An object of mass 0.40 kg is compressed 0.15 m against a horizontal spring of fo

Question

An object of mass 0.40 kg is compressed 0.15 m against a horizontal spring of force constant 520 N/m. The object is then released from rest and moves without friction. After it leaves the spring, it hits a stationary, 0.20-kg object head on. The objects stick together after the collision. a) Find the speed of the 0.40-kg object just before it hits the other object. b) Find the velocities of both objects after the collision. c) Determine how much energy was lost in the collision. Answers are: A)5.41 m/s B) 3.61 m/s C) 1.94 J Please show how you solved the problem. Thanks!

Explanation / Answer

Alright, so for the first part, energy laws can be used because any energy gained in kinetic was lost in elastic energy. We are looking for v1 KE = 1/2mv^2 EE = 1/2kx^2 KE = EE 1/2mvi^2 = 1/2kx^2 Insert quantities as you see fit. Second part: Because they're sticking together and they were not hit at each other from an angle, it is a completely inelastic collision. So, because this is a closed system, we can use momentum m1v1 + m2v2 = (m1+m2)(vf) Because the second mass was stationary, we can get rid of m2v2 so that m1v1 = (m1+m2)(vf) To find work, we need to use the fact that work is equal to the change in kinetic energy so that W = 1/2(m1+m2)(vf)^2 - 1/2(m1)(vi)^2 Have any problems, let me know.

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