Question

A 1800 kg car moving south at 17.5 m/s collide with a 2800 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.22 m/s to the north. Find the velocity of the 2800 kg car before the collision

Answer 1

Answer:

   v₀₂ = -2.67 m / s

Explanation:

Let's use the conservation of the moment, for this we define a system formed by the two cars.

Consider the north direction as positive and the subscript 1 will be used for car 1 and the subscript 2 for the second car

Initial instant. Before the crash

          p₀ = -m₁ v₀₁ + m₂ v₀₂

Final moment. Right after the crash

          p_f = (m₁ + m₂) v

           p₀ = p_f

           -m₁ $v_{o1}$ + m₂ v_{o2} = (m₁ + m2) v_{f}

         v₀₂ = $\frac{(m_1 +m_2) v }{m_2}$+  $\frac{m_1 v_{o1} }{m_{1} }$

 

let's calculate

          v₀₂ = $\frac{(1800 +2800) 5.22 + 1800 (-17.5)}{2800}$

          v₀₂ = -2.67 m / s

the negative sign indicates that the carriage moves in the opposite direction of the temperies