(a) Explain the term net force. (b) Define the principle of conservation of linear momentum and state one example of it. (c) A ball of mass 200 g released f...

(a) Net force is the single resultant force obtained when all the forces acting on a body are combined (added as vectors). It is the force that determines the body's acceleration, \(F_{net}=ma\).

(b) Principle of conservation of linear momentum: in a closed system on which no external force acts, the total linear momentum before an interaction (collision or explosion) equals the total linear momentum after it. Example: the recoil of a gun when a bullet is fired - the forward momentum of the bullet equals the backward momentum of the gun.

(c) Impulse received by the floor.
Mass \(m = 200\,\text{g}=0.2\,\text{kg}\).
Speed on hitting the floor: \(v_1=\sqrt{2gh_1}=\sqrt{2\times10\times2.0}=\sqrt{40}=6.32\,\text{ms}^{-1}\) (downward).
Speed on rebound: \(v_2=\sqrt{2gh_2}=\sqrt{2\times10\times1.8}=\sqrt{36}=6.0\,\text{ms}^{-1}\) (upward).
Taking upward as positive, impulse on the ball

\[ J = m(v_2-(-v_1)) = 0.2\,(6.0+6.32)=0.2\times12.32 = 2.46\,\text{Ns} \]

By Newton's third law the impulse received by the floor is equal and opposite: 2.46 Ns directed downward.

(d) Simple harmonic motion.
\(m=20\,\text{g}=0.02\,\text{kg}\), \(f=5\,\text{Hz}\Rightarrow \omega=2\pi f=2\times3.14\times5=31.4\,\text{rads}^{-1}\).
At \(x=10\,\text{cm}=0.1\,\text{m}\), \(v=200\,\text{cms}^{-1}=2.0\,\text{ms}^{-1}\).

(i) Maximum displacement (amplitude) A. Using \(v=\omega\sqrt{A^2-x^2}\):

\[ v^2=\omega^2(A^2-x^2)\Rightarrow A^2 = x^2 + \frac{v^2}{\omega^2}=0.01+\frac{4}{(31.4)^2}=0.01+0.00406=0.01406 \]\[ A = 0.119\,\text{m}\;(\approx 11.9\,\text{cm}) \]

(ii) Maximum velocity.

\[ v_{max}=\omega A = 31.4\times0.119 = 3.73\,\text{ms}^{-1} \]

(iii) Maximum potential energy (equals the total energy of the oscillation):

\[ P.E._{max}=\tfrac{1}{2}m\omega^2A^2=\tfrac{1}{2}(0.02)(31.4)^2(0.01406)=0.139\,\text{J} \]

So \(P.E._{max}\approx 0.14\,\text{J}\).