# physics

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 1.0-m3 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 430 kg. The water can be directed, by switching a valve, either backward or forward. In both directions, the pressure washer ejects 210 L of water per min with a muzzle velocity of 26.0 m/s.
(a) If the cart starts from rest, after what time should the valve be switched from backward (forward thrust) to forward (backward thrust) for the cart to end up at rest?

(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!)
mass

velocity
magnitude
direction

(c) What is the magnitude of the thrust of this "rocket"?

(d) What is the acceleration of the cart immediately before the valve is switched?
magnitude
direction

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1. The mass decreases while the water jet is blasting, so the acceleration rate constantly increases. Then the thrust direction changes, but the magnitude of the acceleration continues to increase.

The trick to this problem is not to try to answer the questions in the order they are asked.

Let's do (c) first. It's easier that way. The thrust is
|Vexhaust*(dm/dt)|
= 26.0 m/s*(210 kg/60s) = 91 Newtons

The initial mass of the cart, with water, is 1430 kg. When all the water is gone, the final mass is 430 kg.

The velocity change during a period the jet is aimed in one direction is
deltaV = Vexhaust*ln (M1/M2).
Here's why:
Thrust = Vexhaust*dM/dt = M dV/dt
dV = Vexhaust*dM/M
V2 - V1 = Vexhaust*ln(M1/M2)

If you want the velocity change after thrust reversal to equal (but be in the opposite direction of) that before reversal, the intermediate mass M2 must satisfy
M1/M2 = M2/M3
M2 = sqrt(M1*M3) = sqrt(1430*430) = 784 kg. That answers part of (b).
For the velocity then, it is
Vexhaust*lm(M1/M2) = 26.0 ln(1430/784) = 15.6 m/s. So much for part (b)

Since you started with a mass of 1430 kg and dropped to 784 kg, the time to reverse thrust is after you have lost 646 kg at 3.5 kg/s, which means after 185 seconds. So much for part a.

For (d), use a = F/m, with M = 784 kg and F = 91 N
a = 0.116 m/s^2

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