The rule for conservation of momentum is that the value of mass times velocity is conserved for all objects in a closed system. Your equation isn't quite right. It shouldn't have a minus sign on the right hand side. In other words:

m1*u1 + m2*u2 = m1*v1 + m2*v2

The toboggan puzzle is simple in theory but difficult in practice. Let us make a few simplifying assumptions.

1) Assume that the ice is frictionless

2) Assume that there's no drag caused by the air

In this case, Jack doesn't gain any benefit by pushing the box off the side of his toboggan. To get rid of the box, he pushes it at right angles to the direction he's going in. Remember that the direction of travel matters when you measure velocity and acceleration. (Unlike speed.)

Because of convservation of momentum, if the box has mass m2 and is if thrown out at velocity v2 then the toboggan must move at velocity (-m2*v2 / m). Unfortunately, this movement is sideways so it doesn't help him to win the race.

Likewise you could work the same result with Newton's laws. Newton's third law tells us that the box pushes back on Jack when Jack throws it out. Newton's second law tells us that this will cause Jack, and the toboggan, to accelerate away from the box. You need to do some horrible integration to work out how much velocity Jack gains. The maths is _much_ easier with conservation of momentum. :~)

So we know that throwing the box out sideways won't help. However, if Jack throws the box out the back of the toboggan he will gain velocity in the direction that matters. Secondly, the faster the box goes backwards, the more velocity Jack will gain. If he really wants to win, he should use a cannon. :~)

In reality, the relationship between mass, drag and sliding friction means that it's hard to know what'll happen. If wind drag is high compared to the friction from the ice then throwing out the box will slow Jack down. I'll explain some of these issues in more detail if you want me to.

regards,

Gremlin