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Thu 1 Dec, 2005 03:13 pm
Okay you smarties... please help a confused college student out!
You find yourself on a sheet of absolutely frictionless ice, carrying a 5.5 kg bowling ball. You are 18.5 meters from the nearest land and your mass is 75 kg. In order to get off the ice, you throw the ball with a velocity of 3.0 m/sec in a direction directly opposite the land. How long after throwing the ball will you reach land?
Would mv+mv = mv + mv be the right setup? (The first mv being for the 1st mass and its inital velocity. The second mv for the 2nd mass and its initial velocity. The third mv for the 1st mass and its final velocity. The fourth mv for the 2nd mass and its final velocity.)
Doing this I got -.22 m/s. Then I don't know what to do next to find the time.
Can I use rate x time = distance?
Very cool question. I like how they emmited to tell you that at the begining you are STATIONARY therfore u=0
So here is my solution.
Your mass START: 75 + 5.5
= 80.5kg
;; I took ----> as Positive ;;
(m + m)u = mv + mv
(75 + 5.5)(0) = (75)(v) + (5.5)(-3)
0 = 75v - 16.5
16.5 = 75v
0.22m/s = v
Therefore 0.22 m/s to the right. Obviously I assumed that the land is to your right but it doesn't matter.
s = (u + v)/2 (t)
18.5 = (0 + 0.22)/2 (t)
18.5 = 0.11t
168.18s = t
So it will take you 2 minutes and 48.18 seconds to reach the land.
-Reddy Red
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