Solve Momentum Problem: Skateboarder Mass m1=48kg Jumps 2.9kg Board at Rest

[fishert16]

A Skateboarder with mass m1=48kg runs at 6.8m/s and jumps onto her skateboard with mass m2=2.9kg which is at rest at the bottom of a curved ramp. When the skateboarder lands on the board, how high does she and the skateboard go?

Relevant equations

w=[1/2(MVf^s)-1/2(MVi^2)]+[(MGHf-MGHi)]

The attempt at a solution

[(1/2(50.9)(0)^2)-(1/2(50.9)(6.8)^2)]+[(50.9)(9.8)(x)-(50.9)(9.8)(0)]
I found this to work out to
(0-1176.81)+(498.82x-0)
x=2.36m



I have worked out his equation and found Hf to be 2.36m I am just not sure if this is correct or if not could someone enlighten me on another equation or where i went wrong thanks.

 

[neutrino]

It would be better if you could show your work. You may get the right answer using wrong method, too.

 

[e(ho0n3]

There must be more to this problem. For example, how curved is the ramp? What about friction between the board and the ramp? Does the skateboarder jump onto her skateboard with the same running speed?

 

[fishert16]

I have edited to show the work and give a better view to show what i have done.

 

[neutrino]

A Skateboarder with mass m1=48kg runs at 6.8m/s and jumps onto her skateboard with mass m2=2.9kg which is at rest at the bottom of a curved ramp. When the skateboarder lands on the board, how high does she and the skateboard go?
[(1/2(50.9)(0)^2)-(1/2(50.9)(6.8)^2)]+[(50.9)(9.8)(x)-(50.9)(9.8)(0)]
I found this to work out to
(0-1176.81)+(498.82x-0)
x=2.36m

Where is it stated in problem that the person and the skateboard move at 6.8 m/s? You have titled this thread "Momentum problem," but it seems you have not considered momentum at all!

 

[fishert16]

Well that does not help me determine how to fix this problem. Should the velocity used in the problem be found from (MV1+MV2) = M(1+2)V(1+2) and solve for V1+2?

 

[neutrino]

Should the velocity used in the problem be found from (MV1+MV2) = M(1+2)V(1+2) and solve for V1+2?

Yes, that's right.

 

[fishert16]

so i found that to be 6.4m/s and i would plug this into where i intially had 6.8 and this should yield the correct answer

 

[neutrino]

so i found that to be 6.4m/s and i would plug this into where i intially had 6.8 and this should yield the correct answer

Yep. All that you are using is conservation of linear momentum in the first case, and the conservation of mechanical energy to the find the height.

 

[fishert16]

Thanks for the help