Awooden block is cut into two pieces; one is 3 times heavier than the other. a depression is made in both pieces; a firecracker is inserted in the depression and the block is reassembled. the reassembled block is placed on a rough surface and the fuse on the firecracker is lit. when the firecracker explodes, the two pieces of the block slide apart. what is the ratio of the distances traveled by each block?

Let A represent the more massive piece, and B the less massive piece.  Thus

Ma = 3 Mb

In the explosion, momentum is conserved.  We have

Va = V2 = 0

Pin = Pfinal

0 = MaVa + Mb Vb

= 3MbVa + Mb Vb

Va = - 1/3 x Vb

For each block, the kinetic energy gained during the explosion is lost to negative work done by friction on the block.

Wfr = KEf - KEi = -1/2 mbv^2

But work is also calculated in terms of the force doing the work and the distance traveled.

Mfr = Ffr delta x cos 180 degrees

= -Uk Fn x delta x = -Ukmg x delta x

Equate the two work expressions, solve for the distance traveled, and find the ratio of distances.

_-1/2mv^2 = -Uk mg x delta x

Delta x = v^2 / gUx

Delta Xa / delta Xb = (- 1/3 x Vb)^2 / Vb ^2 = 1/9

And so

(Delta x)heavy / (delta x ) light = 1/9

prolly a beacuse the momentuem the car has increases while going down hill