A cart running on frictionless air tracks is propelled by a stream of water expe

Question

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-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 473 kg. The water can be directed, by switching a valve, either backward or forward. In both directions, the pressure washer ejects 205 L of water per min with a muzzle velocity of 21.3 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?

c) What is the velocity of the cart at that time?

d) What is the thrust of this "rocket" before the valve is switched (enter first) and after the valve is switched.

e) What is the acceleration of the cart immediately before the valve is switched?

Explanation / Answer

(a) F = ma F is the thrust, this does not change throughout the problem. Let's turn to the excellent incite Anonymus shared in his answer for part (b) M1/M2 = M2/M3 since the mass will be less, the acceleration will be greater even though the force is the same, so to get the cart to be at rest at the end, we need to use this geometric mean (since eveything is to the first power) So the nozzle will be pointing one way longer since the cart is heavier and needs more impulse to get up to speed, and then pointing the other way for less time since the mass will be less and the cart, therefore, easier to accelerate. one cubic meter of water is (10cm*10)^3. One liter is (10cm)^3 = 1,000ml = 1 kg, so 1 m^3 = 1,000 kg M2 = SQRT(435*1435) = 790.1 kg this is the answer to (b) So, how long does it take to go from 1435kg down to 790.1kg if we are shooting out 161 l per minute or, see above, 161 kg/min distance equals rate times time mass loss equals rates times time t = Massloss/Rate = (1435-790.1)kg/161kg/min = 4.006 min (a) 4.005 min (b) solved above 790.1 kg (c) Anonymus also solved this part. I have nothing to add to his excellent work. V = Vexhaust ln(1435/790.1) = 15.57 m/s (d) Thrust never changes T = (rate of change in mass)*v exhaust Lets get the units to be the same 26.1 m/s is typical 161 kg/min = 161/60 kg/s T = 26.1 m/s*161/60*kg/s = 70.0 N (e) acceleration immediately before flipping the nozzle. We know the mass right then and we always know the Thrust which is the force, so we can easily find the acceleration. F = ma a = F/m a = 70.0N/790.1kg = 0.089m/s^2 Notes: Rocket cars aren't such a great idea, we shot out all of this water and the top speed is 15.6 m/s! Pedal cars seem way more efficient, and can be so much lighter.

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