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# AL PHY

A toy designer has submitted a design for a water pistol with a barrel area 75mm^2 and jet area 1mm^2. The manufacturer required that when the pistol was fired horizontally, the jet should be able to hit a target 3.5m away not more than 1m below the firing line. Given that the average child is able to exert a force of 10N on the plunger, has the designer satisfied the requirements? you may neglect barrel friction and energy loss at the exit jet.

(Atmospheric pressure= 1x 10^5 Pa, density of water= 1x10^3, g=9.8ms^-2)

### 1 Answer

- violinistLv 51 decade agoFavorite Answer
PART 1: to calculate the PRACTICAL length of barrel where the bullet occupies

as the trace of the water bullet ejected is independent of mass,

for simplicity, let us consider a bullet of volume,

V=1*10^ -6 m^3

mass of the bullet, m=V*density

=[1*10^ -6 m^3]*[1000 kg/m^3]

=1*10^ -3 kg

when the bullet is initially stored in the barrel,

it occupies the barrel for a length, l,

=V/ cross sectional area of barrel

= 1*10^ -6 m^3 / 75*10^ -6 m^2

= 0.0133m -------------------------------------------@

PART 2: to calculate the MINIMUM value of length of barrel where the bullet can be accelerate in order to be ejected and reached the target not lower than 1 m.

meanwhile,

consider the trace of the ejected bullet,

vertically:

displacement, s=u*t + 0.5*g*t^2

as the bullet is ejected horizontally, u=0,

1=5*t^2

t=0.447sec.

horizontally:

velocity, w, remains unchanged

and w=horizontal displacement / t

=3.5m / 0.447sec.

=7.826 m/s

speed at the jet equals w=7.826 m/s

consider cross sectional area and velocity of the jet and barrel,

by Bernoulli Principle,

A1v1=A2v2

[75*10^ -6 m^2]*[v]=[1*10^ -6 m^2]*[w]

velocity at the barrel, v=0.104 m/s

by Newton's second law,

Fnet=ma

F-f = m[v - 0]/ T

as initial velocity of the bullet in the barrel = 0 m/s

where F is the force exerted by the child and F= 10 N

f is the force exerted by the atmosphere and

f = P(atmosphere)*A(jet)

=[1*10^5 N/m^2]*[1*10^ -6 m^2]

=0.1 N

T=m*v / [ F-f ]

=[1*10^ -3 kg]*[ 0.104 m/s ] / [ 10 N - 0.1N ]

=1.0505*10^ -5 sec.

v=u+at

a=v/t=0.104/1.0505*10^ -5

=9900m/s^2

Length, s, of barrel where the bullet of 1*10^ -3 cm^3 occupies,

s=[v^2-u^2]/2a

=0.104^2/[2*9900]

=5.463*10^ -7m ----------------------------------------#

CONCLUSION

As @ is smaller than #, the 1*10^ -3 m^3 bullet can reach the target not lower than 1m

Source(s): myself