Elastik toʻqnashuv nima?

Nega biz toʻqnashuvni deyarli mutlaq elastik deb olyapmiz?

Biz mexanikada duch keladigan masalalarning deyarli hech qaysisi mutlaq elastik toʻqnashuv boʻla olmaydi. Bundan ushbu tushuncha amaliy jihatdan juda kam qoʻllangandek tuyulishi mumkin. Biroq amaliyotda u juda ham foydali. Chunki kinetik energiyaning saqlanish sharti tenglamalarimizga qoʻshimcha cheklov qoʻyadi. Shunda biz tenglamalarni yecha olamiz, aks holda nomaʼlumlar koʻpayib ketardi. Odatda natija yetarlicha aniq boʻladi, bunga sabab toʻqnashuv mutlaq elastikka yetarlicha yaqindir.

Tasavvur qiling, gorizontal tekislikda harakatlanayotgan ikkita A va B aravachalar oʻzaro elastik toʻqnashdi. Biz har ikkala aravachaning oxirgi tezligini (f indeks) topishimiz kerak, lekin bizga faqat ularning boshlangʻich tezligi v_{Ai}vAi​v, start subscript, A, i, end subscript va v_{Bi}vBi​v, start subscript, B, i, end subscript berilgan. Impulsning saqlanish qonunini qoʻllasak, bizda ikkita nomaʼlumli, v_{Af}vAf​v, start subscript, A, f, end subscript va v_{Bf}vBf​v, start subscript, B, f, end subscript, bitta tenglama hosil boʻladi.

m_A v_{Ai}+m_B v_{Bi}=m_{A}v_{Af}+m_B v_{Bf}mA​vAi​+mB​vBi​=mA​vAf​+mB​vBf​m, start subscript, A, end subscript, v, start subscript, A, i, end subscript, plus, m, start subscript, B, end subscript, v, start subscript, B, i, end subscript, equals, m, start subscript, A, end subscript, v, start subscript, A, f, end subscript, plus, m, start subscript, B, end subscript, v, start subscript, B, f, end subscript

Shuningdek, kinetik energiyaning ham saqlanishi sababli biz bir vaqtning oʻzida yana bir tenglamaga ega boʻlamiz:

\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_B v_{Bi}^2=\frac{1}{2}m_A v_{Af}^2+\frac{1}{2}m_B v_{Bf}^221​mA​vAi2​+21​mB​vBi2​=21​mA​vAf2​+21​mB​vBf2​start fraction, 1, divided by, 2, end fraction, m, start subscript, A, end subscript, v, start subscript, A, i, end subscript, squared, plus, start fraction, 1, divided by, 2, end fraction, m, start subscript, B, end subscript, v, start subscript, B, i, end subscript, squared, equals, start fraction, 1, divided by, 2, end fraction, m, start subscript, A, end subscript, v, start subscript, A, f, end subscript, squared, plus, start fraction, 1, divided by, 2, end fraction, m, start subscript, B, end subscript, v, start subscript, B, f, end subscript, squared

Biz ikkita nomaʼlumli ikkita tenglamaga ega boʻldik. Bilamizki, biz bu tenglamalar sistemasini yechib tezliklarni topa olamiz.

Bu tenglamalarni yechish biroz zerikarli boʻlgani uchun hozirchalik oxirgi natijani yozib qoʻyamiz:

v_{Af} = \left( \frac{m_{A} - m_{B}}{m_{A}+m_{B}} \right) v_{Ai} + \left( \frac{2m_{B}}{m_{A}+m_{B}} \right) v_{Bi}vAf​=(mA​+mB​mA​−mB​​)vAi​+(mA​+mB​2mB​​)vBi​v, start subscript, A, f, end subscript, equals, left parenthesis, start fraction, m, start subscript, A, end subscript, minus, m, start subscript, B, end subscript, divided by, m, start subscript, A, end subscript, plus, m, start subscript, B, end subscript, end fraction, right parenthesis, v, start subscript, A, i, end subscript, plus, left parenthesis, start fraction, 2, m, start subscript, B, end subscript, divided by, m, start subscript, A, end subscript, plus, m, start subscript, B, end subscript, end fraction, right parenthesis, v, start subscript, B, i, end subscript

v_{Bf} = \left( \frac{2m_{A}}{m_{A}+m_{B}} \right) v_{Ai} + \left( \frac{m_{B} - m_{A}}{m_{A}+m_{B}} \right) v_{Bi}vBf​=(mA​+mB​2mA​​)vAi​+(mA​+mB​mB​−mA​​)vBi​v, start subscript, B, f, end subscript, equals, left parenthesis, start fraction, 2, m, start subscript, A, end subscript, divided by, m, start subscript, A, end subscript, plus, m, start subscript, B, end subscript, end fraction, right parenthesis, v, start subscript, A, i, end subscript, plus, left parenthesis, start fraction, m, start subscript, B, end subscript, minus, m, start subscript, A, end subscript, divided by, m, start subscript, A, end subscript, plus, m, start subscript, B, end subscript, end fraction, right parenthesis, v, start subscript, B, i, end subscript

[Isbotni koʻrish]

Ushbu yechimlarning qiziqarli tomoni shundaki, ular toʻqnashuvning turli xil xususiy hollari uchun cheklangan natijalarni beradi. Bular Nyuton mayatniklarida kuzatiladigan elastik toʻqnashuvlar prinsipini tushunish imkonini beradi.

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