Question

In a water pistol, a piston drives water through a larger tube of radius 1.30 cm into a smaller tube of radius 1.50 mm as in the figure below. (a) If the pistol is fired horizontally at a height of 1.40 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.) (b) If the range of the stream is to be 7.90 m, with what speed must the stream leave the nozzle? (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. (d) What is the pressure at the nozzle? (Give your answer to at least four significant figures.) (e) Use Bernoulli's equation to find the pressure needed in the larger cylinder. Can gravity terms be neglected? 6---Select---YesNo (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.)

Answer 1

Let's break down this water pistol problem step-by-step.

Given Information:

• Radius of larger tube (r₁) = 1.30 cm = 0.013 m

• Radius of smaller tube (r₂) = 1.50 mm = 0.0015 m

• Height of nozzle (y₀) = 1.40 m

• Range (x) = 7.90 m

• Atmospheric pressure (P_atm) = 1.00 atm = 101325 Pa

a) Time to Travel from Nozzle to Ground

This is a projectile motion problem. The horizontal motion does not affect the vertical motion.

• Vertical motion:

  y = y₀ + v₀y*t + (1/2)at²

  where:

  y = 0 (ground level)

  y₀ = 1.40 m

  v₀y = 0 (initial vertical velocity is zero)

  a = g = 9.81 m/s²

  0 = 1.40 + 0t - (1/2)9.81t²

  4.905t² = 1.40

  t² = 1.40 / 4.905

  t = √(1.40 / 4.905)

  t ≈ 0.5338 s

b) Speed to Achieve 7.90 m Range

• Horizontal motion:

  x = v₀x*t

  where:

  x = 7.90 m

  t = 0.5338 s (from part a)

  v₀x = initial horizontal velocity (speed of water leaving nozzle)

  v₀x = x / t

  v₀x = 7.90 m / 0.5338 s

  v₀x ≈ 14.799 m/s

c) Speed of Plunger (Continuity Equation)

• Area of larger tube (A₁) = πr₁² = π(0.013 m)² ≈ 5.309 x 10⁻⁴ m²

• Area of smaller tube (A₂) = πr₂² = π(0.0015 m)² ≈ 7.069 x 10⁻⁶ m²

• Continuity equation: A₁v₁ = A₂v₂

  where:

  v₂ = 14.799 m/s (speed at nozzle)

  v₁ = speed of plunger

  v₁ = (A₂*v₂) / A₁

  v₁ = (7.069 x 10⁻⁶ m² * 14.799 m/s) / 5.309 x 10⁻⁴ m²

  v₁ ≈ 0.1970 m/s

d) Pressure at Nozzle (Bernoulli's Equation)

• Bernoulli's equation: P₁ + (1/2)ρv₁² + ρgy₁ = P₂ + (1/2)ρv₂² + ρgy₂

  where:

  P₁ = pressure at larger tube (unknown)

  v₁ = 0.1970 m/s

  y₁ = 0 (reference level)

  P₂ = pressure at nozzle (unknown)

  v₂ = 14.799 m/s

  y₂ = 1.40 m

  ρ = density of water ≈ 1000 kg/m³

• Since we're looking for the pressure at the nozzle, we can set y₂ ≈ 0 and consider the pressure difference relative to atmospheric pressure.

• Assuming the pressure at the nozzle is close to atmospheric pressure (P_atm), we can write:

  P_nozzle = P_atm + (1/2)ρ(v₁² - v₂²)

  P_nozzle = 101325 Pa + (1/2)1000 kg/m³((0.1970 m/s)² - (14.799 m/s)²)

  P_nozzle ≈ 101325 Pa - 109489 Pa

  P_nozzle ≈ -8164 Pa

  This is a negative pressure, which is not physically possible. There's likely an error in the assumption. The pressure at the nozzle will be close to atmospheric.

  We can assume P_nozzle = P_atm = 101325 Pa.

e) Pressure in Larger Cylinder

• Using Bernoulli's equation from part d):

  P₁ = P₂ + (1/2)ρ(v₂² - v₁²) + ρg(y₂ - y₁)

  P₁ = 101325 Pa + (1/2)1000((14.799)² - (0.1970)²) + 10009.81(1.40 - 0)

  P₁ ≈ 101325 + 109489 + 13734

  P₁ ≈ 224548 Pa

• Gravity terms can be neglected if the height difference is small compared to the pressure difference due to velocity.

  In this case, ρgy₂ ≈ 13734 Pa, while (1/2)ρ(v₂² - v₁²) ≈ 109489 Pa. Therefore, gravity terms are Yes negligible.

f) Force on Trigger

• Pressure difference (ΔP) = P₁ - P_atm = 224548 Pa - 101325 Pa = 123223 Pa

• Force (F) = ΔP * A₁

• F = 123223 Pa * 5.309 x 10⁻⁴ m²

• F ≈ 65.41 N

Answers:

a) Time to travel: 0.5338 s

b) Speed at nozzle: 14.80 m/s

c) Speed of plunger: 0.1970 m/s

d) Pressure at nozzle: 101325 Pa (atmospheric pressure)

e) Pressure in cylinder: 224548 Pa, Gravity terms: Yes

f) Force on trigger: 65.41 N

Answer 2