What Is the Speed of the Bullet?

[tristanmagnum]

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.

 

[phinds]

Show your work

 

[tristanmagnum]

i tried using √164.5/(4.064+.00787) *.0946


i also tried using .5mv^2=.5kx^2

 

[Enigman]

Mech. energy is not conserved in inelastic collisions. Use momentum conservation and then use Energy conservation.

 

[tristanmagnum]

i can't tell if the question want the velocity before the collision or after the collision.

 

[nasu]

It says "muzzle velocity". So it is not the velocity after collision.

 

[tristanmagnum]

so then should i not use the equation i did ?

 

[tristanmagnum]

what equation should i use

 

[nil1996]

You should use energy conservation.
The kinetic energy of the bullet before collision = the potential energy of spring after collision

 

[tristanmagnum]

so 1/2mv2=mgh?

 

[tristanmagnum]

no that's wrong

 

[nil1996]

so 1/2mv2=mgh?

The potential energy of the spring not the gravitational potencial energy(which would stay constant as the system is horizontal)

 

[tristanmagnum]

.5mv^2=.5kx^2?

 

[nil1996]

that's right

 

[tristanmagnum]

would x be the compression of the block after the bullets hit?

 

[nil1996]

Yes,right.x would be the compression after bullet hits it.

 

[tristanmagnum]

mass should be in kg or gr?

 

[nil1996]

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

 

[tristanmagnum]

so then i would use m1v1right?

 

[nil1996]

My answer is coming to be 329m/s.

 

[nil1996]

First ignore the spring and calculate the momentum as
mv =(m+M)v1
Now apply the energy coonservation
(M+m)v1^2=kx^2
find the value of v1 and then put it in first equation.You will get the answer.