I believe you can solve this by recognizing the relationship
[tex]E = \int _a ^b \vec F \cdot \vec ds[/tex]
But don't worry, you don't have to really do any calculus. Since the frictional forces can be approximated as constants, the above equation reduces to
[tex]E = \vec F \cdot \vec s[/tex]
where [tex]E[/tex] is the energy, [tex]\vec F[/tex] is the frictional force, and [tex]\vec s[/tex] is the displacement (i.e. distance) traveled.
Start with the heavier fragment. Calculate its normal force and multiply it by the coefficient of friction to get the frictional force. Use that, and the equation above, to determine the energy converted to frictional heat, associated with the heavier fragment (hint: this is where the 8.2 m comes into play, together with the frictional force which you just calculated).
Now you can use (1/2)mv2 to get the initial velocity of the heavier fragment. Once you know the heavier fragment's velocity you can express its momentum.
Now you can move on to the lighter fragment. Conservation of momentum means that the lighter fragment will have the same initial momentum magnitude as the heavier fragment, but in the opposite direction. If you know its initial momentum, you can calculate its initial velocity (you already know that the mass is 1/7 that of the heavier object, and you know its momentum, so you can solve for velocity). You can also calculate its normal force, and thus its frictional force, and also its initial kinetic energy. That's all the information you need to use in the above equations to calculate its sliding distance.