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A 44 kg child jumps off a 2.2 kg skateboard that was moving at 8.0 m/s. The skateboard comes to a stop as a result. Find the speed at which the child jumped from the board. Show all your work. Assume a frictionless, closed system.

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jorgezarate

Answer:

[tex]v_f=8.4m/s[/tex]    

Explanation:

We need to use the conservation of momentum Law, the total momentum is the same before and after the kid leaves the skate.

In the axis X:

[tex](m_{kid}+m_{skate})*v_{o}=m_{kid}*v_f[/tex]     (1)

[tex]v_f=(m_{kid}+m_{skate})*v_{o}/m_{kid}=(44+2.2)*8/44=8.4m/s[/tex]  

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