Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / Section B / B7

Table of contents
- Solution B7
- i B7
  - Solution i
- ii B7
  - Solution ii
- iii B7
  - Solution iii
- iv B7
  - Solution iv

Solution

For a steady uniform jet, the mass flow rate is

\overset{m}{˙} = ρ A u

. The transported momentum flux and kinetic energy flux of a uniform jet are

\overset{p}{˙} = ρ A u^{2}, \dot{K} = \frac{1}{2} ρ A u^{3} .

(1)

Incompressibility gives piston speed

v_{p} = A u / S

. Equating piston power

F v_{p}

to the jet's kinetic power gives

F = \frac{1}{2} ρS u^{2} .

(2)

This ideal result neglects losses and the kinetic energy of fluid moving slowly inside the large cylinder. Retaining an inlet speed correction would multiply the pressure expression by

1 - (A / S)^{2}

A horizontal impact occurs at the projectile apex. Write

u_{x} = u cos θ

and

u_{y} = u sin θ

. At its height h,

u_{y}^{2} = 2 g h

and the flight time is

t = u_{y} / g

, so

u_{y} = 2 g h, u_{x} = w \frac{g}{2 h}, u = 9.90 m s^{- 1}, θ = 4 5^{\circ}, F = 4.91 N .

(3)

Here

g = 9.81 m s^{- 2}

and

u_{x} = u_{y} = 7.00 m s^{- 1}

. The apex-target condition for a ballistic jet fixes both launch speed and angle.

If the target stops the horizontal momentum of the arriving water, the force is

F_{hit} = \overset{m}{˙} u_{x} = ρ A u u_{x} = 0.277 N, μ m_{t} g = 0.147 N .

(4)

This exceeds the limiting friction, so the target can be dislodged in the ideal stopping-impact model. The jet-area adjustment during ballistic slowing matters: the nozzle area sets

\overset{m}{˙}

; replacing u by

u_{x}

in that mass-flow factor would incorrectly assume unchanged jet area during deceleration. Reflection or a specified deflection would change the transferred momentum. Air drag, breakup and splash losses are excluded from the ballistic estimate.