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ddd [48]
3 years ago
9

A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers

on is 67.0 kg, the mass of the skateboard is 4.10 kg, and the mass of the brick is 2.50 kg. If the person throws the brick forward (in the direction they are facing) with a speed of 23.0 m/s relative to the skateboard and we ignore friction, determine the recoil speed of the person and the skateboard, relative to the ground.
Physics
1 answer:
Inessa [10]3 years ago
6 0

Answer:

    v₁ = -0.8087 m / s

Explanation:

To solve this problem we can use the conservation of momentum, for this we define a system formed by the man, the skateboard and the brick, therefore the force during the separation is internal and the momentum is conserved

Initial instant. When they are united

        p₀ = 0

Final moment. After throwing the brick

        p_{f} = (m_man + m_skate) v1 + m_brick v2

the moment is preserved

        p₀ = p_{f}

        0 = (m_man + m_skate) v₁ + m_brick v₂

        v₁ = -  \frac{ m_{brick}   }{m_{man} + m_{skate}   }  v_{2}

the negative sign indicates that the two speeds are in the opposite direction

let's calculate

        v₁ = - \frac{2.5}{67 + 4.10}  23.0

        v₁ = -0.8087 m / s

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