A cart running on frictionless air tracks is propelled by a stream of water expelled by a gas-powered pressure washer stationed on the cart. There is a 1.00-m^3 water tank on the cart to provide the water for the pressure washer. The mass of the cart, including the operator riding it, the pressure washer with its fuel, and the empty water tank, is 355 kg, the water can be directed, by switching a valve, either backward or forward. In both directions, the pressure washer ejects 179 L of water per min with a muzzle velocity of 24.5 m/s

a) If the cart starts from res, after what time should the valve be switched from backward (forward thrust) to forward (backward thrust) for the cart to end up at res?
b) What is the mass of the cart at that time, and what is its velocity? (Hint: it is safe to neglect the decrease in mass due to the gas consumption of the gas-powered pressure washer!)
c) What is the thrust of this “rocket”?
d) What is the acceleration of the cart immediately before the valve is switched?

a) 112 s
b)m= 694 kg v= 16.4 m/s
c) 145 N
d) 0.209m/s^2

I know that c is supposed to be 81.6 N but I don't know how to get that answer... and that will make my answer for d wrong too... can someone please tell me how to do this?

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1. Let's do (c) first.
Thrust = Ve*dm/dt
where Ve is the exhaust velocity relative to the cart and dm/dt is the mass loss rate, 179kg/60s = 2.983 kg/s.
The thrust is 2.983*24.5 = 73.1 N

The water will be used up in time T given by
10^3 kg/2.983 kg/s = 335 seconds

The car will decelerate faster than it accelerates, since it is heavier when accelerating.

The direction of thrust should be reversed when the mass of the cart with liquid is the geometic mean of initial and final mass. That is because the velocity change equals Vexhaust*ln(Minitial/Mfinal).
You need the same Minitial/Mfinal ratio when pointed in each direction.

I hope you see why.

The time to reverse direction is when
M = sqrt(1355*355) = 693 kg, which corresoonds to a mass loss of 1355-693 = 662 kg. This will require 222 seconds.

The velocity of the cart at that time is
Vexhaust*ln(Minitial/M) = 24.5*ln(1355/662) = 17.5 m/s

(d) a = F/M = 73.1N/693kg = 0.105 g

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2. Can you write the equations that you exactly used for each problem

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3. how did you get the number 1355 for the sqrt of mass

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