How Fast Do Entwined Movie Foes Slide After a Stuntman Swings?

[janiexo]

I've been messing around with this problem for a while now and am getting absolutely nowhere. I was wondering if anyone could help me work it out...

--A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0 kg), who is standing directly under the chandelier. (Assume that the stuntman's center of mass moves downward 5.0 m. He releases the rope just as he reaches the villain.)
A: With what speed do the entwined foes start to slide across the floor?
Take the free fall acceleration to be 9.8
B: If the coefficient of kinetic friction of their bodies with the floor is mu_k= 0.215, how far do they slide?

I don't recall doing anything at all like this in class and can't find any examples like this so I'm really at a loss as to how to approach it.

 

[StatusX]

First find the stuntman's horizontal velocity just before impact. Since he is swinging on a rope, all of his velocity will be horizontal at the bottom of the arc, and you can use conservation of energy to get his velocity. Then use conservation of momentum to get the velocity of the combined stuntman/villain, assuming they stick together. Finally, use the force of friction times distance equals the work done by friction, and find how far they must go for this to equal their starting kinetic energy.

In case you're wondering why you use conservation of energy for the first and last part and momentum for the middle part, I'll explain that real quick. They are both always conserved, but sometimes one of them goes into parts of the problem you aren't interested in. In the first and last part, momentum is conserved because the Earth moves slightly as the guy moves, but you can obviously ignore this effect. You can always ignore momentum conservation when considering friction against the ground and you can ignore the component of momentum in the direction of gravity when you use a constant gravitational field instead of the GMm/r equation. Finally, for the second part, some energy goes to heat in an inelastic collision, so conservation of energy wouldn't help you. I don't know if you wanted all that extra information, but there it is.

 

[janiexo]

Ok I think I understand. I did what you said and ended up with the velocity of the stuntman at the bottom of the arc being 9.9m/s (using conservation of energy: mgh=1/2*mv^2 since initial velocity is 0 and final height is 0). Then used conservation of momentum: m1(9.9)=(m1+m2)v_final and ended up with the velocity being equal to 5.98m/s.

For the 2nd part I came up mu_k*N*d=1/2*m*v^2 with N = mg and got a distance of 6.6
Have I done it correctly?

 

[StatusX]

Looks good.