A block of mass m1 = 2.0 kg slides along a frictionless table with a speed of 10

Question

A block of mass m1 = 2.0 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction with a speed of 3.0 m/s, is a block of mass m2 = 5.0 kg. A massless spring that has a force constant k = 1120 N/M is attached to the second block (a) What is the velocity of the center of mass of the system? (b) During the collision, the spring is compressed by a maximum amount ?x. What is the value of ?x? (c) The blocks will eventually separate again. What are the velocities of the two blocks after they separate?

Explanation / Answer

The definition of elastic vs. inelastic is not really determined by whether the objects stick together or not. It's actually determined by whether the kinetic energy (the sum of all the ½mv² terms) is conserved (i.e. is the same before & after). I think you can assume that some of the KE is _temporarily_ lost as the spring compresses (KE turns into PE); but then as the spring releases, all that PE is turned back into KE; so that ultimately the KE is conserved. But don't focus on whether this collision is elastice or inelastic; that's irrelevant here. Approach it this way: There are actually 3 important "phases" to this interaction: (1) the time before the collision; (2) the moment when the spring is completely compressed; (3) the time after they've separated. The things to notice here are: * Momentum is ALWAYS conserved (whether the collision is elastic or not). That means the momentum will be the same in phase 1, phase 2, and phase 3. * During phase 2, the two blocks are (just for an instant) traveling at the same speed. (Call that speed "Vc" ("c" for "compressed").) * Kinetic energy is lost during Phase 2 (we know that's true because the spring gains PE, and that energy has to come from somewhere) * We can use conservation of momentum to compare the speeds during Phase 1 to those during Phase 2. From the speeds, we can compare the KE between phase 1 and 2; and that will tell us how much KE was lost; and that will tell us how much PE was put into the spring; and that will tell us how far the spring was compressed. Okay. Total momentum during Phase 1: (m1)(v1_initial) + (m2)(v2_initial) Total momentum during Phase 2: (m1)(Vc) + (m2)(Vc) Momentum is conserved, so: (m1)(v1_initial) + (m2)(v2_initial) = (m1)(Vc) + (m2)(Vc) Solve for Vc: Vc = [(m1)(v1_initial) + (m2)(v2_initial)] / (m1+m2) Now look at KE: KE in Phase 1: KE1 = ½m1(v1_initial)² + ½m2(v2_initial)² KE in Phase 2: KE2 = ½m1(Vc)² + ½m2(Vc)² = ½(m1+m2)(Vc)² Substitute the value for Vc that we calculated above: KE2 = ½(m1+m2)([(m1)(v1_initial) + (m2)(v2_initial)] / (m1+m2))² = ½[(m1)(v1_initial) + (m2)(v2_initial)]² / (m1+m2) You are given the values of m1, m2, v1_initial and v2_initial, so you can plug those values to calculate the value of KE1 and of KE2. Note they WON'T be the equal to each other; you can expect KE2 will be less than KE1 because some KE was lost in order to give PE energy to the spring. Decrease in KE = Increase in PE: KE1 - KE2 = PE_spring = ½k(?x)² You just calculated KE1 and KE2, and you are given "k". Solve for "?x". Hope this will get you started. For part (c), I would start by looking at it from the reference frame of the center of mass (which is moving constantly to the right at speed Vc). From the point of view of that reference frame, each object exactly reverses its motion after the collision. Figure out what speeds are required to make that happen, and then convert that to the reference frame of the table.

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