Clay projectile sticking to wooden block with friction

[Sheneron]

[SOLVED] Conservation of Momentum problem

Homework Statement

A 12.5 g wad of sticky clay is hurled horizontally at a 95 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?

Homework Equations

Ki = Kf + loss
Pi = Pf

The Attempt at a Solution

To solve this problem here is what I did:
Ki = Kf + loss
m1vi = (m1 + m2)vf

Using equation 2, I solved for vf. I then plugged that into the vf in the Kf in equation 1. Using this method, however, I did not get a correct answer. I feel like I am missing something at the beginning. Can someone tell me if this is the proper set up or if I am missing something? Thanks.

 

[Doc Al]

Separate the problem into two parts:
(1) The collision itself. Use your momentum conservation equation here, just like you did.
(2) Post collision. What's the KE of the block+clay immediately after the collision? What's the work done by the friction force?

 

[Sheneron]

Would my 2 equations not be this:

1) (m1)(Vbefore) = (m1 + m2)(Vafter)

2) .5m1(Vbefore)^2 = .5(m1 + m2)(Vafter)^2 + loss

 

[Doc Al]

1) (m1)(Vbefore) = (m1 + m2)(Vafter)

Yes.

2) .5m1(Vbefore)^2 = .5(m1 + m2)(Vafter)^2 + loss

No.

Instead use: KE(after) = Work done by friction.

 

[Sheneron]

Ok, that worked.

Im not sure I completely understand why KE(after) = work done by friction.

 

[Doc Al]

Because the work done on the block+clay equals the change in its KE. After the collision, it starts out with KE(after) and ends up with 0 as it comes to rest.

 

[Sheneron]

Simple enough. Thanks