# Velocity of bullet

1. Nov 6, 2013

### tristanmagnum

1. The problem statement, all variables and given/known data

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. Relevant equations

v=√(k/m)*x

3. The attempt at a solution

i tried using that equation and different variations but my mastering physics says i'm incorrect.

2. Nov 6, 2013

3. Nov 6, 2013

### tristanmagnum

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

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

4. Nov 6, 2013

### Enigman

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

5. Nov 6, 2013

### tristanmagnum

i cant tell if the question want the velocity before the collision or after the collision.

6. Nov 6, 2013

### nasu

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

7. Nov 6, 2013

### tristanmagnum

so then should i not use the equation i did ?

8. Nov 6, 2013

### tristanmagnum

what equation should i use

9. Nov 6, 2013

### nil1996

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

10. Nov 6, 2013

### tristanmagnum

so 1/2mv2=mgh?

11. Nov 6, 2013

### tristanmagnum

no thats wrong

12. Nov 6, 2013

### nil1996

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

13. Nov 6, 2013

### tristanmagnum

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

14. Nov 6, 2013

### nil1996

that's right

15. Nov 6, 2013

### tristanmagnum

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

16. Nov 6, 2013

### nil1996

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

17. Nov 6, 2013

### tristanmagnum

mass should be in kg or gr?

18. Nov 6, 2013

### nil1996

Wait i think we cant use the conservation of energy here as the mechanical energy is not conserved.Instead we will use conservation of momentum

19. Nov 6, 2013

### tristanmagnum

so then i would use m1v1right?

20. Nov 6, 2013

### nil1996

My answer is coming to be 329m/s.