Momentum problem -- Two bullets fired into a block

[Zynoakib]

Homework Statement

A 7.00-g bullet, when fired from a gun into a 1.00-kg
block of wood held in a vise, penetrates the block to a
depth of 8.00 cm. This block of wood is next placed on
a frictionless horizontal surface, and a second 7.00-g
bullet is fired from the gun into the block. To what
depth will the bullet penetrate the block in this case?

Homework Equations

The Attempt at a Solution

KE of bullet = 1/2 (0.007)v^2 = 0.0035v^2

Work done during deceleration
W = Fs
0.0035v^2 = F(0.08) = 0.04375v^2

Find the distance of penetration in the second scenario
KE of the bullet = KE of the block and bullet + energy loss due to friction
0.0035v^2 = 1/2 (1 + 0.007)(final velocity of the block and bullet) + 0.04375v^2 (s)

The final velocity of the block and the bullet:
Momenta of the block and the bullet before impact = momenta of the block and the bullet after impact
(0.007)(v) + 0 = (1 + 0.007)(final velocity)
v = 143.857(final velocity)

0.0035v^2 = 1/2 (1 + 0.007)(final velocity of the block and bullet) + 0.04375v^2
0.0035 (143.857(final velocity))^2 = 1/2 (1 + 0.007)(final velocity of the block and bullet) + 0.04375 (143.857(final velocity))^2
72.43 = 0.5035(final velocity)^2 + 905.4 (s)

Then I cannot proceed with the calculation anymore as I cannot solve the two unknowns.

Thanks!

 

[haruspex]

What do you think the relationship might be between the depth of penetration and the work done against friction?

 

[Zynoakib]

What do you think the relationship might be between the depth of penetration and the work done against friction?

The KE of the bullet would all be converted to work done against friction so that the bullet would stop at 8 cm

 

[haruspex]

The KE of the bullet would all be converted to work done against friction so that the bullet would stop at 8 cm

That isn't what I asked. My question would apply equally if the bullet were to be pushed in slowly by a powerful piston.

 

[Zynoakib]

That isn't what I asked. My question would apply equally if the bullet were to be pushed in slowly by a powerful piston.

work done against friction = Frictional force x depth of penetration ?

 

[haruspex]

work done against friction = Frictional force x depth of penetration ?

Right. If you assume that force is constant and the same in both cases, do you see how to solve the problem?

 

[Zynoakib]

=

Right. If you assume that force is constant and the same in both cases, do you see how to solve the problem?

Like this? I did use that. I just don't know how to continue with calculation as there are unsolvable unknowns at the end

Work done during deceleration
W = Fs
0.0035v^2 = F(0.08) = 0.04375v^2

 

[haruspex]

Like this?
0.0035v^2 = F(0.08) = 0.04375v^2
I did use that. I just don't know how to continue with calculation as there are unsolvable unknowns at the end

No, that's taking the work to be the same, which clearly it isn't. The 0.08m penetration is only for the case where the block was held fixed. When the block is allowed to move the penetration will be different, but the force is the same.