(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)
(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)
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Correct Answer: Option n
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\(^{-1}J\)
(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\(^{-1}J\)