What Is the Speed of the Bullet?

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    Bullet Velocity
AI Thread Summary
The discussion focuses on calculating the muzzle velocity of a bullet using a wooden block and a spring system. The method involves applying conservation of momentum and energy principles, as the bullet embeds itself in the block upon impact. Participants clarify that the correct approach is to first use momentum conservation to find the velocity after the collision, then apply energy conservation to relate the kinetic energy of the bullet to the potential energy stored in the spring. The maximum compression of the spring is used to determine the velocity, with the final answer approximated at 329 m/s. The conversation emphasizes the importance of distinguishing between pre- and post-collision velocities in the calculations.
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Homework Statement



Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (the figure below). She fires a bullet into a 4.064-kg wooden block resting on a smooth surface, and attached to a spring of spring constant k = 164.5N/m . The bullet, whose mass is 7.870 g, remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.460cm .
(Figure 1)

http://session.masteringphysics.com/problemAsset/1696765/3/Giancoli7.ch11.p22.jpg

What is the speed v of the bullet?


2. Homework Equations

v=√(k/m)*x

The Attempt at a Solution



i tried using that equation and different variations but my mastering physics says I'm incorrect.
 
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i tried using √164.5/(4.064+.00787) *.0946


i also tried using .5mv^2=.5kx^2
 
Mech. energy is not conserved in inelastic collisions. Use momentum conservation and then use Energy conservation.
 
i can't tell if the question want the velocity before the collision or after the collision.
 
It says "muzzle velocity". So it is not the velocity after collision.
 
so then should i not use the equation i did ?
 
what equation should i use
 
You should use energy conservation.
The kinetic energy of the bullet before collision = the potential energy of spring after collision
 
  • #10
so 1/2mv2=mgh?
 
  • #11
no that's wrong
 
  • #12
tristanmagnum said:
so 1/2mv2=mgh?

The potential energy of the spring not the gravitational potencial energy(which would stay constant as the system is horizontal)
 
  • #13
.5mv^2=.5kx^2?
 
  • #14
that's right
 
  • #15
would x be the compression of the block after the bullets hit?
 
  • #16
Yes,right.x would be the compression after bullet hits it.
 
  • #17
mass should be in kg or gr?
 
  • #18
Wait i think we can't use the conservation of energy here as the mechanical energy is not conserved.Instead we will use conservation of momentum
 
  • #19
so then i would use m1v1right?
 
  • #20
My answer is coming to be 329m/s.
 
  • #21
First ignore the spring and calculate the momentum as
mv =(m+M)v1
Now apply the enrgy coonservation
(M+m)v1^2=kx^2
find the value of v1 and then put it in first equation.You will get the answer.
 
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