Speed Chute
question on the inelastic and elastic collisions?
A package of mass m is released from rest in a warehouse loading dock and ah = 3.0 sliding friction m high drop-out truck. Unfortunately, the truck driver took a break without having removed the previous package, 2m mass from the bottom of the waterfall. (A) Assume the packages attached. What is their common velocity after the collision? "(B) Assume that the collision between packages is elastic. How high is the recovery package of mass m?
(A) v ^ 2 = 2x 9.8 x 3 v = 7.67 m / s befor ecollision Momentum = 7.67 m. after the collision Travel Packages allow bonded with V (2 mm) V = 7.67 V = 7.67 m / 3 = 2.56 m / s (b) mv = 2m + MV1 v2 or v1 v2 = v 2 – ———– 1 and 0.5 mV ^ 2 = 0.5 MV1 ^ 2 + o, 5 x 2 m v2 ^ 2 or v ^ 2 = v1 ^ 2 + 2 v2 ^ 2 ————————- ———- ———- ——– 2 squares an equation v ^ 2 = v1 ^ 2 + ^ 2 + 4V2 4V1 + 2v2 v2 = v1 ^ 2 ^ 2 Form 2 or 2 V2 ^ 2 + 4 = 0 or v1 v2 v2 + 2V1 = 0, then V1 = – V2 / 2 o v2 =- 2V1 the equation of v = V1 – 4V1 or V1 = – v / 3 and v2 = 2v / 3 bounce to the height H 2 g H = v ^ 2 / 9 and we know v ^ 2 = 2 gx 3 or 2 g H = 6 g / cm (9) or H = 3.1 = 33.3 m
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