MT2 F05 w sln - Problem 2 (a) A (b) A B B ∆x ∆x vo H C...

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Unformatted text preview: Problem 2 (a) A (b) A B B ∆x ∆x vo H C v h v (a) Total energy is conserved, since there are no dissipative forces (friction). k ∆x2 EA = and EB = mgH 2 k ∆x2 k ∆x2 EA = EB =⇒ = mgH =⇒ H = 2 2mg (b) The block keeps going up the incline until the wedge and the block reach the same velocity, call it v. When this happens, the block is stationary relative to the wedge. Since there are no dissipative forces, the energy of the system throughout the process is conserved: EA = EB = EC (see Fig. 2). (m + M )v 2 k ∆x2 = mgh + (1) 2 2 Also, since there are no external forces acting in the horizontal direction after the block stops interacting with the spring, the horizontal momentum of the system is conserved: pB = pC . m vo mvo = (m + M )v =⇒ v = m+M Substituting v in equation 1, we get: EA = EC =⇒ k ∆x2 m2 2 − vo (2) 2 2(m + M ) Now we find vo , the speed with witch the block leaves the spring, in terms of the known quantities from conservation of energy from A to B: mgh = EA = EB =⇒ 2 k ∆x2 (m)vo = 2 2 2 k ∆x 2 vo = m Substituting vo in equation 2, we get: mgh = k ∆x2 m k ∆x2 M (1 − ) =⇒ h = · 2 m+M 2mg m + M ...
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This note was uploaded on 03/09/2010 for the course PHYSICS 7A taught by Professor Lanzara during the Fall '08 term at University of California, Berkeley.

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MT2 F05 w sln - Problem 2 (a) A (b) A B B ∆x ∆x vo H C...

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