Question

The figure below show three masses m₁=1.5 kg, m₂=2.7 kg, and m₃=4.6 kg which undergo two successive collisions. The first collision between m₁, which has an initial velocity v=8.6 m/s, and m₂ (which is initially at rest) is completely inelastic. The second collision between the combined mass m₁+m₂ and m₃ (which is initially at rest) is elastic. What is the velocity of m₃ after the second collision?

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Answer 1

Given: m = 1.5kg, m2=2 m2 = 2.7kg, mz= 4.6 kg, v= 8.6m/s. 3 ما به n 1 Since the collision between m, and me is inelostic, Hurs, after the collision m, sticks with me and moves with the combined mass moves with a a velocity (P) The momentum of m, tma" system before the collision is P,= mirt mq 10) = mir 10 m, me system after the is The momentum of u collision P2 = (mit mq) v' Conservation of momentum principle P,= - P2 miw = (ma+ m₂) v or on, w= mi v (1.5) (8.6) (15+2.7) 3.07m/s my + m2 Suppose after the second collision, the speed of mass mz is up and the speed of smitme is vg. The collision is elastic, this, te velocity of approact= the avelocity of separation v'LOE V3- V2 V3 - Vq = v' (i) OT, ΟΥ,

The collision is momentum of the system before the second P = (mitmq) vet m3 (0) = (mi+m2) v' The momentum of the system after the second collision is Pf= (mitm2) V2 + (m3) V3 conservation of momentum principle (m + m2) v = (ma+ m2) v2 + mala (m, & maov'= (m, + m 2) (V3- v') + m3 3 on (from eqh. (i)) on, q (mit mq) v'- (m + m2) v3 + m₂ V3 2 (mitmav' V3 = 2 (1.5+2.7)(3.0 mit mat mz 1.5+2.7 +4.6 O, 2.93 m/s