Question

Problem 3: A bloc of mass m 2.20 Kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M-7.00 Kg which is at rest on a horizontal plane. If the collision is elastic and friction can be ignored, determine: The velocities of the two blocs after the collision, How far back up the incline, the smaller mass will go? 3.6 m M 30°

Answer 1

* Given data m= 2. 2kg M = 7 kg c = 30° h = 3.6 m @ speed of two blocks after collision or speed of m at bottom of incline . & by using conservation of Energy, 1. 1

migh = tme & V, = Jash - Vax9.8* 3.6 . Befor 4 = 8.40 misee y collision. Be for collision :- * & After collision :- » for inelastic collision Relative velocity before & after are same. so, &, &,'t

of mome- ÿ conservation ทำนุ - ตุง - 1 - คง 8.ms, 5 m ง) - m (V, -4) 8, 29, = (7 + m) ง 8. 9 = 29, ( +m) - 4x 2.2 x %.น 4.2 : 9 = 4.62 ราts, from ean O. 2. 9 - 9 - 9 = 8.4 - .02- 2. 9 = 4.38 ชา[seo

of Energy, ② conservation h=6932- 14 2x9.8 :on' = 0.98 m → S = h' - 0.98 sino sin 30 8 s'= 1.96 m up incline