Ngā Tauira Pātai me te Kōrero mō te Nekehanga Whānui a Newton
Pendahuluan
Ko te nekehanga whanaunga he ariā matua i roto i te ahupūngao e whakamārama ana i te tere me te tūranga o tētahi mea ka taea te whakarerekē i runga i te kaimātakitaki. Nā Tā Isaac Newton, me āna ture mō te nekehanga me te kaha ā-papa, i whakatakoto te turanga mō te mārama ki ngā āhuatanga o te nekehanga whanaunga. Ka hipokina e tēnei tuhinga ētahi tauira me ngā matapakinga mō te nekehanga whanaunga a Newton. Ka whakamāramahia e mātou ēnei raruraru me ngā mahi whakaoti taipitopito kia māmā ai te mārama.
Te Ariā Taketake a Newton mō te Nekehanga Whānui
I roto i te ahupūngao Newtonian, ka inehia tonutia te nekehanga o tētahi mea e pā ana ki tētahi anga tohutoro. Mena he rua ngā anga tohutoro e neke ana tetahi ki tetahi i te tere v, ka kitea te tūranga me te tere o tētahi mea i roto i ngā anga e rua i roto i ngā anga e rua. Ko ētahi ariā hei mārama ko:
1. Anga Tohutoro Ā-Inertia: He anga tohutoro e neke ai te mea i te tere pumau mēnā kāore he kaha e pā ki te mea.
2. Tere Whakawhanaunga: Te tere o tētahi mea i inehia e pā ana ki tētahi atu anga tohutoro.
3. Nekehanga Ā-Whārite: Te rerekētanga o te tūranga i waenga i ngā mea e rua, i te mea kotahi rānei me ngā anga tohutoro rerekē e rua.
I tino mārama te whakaahuatanga a Newton i te nekehanga whanaunga mā roto i ana ture nekehanga, ā, ka taea e tātou te whakamahi i ngā panonitanga Kariri hei whakawhiti i waenga i ngā anga tohutoro korekore e rua.
Tauira Pātai Kōrero
Pātai 1: Nekehanga Āhuatanga
Pātai:
E rua ngā kaipuke, ko te kaipuke A me te kaipuke B, kei roto i te moana nui. Kei te neke te kaipuke A ki te rawhiti me te tere o te 20 m/s, ko te kaipuke B ia kei te neke ki te raki me te tere o te 30 m/s. Tātaihia te tere o te kaipuke B e pā ana ki te kaipuke A.
Kōrero:
Hei whakaoti i tēnei raruraru, ka whakamahia e mātou te ariā o te tere whanaunga. Ka taea te tatau i te tere whanaunga o te kaipuke B e pā ana ki te kaipuke A mā te whakamahi i te tikanga whārite.
1. Whakaaturia ngā tere o te kaipuke A (\(\vec{v_A}\)) me te kaipuke B (\(\vec{v_B}\)) hei whākapū.
\[
\vec{v_A} = 20 \, \text{m/s ki te rawhiti} \implies \vec{v_A} = 20 \hat{i} \, \text{m/s}
\]
\[
\vec{v_B} = 30 \, \text{m/s ki te raki} \implies \vec{v_B} = 30 \hat{j} \, \text{m/s}
\]
2. Ka tatauhia te tere whanaunga o te kaipuke B ki te kaipuke A (\(\vec{v_{BA}}\)) mā te:
\[
\vec{v_{BA}} = \vec{v_B} – \vec{v_A}
\]
Whakakapia ngā uara o \(\vec{v_A}\) me \(\vec{v_B}\):
\[
\vec{v_{BA}} = 30 \hat{j} \, \text{m/s} – 20 \hat{i} \, \text{m/s}
\]
\[
\vec{v_{BA}} = -20 \hat{i} + 30 \hat{j} \, \text{m/s}
\]
3. Hei kimi i te rahi o te tere whanaunga, whakamahia te ariā Pythagorean:
\[
|\vec{v__{BA}}| = \sqrt{(-20)^2 + (30)^2}
\]
\[
|\vec{v__{BA}}| = \sqrt{400 + 900}
\]
\[
|\vec{v_{BA}}| = \sqrt{1300} = 10 \sqrt{13} \, \text{m/s}
\]
Nō reira, ko te tere o te kaipuke B e pā ana ki te kaipuke A ko \(10 \sqrt{13}\) m/s.
Pātai 2: Te Nekehanga Ā-Whānui i roto i tētahi Pūnaha Taunga
Pātai:
Kei te neke te kaihaerere ki te raki i te 5 m/s i runga ake i te tereina e neke ana ki te rawhiti i te 20 m/s. Tātaihia te tere o te kaihaerere e pā ana ki te whenua.
