(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)