Review | Constants Block 1, of mass m1 = 6.90 kg , moves along a frictionless air track with speed vi = 17.0 m/s. It collides with block 2, of mass m2 = 53.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1) Part C What is the change AK = Kfinal - Kinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules. Figure 1 of 1 • View Available Hint(s) Before collision: m2 m AK = 938.348 J Submit Previous Answers After collision: X Incorrect; Try Again; 4 attempts remaining Provide Feedback Next > 國

Please Asap

Part A A block of mass 2.55 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass 7.60 kg which is at rest on a horizontal surface. (Assume a smooth transition at the bottom of the incline.) (Figure 1) If the collision is elastic, and friction can be ignored, determine the speed of the block with mass 2.55 kg after the collision. v'm = ΕΠΙ ΑΣΦ Submit Request Answer Part B ? m/s If the collision is elastic, and friction can be ignored, determine the speed of the blockwith mass 7.60 kg after the collision. = 跖 ΑΣΦ Submit Request Answer ་ Part C How far back up the incline the smaller mass will go. ΜΕ ΑΣΦ d = Submit Request Answer ? ? m m/s

Need help checking a practice problem. Image attached, thanks!