Question

Chapter 07, Problem 63 Two ice skaters have masses m, and m2 and are initially stationary. Their skates are identical. They push against one another, as in Figure 7.10, and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater 2. What is the ratio m/m2 of their masses? Number Units the tolerance is +/-2%

Answer 1

* From Mewton's Law of nioton] V2-43 + 2*ars --- ---- () ve final speed u = initial speaal a: Acceleration s = displacement thence for Skate 1!- Ví? = 0 +2+Q2 S for skate ai 12 + 2x 42 x Sz ť from eqn ci) 1 eqnwis} 2x91x5, 2xQ2* S2 ---- Lill a =92a ) S, : 252 13- 2x2x252 29x S2 = 2 12 liv) from Conservation of moonutuna meu,+Mauz = maria mara . ..(V) o+0= mat(V1) + m2* (V2) Lo cover sign, because they Start moving in opposite olimection -Vi mirea Mare a warn] =67] - ml 0170- 7 mz A (NO UNZT) (UNZTLESS)