Solve Momentum Problem: Skater w/60kg Mass+3kg Skateboard

[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?

 

[Doc Al]

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.

OK.

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?

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!)

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

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]

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!)

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...

Assume the same conditions apply for the second throw. Note that the velocity of the thrown weight is given with respect to the skater.

Ok thanks !

 

[Doc Al]

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

 

[Nickie]

I would approach this problem by first clarifying some assumptions and variables. Is the ground frictionless or is there some resistance that could affect the velocity of the skater and skateboard? Is the skater on a flat surface or is there an incline? These factors could affect the outcome of the problem.

Assuming that the ground is frictionless and the skater is on a flat surface, we can apply the principles of conservation of momentum to solve this problem. The initial momentum of the system (skater+skateboard+weights) is zero since it is at rest. After the first throw, the momentum of the first weight is 5 kg x 8 m/s = 40 kg*m/s in the positive direction. According to the law of conservation of momentum, the momentum of the skater+skateboard must also be 40 kg*m/s in the negative direction in order to balance out the system's momentum to zero.

Therefore, after the first throw, the skater and skateboard will have a total momentum of -40 kg*m/s. Now, when the second weight is thrown, the total momentum of the system must still be conserved. The second weight will have a momentum of 5 kg x 8 m/s = 40 kg*m/s in the positive direction. Therefore, in order to balance out the system's momentum to zero, the skater and skateboard must have a momentum of -40 kg*m/s in the negative direction.

This means that the skater's velocity after the second throw will be in the negative direction at the same speed as the first throw, which is 8 m/s. The skateboard will also have the same velocity as the skater since they are connected and there is no external force acting on the skateboard.

In conclusion, the velocity of the skater after the second throw will be -8 m/s, assuming the ground is frictionless and the skater is on a flat surface. However, if there are other factors involved, such as friction or incline, the velocity could be different. More information or clarification on these variables would be needed to accurately solve the problem.