Two hockey pucks approach each other as shown in the figure below. Puck 1 has an

Question

Two hockey pucks approach each other as shown in the figure below. Puck 1 has an initial speed of 21 m/s, and puck 2 has an initial speed of 10 m/s. They collide and stick together. (a) If the two pucks form a "system," is the momentum of this system along x or y conserved? (b) Find the components along x and y of the initial velocities of both particles. v1xi = m/s v1yi = m/s v2ix = m/s v2iy = m/s (c) Express the conservation of momentum condition for motion along x and y. (Do this on paper. Your instructor may ask you to turn in this work.) (d) Is this collision elastic or inelastic? (e) Find the final velocity of the two pucks after the collision. vf = m/s ? =

Explanation / Answer

Q. Two hockey pucks approach each other as shown in the figure below. Puck 1 has an initial speed of 30 m/s, and puck 2 has an initial speed of 6 m/s. They collide and stick together. (a) If the two pucks form a "system," is the momentum of this system along x or y conserved? I found YES they form a system X is horizontal and Y is Vertical (b) Find the components along x and y of the initial velocities of both particles. v1xi = v1yi = v2ix = v2iy = Ok, here the obvious would be that v2xi would be 0 and v2xy would be 6 m/s I don't understand how to find the v1xi and v1xy with the angle and speed? Could someone please show me? (c) Express the conservation of momentum condition for motion along x and y. (Do this on paper. Your instructor may ask you to turn in this work.) I drew a diagram of what would happen and what would happen to each puck... I think this is alright? (d) Is this collision elastic or inelastic? It would be inelastic collision because there wasn't a bounce, right? (they stuck together) (e) Find the final velocity of the two pucks after the collision. vf = ? = ° counterclockwise from the +x-axis Not really sure what I can do with this one.... Especially, not without the V1i and v1f (f) What fraction of the initial kinetic energy is lost in the collision? Please help me, I have been struggling all day on this problem and it really would make my night if someone would give me a hand here. Thank you. (f) What fraction of the initial kinetic energy is lost in the collision? ANS: a) Correct. You're off to a good start! b) Treat the x & y directions as independent from each other. In general, Vx = V*cosT and Vy = V*sinT, where T is the CCW direction from the + x axis. With this in mind, V1xi = 30*(cos-30°) = 25.98 m/s V1yi = 30*(sin-30°) = -15 m/s V2ix = 6*cos90° = 0 V2iy = 6*sin90° = 6 c) SPxi = SPxf.........SPyi = SPyf d) Correct. When they stick together, it's inelastic. e) Find the x & y components of V and combine them with the Pythagorean theorem: Since the masses are equal, they cancel out, and we can just assume they are 1.0: From (c), SPxi = 25.98 + 0 = 2*Vfx ? Vfx = 12.99 m/s SPyi = -15 + 6 = 2Vfy ? Vfy = -4.5 m/s The '2' is because the final mass is twice the original individual masses So, Vf = v[Vfx²+Vfy²] = 13.747 m/s T = arctan[Vfy/Vfx] = arctan[-4.5/12.99] = -19.11° f) Flost = 1.0 - KEf/KEi = 1.0 - (2*13.747²)/(30² + 6²) = .60

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