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  • one year ago

Initially, the block whose mass m = 1.0 kg and the block whose mass is M are both at rest on a frictionless inclined plane at 30 degrees from the ground. The block of mass M rests against a spring that has a spring constant of 11,000 N/m. The distance along the plane between the two blocks is 4.0 m. The block of mass m is released, makes an elastic collision with the block of mass M, and rebounds a distance of 2.56 m back up the inclined plane. The block whose mass is M comes to rest momentarily 4.0 cm from its initial position. Find M.

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  1. Shadowys
    • one year ago
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    Considering that there is no friction, and there is obvious loss in potential energy of m, and so the energy will be converted to the only available place to store energy, the spring, thus spring potential energy. with that in mind, the total change in mechanical energy(for elastic collision) \(mg \Delta h_1 = \frac{1}{2} k \Delta x^2 +Mg\Delta h_2 \) where \(\Delta h =\Delta d \sin \theta\), assuming that 4cm that M moved is downwards, so, \((1)(9.8)(4-2.56)(\sin 30^o)=\frac{1}{2}(11000)(0.04)^2 + M(9.8)(-0.04)\) find M.

  2. Shadowys
    • one year ago
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    sorry, it's \(M(9.8)(-0.04)(\sin 30^o)\) for the last term.

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