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A 48.0-kg boy, riding a 3.30-kg skateboard at a velocity of 5.10 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 6.00 m/s, 7.80° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

Respuesta :

lublana

Answer:

−7.2 m/s

Explanation:

We are given that

[tex]m_1=48 kg[/tex]

[tex]m_2=3.3 kg[/tex]

[tex]u=5.1 m/s[/tex]

[tex]v_1=6 m/s[/tex]

[tex]\theta=7.8^{\circ}[/tex]

We have to find the skateboards's velocity relative to the sidewalk at this instant.

Along x- direction

According to law of conservation of momentum

[tex](m_1+m_2)u=m_1v_1cos\theta+m_2v_2[/tex]

Substitute the values

[tex](48+3.3)\times 5.1=48\times 6cos(7.8)+3.3v_2[/tex]

[tex]261.63=285.40+3.3v_2[/tex]

[tex]3.3v_2=261.63-285.40[/tex]

[tex]3.3v_2=-23.77[/tex]

[tex]v_2=\frac{-23.77}{3.3}=-7.2m/s[/tex]

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