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Solutions11 - Physics 140 Discussion 11 Solutions 8.47...

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Physics 140 - Discussion 11 - Solutions 8.47. I DENTIFY : When the spring is compressed the maximum amount the two blocks aren’t moving relative to each other and have the same velocity V r relative to the surface. Apply conservation of momentum to find V and conservation of energy to find the energy stored in the spring. Since the collision is elastic, Eqs. 8.24 and 8.25 give the final velocity of each block after the collision. S ET U P : Let x be the direction of the initial motion of A . E XECUTE : (a) Momentum conservation gives (2 00 kg)(2 00 m/s) (12 0 kg) V . . = . and 0 333 m/s. V = . Both blocks are moving at 0.333 m/s, in the direction of the initial motion of block A . Conservation of energy says the initial kinetic energy of A equals the total kinetic energy at maximum compression plus the potential energy b U stored in the bumpers: 2 2 1 1 b 2 2 (2 00 kg)(2 00 m/s) (12 0 kg)(0 333 m/s) U . . = + . . and b 3.33 J. U = (b) 2 1 2 00 kg 10 0 kg (2 00 m/s) 1 33 m/s. 12 0 kg A B A x A x A B m m v v m m - . - . = = . = - . + . Block A is moving in the x - direction at 1.33 m/s.
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