Having trouble on this energy and momentum question

[garr6120]

Homework Statement

A rifle bullet of mass 10.0 g strikes and becomes embedded in a wooden block of mass 490 g, which is at rest on a horizontal, frictionless surface and is attached to a spring bumper.
The impact compresses the spring, whose force constant is 100 N/m, by 20 cm.
(a) what is the maximum potential energy of the spring?
(b) Determine the velocity with which the block and bullet first begin to move.
(c) What was the initial velocity of the bullet?
(d) What was the initial kinetic energy of the bullet?

Homework Equations

E = 1/2*mv^2 + 1/2*kx^2

The Attempt at a Solution

I first converted the 490 g into 0.490 kg, 10.0 g into 0.01 kg, 20 cm into 0.2 m.
Next, I solved for maximum potential energy by only using the equation E = 1/2*kx^2
And then I get stuck trying to find the speed of the block and bullet and the other questions.

 

[garr6120]

could someone try to help me i am new here

 

[gneill]

Start by identifying the different stages in the progress of the system. I'll start you off:

1. A bullet is fired at some velocity v1.
2. The bullet strikes and is embedded in a block (What type of collision is that? What's conserved?)
...carry on

 

[garr6120]

The bullet strikes then gets embedded in the block giving you completely inelastic collision therefore momentum is conserved and kinetic energy will be lost due to the sound of the bullet hitting the block.
The impact then compresses the spring, giving the spring elastic potential energy
The spring then extends giving the block kinetic energy
Therefore, could you say that 1/2*kx^2 is equal to 1/2mv^2' (' means after the collision, and m is the total mass of the block and bullet)?
So 1/2*kx^2 would then equal the total kinetic energy?

 

[gneill]

The bullet strikes then gets embedded in the block giving you completely inelastic collision therefore momentum is conserved and kinetic energy will be lost due to the sound of the bullet hitting the block.

Yes, sound and heat due to friction and deformation of the bullet and block.

The impact then compresses the spring, giving the spring elastic potential energy

Well, the impact doesn't do it so much as the kinetic energy remaining in the block after the collision, right?

The spring then extends giving the block kinetic energy
Therefore, could you say that 1/2*kx^2 is equal to 1/2mv^2' (' means after the collision, and m is the total mass of the block and bullet)?
So 1/2*kx^2 would then equal the total kinetic energy?

Well technically the problem doesn't cover the rebounding of the block from the compressed state. However, having gone through the steps in the "forward" direction you can certainly work them backwards starting with the compressed spring imparting kinetic energy to the bullet+block. As you say, the spring's PE goes into the bullet+block, so you can determine their speed at the moment after the collision. You should be able to work the collision math backwards to find the bullet's initial speed.