(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 from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 in. Calculate the impulse received by the floor. (g = 10 ms\(^{-2}\)). (d) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms\(^{-1}\). Calculate its: (i) maximum displacement from the mean position; (ii) maximum velocity; (iii) maximum potential energy. (g = 10 ms\(^{-2}\) \(\pi\) = 3.14)
Explanation
(a) Net force is the effective or resultant force resulting from the actions of a system of forces on a body.
(b) The principle ofconservation of linear momentum explains that in an isolated or closed system of colliding bodies the total linear momentum in a fixed direction remains constant, e.g. the recoil of gun, colliding trolleys, rocket propulsion, etc.
(c) V\(_1^2\) = \(U_1^2 + 2gh_1\) = 0 + 2 x 10 x 2 = 40 \(V_1 = \sqrt{40}\) = 6.325m/s \(v_2^2 = u_2^2 - 2gh_2\) O = \(U^2_2\) - 2 x 10 x 1.8 u = \(\sqrt{36}\) = 6ms\(^{-1}\) Impulse Change in momentun mv\(_1\) - (-mu\(_2\)) = m (v\(_1\) + u\(_2\)) = 0.2(6.325 + 6.000) = 2.46 Ns
(d) v = \(\omega \sqrt{r^2 - y^2}\) 2\(\pi\)f\(\sqrt{r^2 - y^2}\) 2 = 2 x 3.14 x 5 x \(\sqrt{r^2 - 0.1^2}\) r = 0.12m (ii) Vmax = \(\omega\)r = 2\(\pi\)fr = 2 x 3.142 x 5 x 0.12 = 3,77 m/s (iii) P.E = \(\frac{1}{2}m \omega^2e^2 = \times \frac{1}{2}m(2 \pi f)^2r^2\) = \(\frac{1}{2} \times 0.02(2 \times 3.142 \times 5)^2 \times (0.12)^2\) = 1.42 x 10\(^{-1}J\)