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A 10.0kg block of ice is sliding due east at 8.00 m/s when it collides elastically with a 6.00kkg block of ice that is sliding in the same direction at 4.00 m/s. Determine the velocities of the blocks of ice after the collision?
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Added Mon, 22 Jun '15
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Conserve momentum: initial p = final p
10kg * 8m/s + 6kg * 4m/s = 10kg * u + 6kg * v
104 = 10u + 6v

For an elastic, head-on collision, we know (from conservation of energy), that
the relative velocity of approach = relative velocity of separation, or
8m/s - 4m/s = v - u
v = u + 4m/s
Plug that into the momentum equation:
104 = 10u + 6(u + 4) = 16u + 24
u = 5 m/s  ( 10 kg block )
v = u + 4m/s 
v = 9m/s  ( 6 kg block )
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Added Mon, 22 Jun '15
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