ch15-p095 - 95. (a) By Eq. 15-13, the mass of the block is...

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T mm k pb = + = 20 π .44 . s (b) The speed before the collision (since it is at its maximum, passing through equilibrium) is v 0 = x m ω 0 where 0 = 2 π / T 0 ; thus, v 0 = 3.14 m/s. Using momentum conservation (along the horizontal direction) we find the speed after the collision. = + =2.61 . 0 Vv m b m/s The equilibrium position has not changed, so (for the new system of greater mass) this
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