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noname [10]
3 years ago
7

A 50 kg bumper car with a 40 kg child and it is at rest when a 60 kg child in her own bumper car slams into it the collision las

t point1 seconds and the rubber bumpers are perfectly bouncy with no heating from the collision what is the force experienced by the first car
Physics
1 answer:
IrinaK [193]3 years ago
4 0

Answer:

 F = 99 v₂₀

v₂₀ = 1 m / s,        F = 99 N

Explanation:

In this exercise it is asked to find the force during the collision, for this we use the relationship between the momentum and the momentum of car 1

            I = Δp

            F t = p_f- p₀

            F t = m (v_f -v₀)                        (1)

We must find the final speed of car 1, for this we define a system formed by the two cars, in this case the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

        p₀ = 0 + m₂ v₂₀

         

final instant. After the crash

        p_f = m₁ v₁ + m₂ v_{2f}

the moment is preserved

        p₀ = p_f

        m₂ v₂₀ = m₁ v_{1f} + m₂ v_{2f}           (2)

        m₂ (v₂₀ - v_2f}) = m₁ v_{1f}

as the collision is elastic the kinetic energy is also conserved

        K₀ = K_f

        ½ m₂ v₂₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

        m₂ (v₂₀² -v_{2f}²) = m₁ v_{1f}²

let's write our system of equations, using

         a² - b² = (a + b) (a-b)

         m₂ (v₂₀ - v_{2f}) = m₁ v_{1f}

         m₂ (v₂₀ -v_{2f}) (v₂₀ + v_{2f}) = m₁ v_{1f}²

to solve we divide the equations

       v₂₀ + v_{2f} = v_{1f}

with this we substitute in equation 2 and find the speed of each car, in this case we need the speed of car 1

         m₂ v₂₀ = m₁ v_{1f} + m₂ (v_{1f}-v₂₀)

         2m₂ v₂₀ = (m₁ + m₂) v_{1f}

          v_{1f} = \frac{2m_2}{m_1+m_2}  v_{2o}

We substitute in the drive ratio of car 1

            F t = m (v_f -v₀)

            F = m₁ (\frac{2m_2}{m_1+m_2}  v_{2o} - 0) / t

            F = \frac{2m_1 m_2 }{m_1+m_2}   \   \frac{v_{2o}}{t}

the mass of each car is the mass of the car plus the mass of the boy

           m₁ = 50 +40 = 90 kg

           m₂ = 50 +60 = 110 kg

     

time is t = 1

         

we substitute the values

           F = \frac{ 2\  90 \ 110}{90+110}  \ \frac{v_{2o}}{1}2 90 100/90 + 110 vo2 / 1

           F = 99 v₂₀

The value of the initial velocity of car 2 is not indicated in the problem, if this velocity is known it can be included and the force value is obtained, suppose that the initial velocity v₂₀ = 1 m / s

           F = 99 N

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