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A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash. If we consider this event to be a perfectlyinelastic collision, what is the speed and direction of the cars after the impact?a

Sagot :

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

[tex]\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}[/tex]

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

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