Collision, Bullet fired into block of wood

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Homework Help Overview

The problem involves a bullet colliding with a block of wood, where the bullet embeds itself in the block. The subject area includes concepts of momentum conservation and collision types, specifically focusing on inelastic collisions.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and kinetic energy, questioning the validity of using kinetic energy in an inelastic collision. There are attempts to clarify the nature of the collision and its implications on energy transfer.

Discussion Status

Some participants have provided guidance on the inelastic nature of the collision and the implications for energy conservation. There is ongoing exploration of definitions and conditions for elastic versus inelastic collisions, with no explicit consensus reached.

Contextual Notes

Participants are considering the effects of friction and energy loss in the context of the collision, as well as the assumptions made about energy transfer during the event.

synergix
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Homework Statement


A 20.0 g bullet moving horizontally at 70.0 m/s strikes a 9.00 kg block of wood lying at rest on the pavement. The bullet embeds itself in the block of wood during the collision. Find
a. the speed of the block of wood immediately after the collision,
b. the frictional force between the pavement and the block of wood if the block of wood moves 2.50 m before stopping.

Homework Equations


conservation of momentum and kinetic energy

The Attempt at a Solution


Using conservation of momentun:
A)MbVb+MwVw=(Mb+Mw)V
MbVb+0=(Mb+Mw)V
(.0200kg)70m/s=(9.00kg+.0200kg)V
V=[(.0200kg)(70m/s)]/(9.02kg)
V=.1552106...m/s
Using conservation of Kinetic energy:
.5MbVb^2+0=.5(Mb+Mw)V^2
(.0100kg)(70m/s)^2=4.51kgV^2
V^2=[.0100kg(4900m/s)]/4.51kg
V^2=10.86474501...
V=3.29617...m/s

Should'nt these values be the same have I done something wrong? I used both to check that my answers were correct and then I got this result. the second part should be easy once we figure ths out.
 
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Conservation of kinetic energy won't apply because there is an inelastic collision.

Momentum is conserved, and the initial velocity is the V that you would use to solve the second part of the question.

V2 = 2*a*x

With you acceleration (deceleration in this case and the distance covered) then you should be able to figure the retarding force of the pavement and hence the coefficient of kinetic friction.
 
Excellent! ok so I have one more quick question is this an inelastic collision because the bullet was embedded into the wood and so the wood absorbed all the energy of the bullet?
 
Yes the wood was deformed and heated by the bullet
 
synergix said:
Excellent! ok so I have one more quick question is this an inelastic collision because the bullet was embedded into the wood and so the wood absorbed all the energy of the bullet?

No. Some of the energy was converted to moving the block the 2.5 m. Friction brought it to rest. At that point the kinetic energy budget got drawn to 0.
 
what I meant was all the kinetic energy of the bullet was directly transferred to the block. ignoring the production of heat and noise and what not else. I am not forgetting that the block was set in motion as a result. would the collision have been elastic if the pavement had been frictionless. or would it have been elastic if they bounced off each other? I am just trying to get a clear picture/definition in my head of an inelastic collision using this example.
 
synergix said:
what I meant was all the kinetic energy of the bullet was directly transferred to the block. ignoring the production of heat and noise and what not else. I am not forgetting that the block was set in motion as a result. would the collision have been elastic if the pavement had been frictionless. or would it have been elastic if they bounced off each other? I am just trying to get a clear picture/definition in my head of an inelastic collision using this example.

Basically you have it. If the coefficient of restitution is not 1, then it will have been an inelastic collision.

Perhaps these would help?
Inelastic collision:
http://en.wikipedia.org/wiki/Inelastic_Collision

Elastic collision:
http://en.wikipedia.org/wiki/Elastic_Collision#Equations

Coefficient of restitution:
http://en.wikipedia.org/wiki/Coefficient_of_restitution#Further_details
 
Ty! :)
 

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