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.
