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?
My answers are:
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?
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
Can you write the equations that you exactly used for each problem
how did you get the number 1355 for the sqrt of mass
To solve this problem, we need to consider the principles of conservation of momentum and fluid mechanics. Let's break down the problem and go through each step:
a) To determine when to switch the valve from backward to forward, we need to find the time it takes for the cart to come to rest. We can use the principle of conservation of momentum. Initially, the cart is at rest, so the total momentum of the system is zero. When the valve is switched from backward to forward, the water expelled in the backward direction imparts a backward momentum to the system. This momentum needs to be canceled out by the forward momentum of the expelled water. The formula for momentum is:
Total momentum = mass of cart x velocity of cart + mass of expelled water x velocity of expelled water
Since the water is expelled at a muzzle velocity of 24.5 m/s, we know that the mass of expelled water x velocity of expelled water is constant. So, the mass of the cart x velocity of the cart must be equal to the negative of the mass of expelled water x velocity of expelled water.
Therefore:
(mass of cart)x(initial velocity of cart) = -(mass of expelled water)x(muzzle velocity of expelled water)
Given that the initial velocity of the cart is zero, we can solve for the time it takes for the cart to come to rest:
(mass of cart)x(initial velocity of cart) = -(mass of expelled water)x(muzzle velocity of expelled water)
(355 kg)(0 m/s) = -(179 kg)(24.5 m/s) x time
0 = -4385.5 x time
time = 0
This means that the cart will not come to rest and will continue to move backward indefinitely. Therefore, we need to reevaluate our initial answer for part (a).
b) Since the cart continues to move backward indefinitely, it will not return to rest. Its velocity will keep increasing in the backward direction. Therefore, our initial answer for the mass of the cart and velocity at rest is incorrect.
c) The thrust of the "rocket" can be calculated by multiplying the mass of expelled water by the velocity of expelled water:
Thrust = (mass of expelled water) x (muzzle velocity of expelled water)
Thrust = (179 kg) x (24.5 m/s)
Thrust ≈ 4,385.5 N
d) The acceleration of the cart immediately before the valve is switched can be calculated using Newton's second law: F = ma, where F is the thrust and m is the mass of the cart.
Acceleration = Thrust / mass of cart
Acceleration ≈ 4,385.5 N / 355 kg
Acceleration ≈ 12.36 m/s^2
Therefore, the correct answers are:
a) The cart will not come to rest.
b) The mass of the cart and velocity at that time cannot be determined.
c) The thrust of the "rocket" is approximately 4385.5 N.
d) The acceleration of the cart immediately before the valve is switched is approximately 12.36 m/s^2.