Question

A mass of 5 kg is moving at 14 m/s in the +x direction and it collides inelastically with a mass of 7 kg moving in the -x direction. The collision takes places in 0.38 seconds and the average force on the mass moving in the +x direction is 232 Newtons in the -x direction. How much total kinetic energy is lost in the collision in Joules? Answer is a positive number since I'm asking how much is lost.

Answer #1

**m1 = 5 kg**

**v1x = 14 m/s**

**m2 = 7 kg**

**v2x = ?**

**after collsion**

**v1'x = ?**

**v2'x = ?**

**Force applied on m1**

**F1 = m1*(v1'x-v1x)/t**

**-232 = 5*(v1'x - 14)/0.38**

**v1'x = -3.632 m/s**

**final velocity of m1 , v1'x = -3.632 m/s**

**final velocity of m2 , v2'x = -3.632 m/s**

**v1'x = v2'x = v = -3.632 m/s**

**from momentum conservation**

**mometum before collision = momentum after
collision**

**m1*v1x + m2*v2x = (m1 + m2)*v2'x**

**(5*14) + (7*v2x) = (12)*(-3.632)**

**v2x = -16.23 m/s**

**initial energy Ki = (1/2)*m1*v1x^2 +
(1/2)*m2*v2x^2**

**Ki = (1/2)*5*14^2 + (1/2)*7*16.23^2 = 1412 J**

**final energy Kf = (1/2)*(m1+m2)*v^2**

**Kf = (1/2)*12*3.632^2 = 9.15 J**

**energy lost = 1332.85 J**

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