- #1
Ocata
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- 5
Homework Statement
Block m1 has mass 10kg and 10m/s
Block m2 has mass 5kg and 0m/s
Energy loss is 0%
Homework Equations
Conservation of momentum
Conservation of energy
Quadratic Equation
The Attempt at a Solution
(mv)1i + (mv)2i = (mv)1f + (mv)2f
10(10) + 5(0) = 10(v)1f + 5(v)2f
[itex] v_{2f} = 20 - 2(v_{1f})[/itex] Then substituting v2f into the following...
Then
[itex]E_{i} = E_{f}[/itex]
[itex](\frac{1}{2}mv^{2})_{1i} + (\frac{1}{2}mv^{2})_{2i} = (\frac{1}{2}mv^{2})_{1f} + (\frac{1}{2}mv^{2})_{2f}[/itex]
[itex] 1000 = 10v^{2}_{1f} + 5v^{2}_{2f} [/itex]
[itex] 1000 = 10v^{2}_{1f} + 5(20 - 2v_{1f})^{2} [/itex]
[itex] 0 = 30v^{2}_{1f} - 400v_{1f} + 1000[/itex]
by quadratic formula:
[itex] v_{1f} = 10m/s [/itex] or 3.33m/s
Then [itex] v_{2f} = 20 - 2(3.33m/s) = 13.333333[/itex]
or
[itex] v_{2f} = 20 - 2(10m/s) = 0m/s[/itex]
So, the two possible sets of final velocities are:
Possibility 1) velocity of [itex] m_{1f} = 10m/s [/itex] and [itex] m_{2f} = 0m/s[/itex]
or
Possibility 2) velocity of [itex] m_{1f} = 3.333m/s [/itex] and [itex] m_{2f} = 13.333m/s[/itex]Neither case makes sense to me. In the first case, how can something traveling at 10m/s hit a stationary object and continue traveling at 10m/s while the stationary object remains stationary? Maybe I have it mixed up or something? Maybe the object traveling at 10m/s actually becomes stationary while the stationary object is propelled to 10m/s?
In the second case, how can an object traveling at 10m/s collide with an object and cause the object to travel faster than it? Is that even possible?
Thanks