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## Homework Statement

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.

## Homework Equations

Momentum

## The Attempt at a Solution

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System studied: cart A and the chute.

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 ?