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nadya68 [22]
3 years ago
14

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 th

e 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?
Physics
2 answers:
Natalija [7]3 years ago
7 0

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

m_1v_1 + m_2v_2 = (m_1 + m_2) v

now we have

45 (2.5) + 4(0) = (45 + 4) v

now we have

v = \frac{45}{49} (2.5)

v = 2.29 m/s

igomit [66]3 years ago
4 0

Answer:

V = 2.29 m/s

Explanation:

Given that,

Mass of the boy, m_1=45\ kg

Mass of the skateboard, m_2=4\ kg                            

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 :

45\times 2.5=(45+4)V

V=\dfrac{45\times 2.5}{(45+4)}

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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