Kōrero:
Hei whakatau i te tere o te hikoi e pā ana ki te whenua, ka whakamahia anō e mātou te ariā o te tāpiritanga whārite.
1. Whakaaturia te tere o te kaihaerere (\(\vec{v_P}\)) me te tere o te tereina (\(\vec{v_K}\)) hei whārite.
\[
\vec{v_P} \text{ e pā ana ki te tereina} = 5 \hat{j} \, \text{m/s}
\]
\[
\vec{v_K} \text{ e pā ana ki te whenua} = 20 \hat{i} \, \text{m/s}
\]
2. Ko te tere o te hikoi e pā ana ki te whenua (\(\vec{v_{PT}}\)) ko te tapeke whārite:
\[
\vec{v_{PT}} = \vec{v_P} + \vec{v_K}
\]
Whakakapia ngā uara o \(\vec{v_P}\) me \(\vec{v_K}\):
\[
\vec{v_{PT}} = 5 \hat{j} \, \text{m/s} + 20 \hat{i} \, \text{m/s}
\]
3. Hei kimi i te rahi o te tere whanaunga:
\[
|\vec{v__{PT}}| = \sqrt{(20)^2 + (5)^2}
\]
\[
|\vec{v__{PT}}| = \sqrt{400 + 25}
\]
\[
|\vec{v_{PT}}| = \sqrt{425} = 5 \sqrt{17} \, \text{m/s}
\]
Nō reira, ko te tere o te kaihaerere e pā ana ki te whenua ko \(5 \sqrt{17}\) m/s.
Pātai 3: Te Nekehanga Whānui i runga i te Papa Piko
Pātai:
Ka whiua he pōro me te tere o \( \vec{u} = 10 \hat{i} + 10 \hat{j} \) m/s e pā ana ki tētahi kāreti e neke ana i te tere pumau o te 15 m/s. Tātaihia te tere o te pōro e pā ana ki te whenua.
Kōrero:
Whakamahia te mātāpono kotahi i roto i te tāpiritanga whārite.
1. Whakaaturia te tere o te pōro (\(\vec{u}\)) e pā ana ki te kāreti me te tere o te kāreti (\(\vec{v_K}\)) hei whārite.
\[
\vec{u} = 10 \hat{i} + 10 \hat{j} \, \text{m/s}
\]
\[
\vec{v_K} = 15 \hat{i} \, \text{m/s}
\]
2. Ko te tere o te pōro e pā ana ki te whenua (\(\vec{v_{BT}}\)) ko:
\[
\vec{v_{BT}} = \vec{v_K} + \vec{u}
\]
\[
\vec{v_{BT}} = 15 \hat{i} + (10 \hat{i} + 10 \hat{j})
\]
\[
\vec{v_{BT}} = (15 + 10) \hat{i} + 10 \hat{j}
\]
\[
\vec{v_{BT}} = 25 \hat{i} + 10 \hat{j}
\]
3. Hei kimi i te rahi o te tere whanaunga:
\[
|\vec{v_{BT}}| = \sqrt{(25)^2 + (10)^2}
\]
\[
|\vec{v_{BT}}| = \sqrt{625 + 100}
\]
\[
|\vec{v_{BT}}| = \sqrt{725} = 5 \sqrt{29} \, \text{m/s}
\]
Nō reira, ko te tere o te pōro e pā ana ki te whenua ko \(5 \sqrt{29}\) m/s.
Whakamutunga
Ko te ariā a Newton mō te nekehanga whanaunga he tūāpapa matua o te ahupūngao matarohia. Mā te whakamahi i ngā mātāpono taketake pēnei i te tāpiritanga whārite, ka taea e tātou te whakatau i te tere whanaunga me te nekehanga o tētahi mea e pā ana ki tētahi atu, ki ngā anga tohutoro rerekē rānei. Mā ngā tauira i runga ake nei ka whakaatu me pēhea te whakamahi i tēnei ariā i roto i ngā horopaki rerekē, ka hoatu he māramatanga hohonu ake mō te nekehanga whanaunga.
Mā te mārama me te mahi i ēnei ariā, ka taea e tātou te mārama ake ki ngā ture nekehanga a Newton me te pānga ki te ao tūroa. Ehara i te mea ka āwhina noa tēnei mātauranga ki te whakaoti rapanga ahupūngao engari ka whakarato hoki i ngā māramatanga hohonu ki te mahi a te ao whānui.
