Question

In a water pistol, a piston drives water through a larger tube of radius 1.30 cm into a smaller tube of radius 1.10 mm as in the figure below. A diagram shows the interior of a water pistol pointing to the right, and the flow of water through it. The left end of the trigger is connected to the piston on the right end of a cylinder. The left end of the cylinder is connected to a tube, such that water within the cylinder can exit through the tube. The tube starts at the cylinder, extends to the left, curves upward and to the right, and extends horizontally to the right until the end of the tube reaches the nozzle of the pistol. The cross-sectional areas of the cylinder and tube are labeled A1 and A2, respectively. An arrow labeled vector F to the right of the trigger points to the left toward the trigger. The velocities of water within the cylinder and water ejected from the nozzle are labeled vector v1 and vector v2, respectively.

(a) If the pistol is fired horizontally at a height of 1.10 m, use ballistics to determine the time it takes water to travel from the nozzle to the ground. (Neglect air resistance and assume atmospheric pressure is 1.00 atm. Assume up is the positive y-direction. Indicate the direction with the sign of your answer.) s

(b) If the range of the stream is to be 7.60 m, with what speed must the stream leave the nozzle? m/s

(c) Given the areas of the nozzle and cylinder, use the equation of continuity to calculate the speed at which the plunger must be moved. m/s (

d) What is the pressure at the nozzle? (Give your answer to at least four significant figures.) Pa (

e) Use Bernoulli's equation to find the pressure needed in the larger cylinder. Pa

Can gravity terms be neglected? Yes No

(f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure. Enter magnitude.) N

Answer 1

(a) As pistol is fired horizontally, there is no vertical velocity component.

so,

h = 1/2gt²

t = sqrt ( 2h / g)

t = sqrt ( 2 * 1.10 / 9.8)

t = 0.4738 seconds

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(b) use

d = vt

v = d / t

v = 7.60 / 0.4738

v = 16.04 m/s

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(c) equation of continuity is

A1v1 = A2v2

where A is area of cross section and v is velocity

we need to find v1 here

v1 = A2v2 / A1

v1 = r2² v2 / r1²

v1 = 1.10e-3² * 16.04 / 1.30e-2

v1 = 0.1148 m/s

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(d) Here, we can use Bernoulli's equation

P1 = P2 + 1/2 ( v₂² - v₁²)

here P2 = 1 atm or 1.013e5 Pa (given in question)

P1 = 1.013e5 + 1/2 * 1000 * ( 16.04² - 0.1148²)

P1 = 2.3e5 Pa

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(e) same as part (d)

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(f) F = ( 2.3e5 - 1.013e5) * pi * 1.3e-2²

F = 68.33 N