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A 45 kg boy runs at a rate of 2.5 m/s and jumps on top of a stationary skateboard that has a mass of 4 kg. After jumping onto the board, the boy and the board travel off together. Assuming that momentum is conserved, what is the final speed of the boy and the board?

Respuesta :

Answer:

v = 2.29 m/s

Explanation:

As we know that the external force on the system of mass of boy + board is ZERO

So here we can use momentum conservation

now we have

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

now we have

[tex]45 (2.5) + 4(0) = (45 + 4) v[/tex]

now we have

[tex]v = \frac{45}{49} (2.5)[/tex]

[tex]v = 2.29 m/s[/tex]

Answer:

V = 2.29 m/s

Explanation:

Given that,

Mass of the boy, [tex]m_1=45\ kg[/tex]

Mass of the skateboard, [tex]m_2=4\ kg[/tex]                            

Initial speed of the boy, v = 2.5 m/s

Let V is the final velocity of the boy and the board. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :

[tex]45\times 2.5=(45+4)V[/tex]

[tex]V=\dfrac{45\times 2.5}{(45+4)}[/tex]

V = 2.29 m/s

So, the velocity of the boat after Batman lands in it 2.29 m/s. Hence, this is the required solution.                                                                

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