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Introductory Physics Homework Help
Mass flow and rolling carts (Kleppner 3.11)
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[QUOTE="geoffrey159, post: 4944518, member: 532398"] [h2]Homework Statement [/h2] Material is blown into cart A from cart B at a rate b kilograms per second. The material leaves the chute vertically down- ward, so that it has the same horizontal velocity u as cart B. At the moment of interest, cart A has mass M and velocity v. Find dv/dt, the instantaneous acceleration of A. [h2]Homework Equations[/h2] Momentum [h2]The Attempt at a Solution[/h2] [/B] System studied: cart A and the chute. [B][/B] Let's put that ##M_A## is the mass of cart A (unloaded), ##m(t)## is the material's mass in cart A at time ##t##, and ## \triangle m = m(t+\triangle t) - m(t)## is a small amount of material falling from the chute into cart A in a ##\triangle t## seconds. At a given time ##t##, the horizontal momentum will be: ## P(t) = (M_A + m(t)) v(t) + \triangle m \ u(t) ## ## P(t+\triangle t ) = (M_A + m(t) + \triangle m) v(t+\triangle t) ## So that in time ##t##: ## \frac{dP}{dt} = (M_A + m) \frac{dv}{dt} + \frac{dm}{dt} (v - u)## In this system, only friction from the wheels of cart A contribute to horizontal external force : ##f_{ext} (t) = -\mu g (M_A +m(t)) ## Since at time of interest ##t_i##, we are given: ##M_a + m(t_i) = M##, ## v(t_i) = v##, ##u(t_i) = u##, ## \frac{dm}{dt} = b##, the acceleration should be: ##\frac{dv}{dt}(t_i) = \frac{1}{M_A + m(t_i)}(f_{ext}(t_i) + \frac{dm}{dt} (u - v) = -\mu g +\frac{b}{M} (u-v) ##Do you agree with this solution ? [/QUOTE]
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Introductory Physics Homework Help
Mass flow and rolling carts (Kleppner 3.11)
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