colisons 2016 Chega study lauded solnors and study Help lohegg A block with mass
ID: 1569864 • Letter: C
Question
colisons 2016 Chega study lauded solnors and study Help lohegg A block with mass mi o sso kg is released from rest on a treek at a distance 2.95 m above the top eratable.mthen eo deseancalv-n tables shown in the figure below. tretionless (a) Determine the vetooties of the two objects Just after the reaison. (Enter the megnt of How high up the track does the 0.550 object travel back after the colision? How far away from the bottom of the table does the 1.10 kg object land, given that the height the table is na as m? (d) How far away from the bottom ofthe table does the o sso-kg object eventualv ann Need Help?Explanation / Answer
velocity of m1 at bottom of ramp = sqrt(2 gh )= sqrt(2 x 9.8 x 2.95)
v0 = 7.60 m/s
For elastic collision,
velocity of approach = velocity of separation
v0 = v2 + v1
v2 = 7.60 - v1
Applying momentum conservation for the collision,
0.550 x 7.60 = 1.10 v2 - 0.550 v1
0.550 x 7.60 = 1.10(7.60 - v1) - 0.550 v1
v1 = 2.53 m/s
v2 = 5.07 m/s
(b) h = v^2 / 2 g
h = (2.53^2) / (2 x 9.8) = 0.327 m
(c) h2 = g t^2 / 2
t = sqrt(2 x 1.85 / 9.8) = 0.614 s
d = 5.07 x 0.614 = 3.12 m
(d) d = 2.53 x 0.614 = 1.55 m
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