(a) A particle moves on a straight path with an initial speed u and final speed v in time t. Show that the total distance X covered by the particle is given by
(iii) the law of floatation.
(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\(^{-3}\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\(^{-3}\)? [ g = 10 ms\(^{-2}\)]
(a) Derivation of \(x = ut + \tfrac{1}{2}at^2\)
For uniform acceleration \(a\), initial speed \(u\), final speed \(v\) after time \(t\), the distance is the area under the velocity-time graph (a trapezium):
\[ x = \left(\frac{u+v}{2}\right)t \]
But \(v = u + at\). Substituting:
\[ x = \left(\frac{u + (u+at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t = ut + \tfrac{1}{2}at^2 \quad\text{(shown)} \]
(b)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the applied (resultant) force and takes place in the direction of the force; hence \(F = ma\).
(b)(ii) Principle of conservation of energy: Energy can neither be created nor destroyed but only changed from one form to another; the total energy of an isolated system remains constant.
(b)(iii) Law of floatation: A floating body displaces its own weight of the fluid in which it floats.
(c) Volume of the balloon when it just begins to rise
The balloon just begins to rise when the upthrust from the air equals the total weight (of the balloon fabric plus the gas):
\[ \rho_{\text{air}}\,V g = (m + \rho_{\text{gas}}\,V)g \]
\[ 1.29\,V = 0.030 + 0.54\,V \]
\[ (1.29 - 0.54)V = 0.030 \;\Rightarrow\; 0.75\,V = 0.030 \]
\[ V = \frac{0.030}{0.75} = 0.04\,\text{m}^3 \]
The volume of the balloon is \(0.04\,\text{m}^3\).