Question

The plunger on a loaded syringe is plunged at 20 cm/sec.

Assume the water is inviscid, and ignore air resistance. (Answer a-i)

Take force of gravity = 10 m/(s^2)

Cross sectional Area of main body 2cm, Corss sectional area of tip/nozzle is 2mm

With 5 mL of water in the syringe.

a) What is the velocity of the water as it exits the syringe?

b) If aimed vertically, how high will the water go?

c) If the syringe is tipped at an angle what is the maximum horizontal range possible?

d) What force must be applied to the plunger to achieve this 20 cm/sec plunging speed?

e) During the plunge what is the pressure inside the syringe?

f) How much power must be applied to accomplish the plunge?

g) What is the total energy expended over the entire plunge?

h) How much thrust does this plunging achieve?

i) If mercury were used instead of water, how would it change (a-h)?

Answer 1

GIven is:-

velocity of the plunger = 20cm/s or 0.20m/s

Cross sectional area of main body is

cross sectional area of nozzle is    or   

part-a

According to the equation of contunuity

hence the velocity of the water exits from the syringe is 20m/s

part-b

we know that

by putting all the values in it, we get

thus the water will go 20 meters in vertical direction

part-c

Maximum horizontal range is possible, when the projectile is projected at 45 degrees from the horizontal.

THus the range is

  

at    the range will be

Thus the horizontal range will be 40 meters if the syringe is tipped at an angle of 45 degrees from the horizontal.