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

  1. 👍 0
  2. 👎 0
  3. 👁 201
  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

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. Physics

    The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.45 m/s at an angle of 53.0 degrees

  2. Science

    A 0.5kg cart connected to light spring for which the force constant is 20 Nm on horizontal frictionless air track. Calculate the total energy of the system and the maximum speed of the cart if the amplitude of the motion is 3.0cm

  3. physics

    A 0.054-kg pet lab mouse sits on a 0.35-kg air-track cart, as shown in (Figure 1) . The cart is at rest, as is a second cart with a mass of 0.25 kg. The lab mouse now jumps to the second cart. After the jump, the 0.35-kg cart has

  4. chem

    In an industrial process to make alcohol, bacteria, sugar, and water are fed into a bioreactor. The bacteria make alcohol, and water as well as leftover sugar. We desire to remove all the cells from the process stream so we may

  1. chem

    In an industrial process to make alcohol, bacteria, sugar, and water are fed into a bioreactor. The bacteria make alcohol, and water as well as leftover sugar. We desire to remove all the cells from the process stream so we may

  2. Physics Help

    A small, 200 g cart is moving at 1.50 m/s on an air track when it collides with a larger, 2.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. Q: What is the speed of the large cart after the collision?

  3. physics

    A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of + 17.4 m/s, while the exiting water stream has a velocity of – 15.0 m/s. The mass of

  4. Physics

    A flat-topped toy cart moves on frictionless wheels, pulled by a rope under tension T. The mass of the cart is m1. A load of mass m2 rests on top of the cart with the coefficient of static friction μs between the cart and the

  1. physics

    A fireman d = 35.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of θi = 28.0° above the horizontal. If the speed of the stream as it leaves the hose is vi = 40.0 m/s, at what

  2. physics

    A fireman 46.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of 30.0° above the horizontal. If the speed of the stream as it leaves the hose is 40.0 m/s, at what height will

  3. physics

    A cart is propelled over an xy plane with acceleration components ax = 5.1 m/s2 and ay = -2.9 m/s2. Its initial velocity has components v0x = 8.2 m/s and v0y = 11.2 m/s. In unit-vector notation, what is the velocity of the cart

  4. Physics

    A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m:

You can view more similar questions or ask a new question.