Momentum problem

assaftolko

A skater with a mass of 60 Kg is on a skateboard with a mass of 3 kg. The skater holds in his hands two weights each of them with a mass of 5 kg. The system is at rest and at some moment the skater throws horizontaly one of the weights so that it's velocity in respect to the skater is 8 m/s. After some more time he throws the second weight in the same direction.

What is the velocity of the skater after the second throw?

Well from conservation of momentum you get that for the first throw the momentum of the skater+skateboard after the throw minus the momentum of the first weight after the throw equals to 0. But why is it so clear that the skater and the skateboard will move in the same velocity? They don't say anything about any friction between the skater and the skateboard and if there's no friction then when the skater throws the weight he should move to the other direction in respect to the ground and to the skateboard. The skateboard should stay put because there is no horizontal force acting upon him. Am I wrong?

Also - isn't there something missing about the velocity of the second weight in order to solve the question?

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Doc Al

OK.
Why would you think that there's no friction between skateboard and skater? You are supposed to assume that they move together, just like they normally would. (But you're correct--if the skateboard where frictionless, it would just stay put. But it would be a pretty useless skateboard!)
Assume the same conditions apply for the second throw. Note that the velocity of the thrown weight is given with respect to the skater.

assaftolko

Well I don't know I mean you can solve the problem even if it's friction free and usually if there's friction they say it... Also if there's static friction I think you need to check if the max value of it can hold the skater in place and so you'll need to have the friction coefficient mue s... So if there's no mentioning of friction by the word or by stating the value of mue s I figgured there isn't any friction....

Ok thanks !

Doc Al

Don't make the problem harder than it is. (But good thinking though.)