An amoeba of mass 1 ｘ10⁻¹² kg propels itself through water by blowing a jet of water through a tiny orifice. The amoeba ejects water with a speed of 1 ｘ10⁻⁴ m s⁻¹
at a rate of 1 ｘ10⁻¹³ kg s⁻¹. Assume that the water is continuously replenished so that the mass of the amoeba remains the same.
(a) If there were no force on amoeba other then the reaction force caused by the emerging jet, what would be the acceleration of the amoeba?
(b) If amoeba moves with constant velocity through water, what is force of surrounding water (exclusively of jet) on the amoeba? 
(Ans: 1.0 ｘ 10⁻⁵ m s⁻², 1.0 ｘ 10⁻¹⁷ N)


Given:

Mass of an amoeba = M = 1 ｘ10⁻¹² kg

speed of the water an amoeba ejects = v = 1 ｘ10⁻⁴ m s⁻¹

Rate of water ejects = mt = 1 ｘ10⁻¹³ kg s⁻¹


To Find:

(a) Accelerations = a = ?

(b) Force = F = ?

Solution:

(a) Accelerations = a = ?

We know that rate of change of momentum is called force ie. 

F = Pt  

Ma = mvt  

or

a  = mvtM 

or

a  = mtｘvM 

by putting the corresponding values

a = 1 ｘ10⁻¹³ kg s⁻¹ ｘ`\frac {1 ｘ10⁻⁴ m s⁻¹}{1 ｘ10⁻¹² kg}`

a = 1 ｘ10⁻¹³ ｘ1 ｘ10⁸ m s⁻² 

a = 1 ｘ10⁻⁵ m s⁻²  ------------------ Ans. (1)

(b) Force = F = ?

we know that 

F = ma

by putting values

F = 1 ｘ10⁻¹² kg ｘ 1 ｘ10⁻⁵ m s⁻²

F = 1 ｘ10⁻¹⁷ N  -------------Ans. (2)

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