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Fall 2008 - Spelio (solutions)

# Fall 2008 - Spelio (solutions) - πn ° 2 m k x t = A °...

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1 Problem 3 v is a velocity of the system after the collision : v = v 0 2 x = A sin ( ω t + φ ) If x = 0 - the initial position of blocks at t = 0 then φ = 0, ω = k 2 m 2 m v 0 2 2 2 = kA 2 2 A = mv 2 0 2 k x = mv 2 0 2 k sin ( k 2 m t ) b) to stop the system mv 0 + 2 mv = 0 v = v 0 2 v = ˙ x ( t ) = mv 2 0 2 k k 2 m cos k 2 m T = v 0 2 cos k 2 m T = 1 k 2 m T = π + 2 π n T = ( π + 2 π n ) 2 m k It also can be easily seen from the following fact: since the speed should be v 0 2 then the point of contact of all blocks should be at x = 0 and two blocks should move to the left. The first possible strike is after the half of the period, the second after T 2 + T and so on... Since T = 2 π ω then T strike = ( π + 2 π n ) 2 m k x ( t ) = 0 for t > T c c) The maximum speed of the system is when there is the maximum initial momentum. The maximum momentum corresponds to the case when two carts are moving to the right (through the x = 0). So T

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Unformatted text preview: πn ° 2 m k x ( t ) = A ° sin ω ° t 2 ω ° = ° k 3 m mv + 2 m v 2 = 3 mv f v f = 2 3 v 3 m ( 2 3 v ) 2 2 = kA ° 2 2 A ° = 2 v ° m 3 k x ( t ) = 2 v ° m 3 k sin ± ° k 3 m t ² 1 Problem 5 γ mM r 2 1 − T = mω 2 r 1 γ mM r 2 2 + T = mω 2 r 2 γ mM r 3 1 − T r 1 = γ mM r 3 2 + T r 2 T = γmM r 3 2 − r 3 1 ( r 1 + r 2 ) r 2 1 r 2 2 r = r 1 + r 2 2 , r 1 = r − l 2 , r 2 = r + l 2 T = γmM r 3 ° (1 + l 2 r ) 3 − (1 − l 2 r ) 3 ± 2 r 5 (1 − l 2 r ) 2 (1 + l 2 r ) 2 T = γmM ° 1 + 3 l 2 r − 1 + 3 l 2 r ± (1 + 2 l 2 r )(1 − 2 l 2 r ) 2 r 2 T = 3 γmMl 2 r 3 (1 − l 2 r 2 ) T = 3 γmMl 2 r 3...
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