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Unformatted text preview: Homework 9 Solution CHANG 1 Take both blocks A and B together as one system so that the spring force becomes internal and thus has no effect on the momentum of the system. The total momentum is then constant because there is no total force acting on the system. Also notice there is no friction so the total energy remains unchanged as well. Apply both conservation laws: COM: m A A m B B m A A m B B COE: 1 2 m A A 2 1 2 m B B 2 1 2 kx 2 1 2 m A A 2 1 2 m B B 2 The two blocks were held stationary initially, so A B 0, and there is no energy in the spring after the blocks have separated from each other. Plugging in the given data leads to the following simultaneous equations: m A 1.2 m B 4.8 1 2 1600 0.3 2 1 2 m A 1.2 2 1 2 m B 4.8 2 m A 20 Kg, m B 5 Kg Y&F 8.80 Consider the completely inelastic collision between the dart d and the sphere s . From conservation of momentum we have m d d m s s m d m s c We need to find the common velocity from the motion of the dart-sphere system after collision. For the combination to complete the loop, the tension in the wire has to be greater than zero, otherwise the wire goes slack and the sphere will not be able to continue the circular motion. The minimum velocity at the top of the loop can be found by making tension in the wire zero at the top of the loop, and with the weight as the only force producing centripetal acceleration, we have from Newtons 2nd law mg m 2 / R gR at the top. This is the same result that we studied for a particle completing a vertical loop along the circular track. Therefore the general conclusion is that for a particle to complete a vertical circular looptrack....
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