A 2.0-g particle moving at 5.2 m/s makes a perfectly elastic head-on collision w

Question

A 2.0-g particle moving at 5.2 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. 2.0 g particle m/s 1.0 g particle m/s (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. 2.0 g particle m/s 10.0 g particle m/s (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b). KE in part (a) J KE in part (b) J In which case does the incident particle lose more kinetic energy? case (a) case (b)

Explanation / Answer

(a) conservation of momentum: pi = pf [m1v1 + m2v2]i = [m1v1 + m2v2]f (.002)(5.2) + .001(0) = (.002v1) + (.001)(v2) .0104 = .002v1 + .001v2 .002v1 = .0104 - .001v2 v1 = 5.2 - .5v2 conservation of KE: KEi = KEf 0.5m1v1^2 + 0.5m2v2^2 = 0.5m1v1^2 +0.5m2v2^2 0.5(.002)(5.2^2) + 0.5(.001)(0) = 0.5(.002)(v1^2) + 0.5(.001)(v2^2) .02704 = .001(5.2-.5v2)^2 + (5e-4)(v2^2) .02704 = .001(27.04 - 5.2v2 + .25v2^2) + 5e-4(v2^2) v2 = 6.933 m/s v1 = 1.85 m/s speed of 2g particle: 1.85 m/s speed of 1g particle = 6.93 m/s (b) conservation of momentum: pi = pf [m1v1 + m2v2]i = [m1v1 + m2v2]f (.002)(5.2) + .01(0) = (.002v1) + (.01)(v2) .0104 = .002v1 + .01v2 .002v1 = .0104 - .01v2 v1 = 5.2 - 5v2 conservation of KE: KEi = KEf 0.5m1v1^2 + 0.5m2v2^2 = 0.5m1v1^2 +0.5m2v2^2 0.5(.002)(5.2^2) + 0.5(.01)(0) = 0.5(.002)(v1^2) + 0.5(.01)(v2^2) .02704 = .001(5.2-5v2)^2 + (5e-3)(v2^2) .02704 = .001(27.04 - 52v2 + 25v2^2) + 5e-3(v2^2) v2 = 6.41e-4 m/s v1 = 5.20 m/s speed of 2g particle: 5.20 m/s speed of 10g particle = 6.41e-4 m/s (c) KE(a) = .027 J KE(b) = .027 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.