Collision, Bullet fired into block of wood

In summary, 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.
  • #1
synergix
178
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
Yes the wood was deformed and heated by the bullet
 
  • #5
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.
 
  • #6
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.
 
  • #7
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
 
  • #8
Ty! :)
 

1. How does the speed of the bullet affect the collision with the block of wood?

The speed of the bullet plays a crucial role in the collision with the block of wood. The higher the speed of the bullet, the more force it will exert on the wood, resulting in a deeper penetration and potentially more damage to the wood.

2. What factors determine the depth of the bullet's penetration into the block of wood?

The depth of penetration depends on several factors, including the speed and mass of the bullet, the density and thickness of the wood, and the angle at which the bullet strikes the wood. Additionally, the type and shape of the bullet can also affect the depth of penetration.

3. How does the density of the wood affect the force of impact in a collision with a bullet?

The density of the wood plays a crucial role in determining the force of impact in a collision with a bullet. The denser the wood, the more force is required to penetrate it. This means that a bullet will have a more significant impact on denser woods compared to less dense ones.

4. What happens to the energy of the bullet after it collides with the block of wood?

After the collision, the bullet's energy is transferred to the block of wood, causing it to deform and potentially break apart. Some of the energy is also dissipated as heat and sound. The remaining energy will continue to push the wood until it comes to a complete stop.

5. Can the angle at which the bullet strikes the wood affect the outcome of the collision?

Yes, the angle at which the bullet strikes the wood can significantly impact the collision's outcome. A bullet that strikes the wood at a perpendicular angle will have a more direct force, while a bullet that strikes at an angle will have a more glancing blow and may not penetrate as deeply.

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