #3
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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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