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A 50 kg boy jumps off the front of a 1.5 kg skateboard moving forward. Find the skateboard’s velocity immediately after the boy jumps, assuming that the skateboard’s initial velocity is 3.5 m/s and the boy’s velocity when jumping off the front is 5 m/s.

Respuesta :

Here in this case we can say that there is no external force on skate board and boy as a system

so here we will use the theory of momentum conservation

So we will say

initial momentum of boy + board = final momentum of boy + board

[tex](m_1 + m_2)v_i = m_1v_1 + m_2v_2[/tex]

[tex](50 + 1.5)(3.5) = 50(5 + v_2) + 1.5v_2[/tex]

[tex]180.25 = 250 + 51.5 v_2[/tex]

[tex]180.25 - 250 = 51.5 v_2[/tex]

[tex]-69.75 = 51.5 v_2[/tex]

[tex]v_2 = -1.35 m/s[/tex]

so here the skateboard will move off in opposite direction with speed 1.35 m/s


krlosdark12

Answer:

the skate board velocity is -1.35 m/s

Explanation:

Here we are goin to use the principle of conservation of linear momentum that states:

[tex]m1*vo1+m2*vo2=m1*vf+1m2*vf2[/tex]

that is:

[tex](50kg+1.5kg)*3.5m/s=50kg*Vf1+1.5Kg*Vf2[/tex]

the velocity of the boy is 5m/s relative to the skate:

[tex]Vbs=Vb-Vs\\Vb=Vbs+Vs=5m/s+Vf2[/tex]

[tex](50+1.5)*3.5=50*(5 +vf2)+1.5*Vf2\\180.25=250+(51.5)Vf2\\vf2=\frac{-69.75}{51.5}=-1.35m/s[/tex]

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