Respuestas
Respuesta dada por:
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Partimos de los datos:
![m_{a}=m \\ m_{b}=3m \\ v_{0a}=i \\ v_{0b}=0 m_{a}=m \\ m_{b}=3m \\ v_{0a}=i \\ v_{0b}=0](https://tex.z-dn.net/?f=+m_%7Ba%7D%3Dm+%5C%5C++m_%7Bb%7D%3D3m+%5C%5C++v_%7B0a%7D%3Di++++%5C%5C++v_%7B0b%7D%3D0+)
Llamaremos a las velocidades finales de cada cuerpo:
![v_{fa}= f_{a} \\ v_{fb}= f_{b} v_{fa}= f_{a} \\ v_{fb}= f_{b}](https://tex.z-dn.net/?f=+v_%7Bfa%7D%3D+f_%7Ba%7D+++%5C%5C++v_%7Bfb%7D%3D+f_%7Bb%7D++)
Sabemos que en la colisión se conserva el vector cantidad de movimiento lineal:
![\overrightarrow{ p}_{0}=\overrightarrow{ p}_{f} \\ \\ m_{a} \overrightarrow{v}_{oa} + m_{b} \overrightarrow{v}_{oa} = m_{a} \overrightarrow{v}_{fa} + m_{b} \overrightarrow{v}_{fa} \\ \\ (m)(i)+(3m)(0)=(m)( f_{a})+(3m)( f_{b}) \\ \\ (m)(i)=(m)( f_{a})+(3m)( f_{b}) \\ \\ ( \diagup\!\!\!\!\!m)(i)=(\diagup\!\!\!\!\!m)( f_{a})+(3\diagup\!\!\!\!\!m)( f_{b}) \\ \\ i= f_{a}+3 f_{b} \\ \\ f_{a}=i-3f_{b} \ \ \ (1) \overrightarrow{ p}_{0}=\overrightarrow{ p}_{f} \\ \\ m_{a} \overrightarrow{v}_{oa} + m_{b} \overrightarrow{v}_{oa} = m_{a} \overrightarrow{v}_{fa} + m_{b} \overrightarrow{v}_{fa} \\ \\ (m)(i)+(3m)(0)=(m)( f_{a})+(3m)( f_{b}) \\ \\ (m)(i)=(m)( f_{a})+(3m)( f_{b}) \\ \\ ( \diagup\!\!\!\!\!m)(i)=(\diagup\!\!\!\!\!m)( f_{a})+(3\diagup\!\!\!\!\!m)( f_{b}) \\ \\ i= f_{a}+3 f_{b} \\ \\ f_{a}=i-3f_{b} \ \ \ (1)](https://tex.z-dn.net/?f=%5Coverrightarrow%7B+p%7D_%7B0%7D%3D%5Coverrightarrow%7B+p%7D_%7Bf%7D+%5C%5C+%5C%5C+m_%7Ba%7D+%5Coverrightarrow%7Bv%7D_%7Boa%7D+%2B+m_%7Bb%7D+%5Coverrightarrow%7Bv%7D_%7Boa%7D+%3D+m_%7Ba%7D+%5Coverrightarrow%7Bv%7D_%7Bfa%7D+%2B+m_%7Bb%7D+%5Coverrightarrow%7Bv%7D_%7Bfa%7D+%5C%5C+%5C%5C+%28m%29%28i%29%2B%283m%29%280%29%3D%28m%29%28+f_%7Ba%7D%29%2B%283m%29%28+f_%7Bb%7D%29+%5C%5C+%5C%5C+%28m%29%28i%29%3D%28m%29%28+f_%7Ba%7D%29%2B%283m%29%28+f_%7Bb%7D%29+%5C%5C+%5C%5C+%28+%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21m%29%28i%29%3D%28%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21m%29%28+f_%7Ba%7D%29%2B%283%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21m%29%28+f_%7Bb%7D%29+%5C%5C+%5C%5C+i%3D+f_%7Ba%7D%2B3+f_%7Bb%7D+%5C%5C++%5C%5C++f_%7Ba%7D%3Di-3f_%7Bb%7D+++%5C+%5C+%5C+%281%29)
Esa será la ecuación (1). Por otro lado, si el choque es elástico significa que la energía cinética del sistema se conserva:
![E_{ko}= E_{kf} \\ \\ \dfrac{1}{2} m_{a} v_{oa}^{2} + \dfrac{1}{2} m_{b} v_{ob}^{2}=\dfrac{1}{2} m_{a} v_{fa}^{2}+\dfrac{1}{2} m_{b} v_{fb}^{2} \\ \\ \diagup\!\!\!\!\!\!\!\dfrac{1}{2} m \cdot i^{2}=\diagup\!\!\!\!\!\!\!\dfrac{1}{2} m \cdot f_{a} ^{2}+\diagup\!\!\!\!\!\!\!\dfrac{3}{2} m \cdot f_{b} ^{2} \\ \\ i^{2}= f_{a} ^{2}+ 3 f_{b}^{2} \ \ \ (2) E_{ko}= E_{kf} \\ \\ \dfrac{1}{2} m_{a} v_{oa}^{2} + \dfrac{1}{2} m_{b} v_{ob}^{2}=\dfrac{1}{2} m_{a} v_{fa}^{2}+\dfrac{1}{2} m_{b} v_{fb}^{2} \\ \\ \diagup\!\!\!\!\!\!\!\dfrac{1}{2} m \cdot i^{2}=\diagup\!\!\!\!\!\!\!\dfrac{1}{2} m \cdot f_{a} ^{2}+\diagup\!\!\!\!\!\!\!\dfrac{3}{2} m \cdot f_{b} ^{2} \\ \\ i^{2}= f_{a} ^{2}+ 3 f_{b}^{2} \ \ \ (2)](https://tex.z-dn.net/?f=E_%7Bko%7D%3D+E_%7Bkf%7D+%5C%5C++%5C%5C++%5Cdfrac%7B1%7D%7B2%7D+m_%7Ba%7D+v_%7Boa%7D%5E%7B2%7D++%2B+%5Cdfrac%7B1%7D%7B2%7D+m_%7Bb%7D+v_%7Bob%7D%5E%7B2%7D%3D%5Cdfrac%7B1%7D%7B2%7D+m_%7Ba%7D+v_%7Bfa%7D%5E%7B2%7D%2B%5Cdfrac%7B1%7D%7B2%7D+m_%7Bb%7D+v_%7Bfb%7D%5E%7B2%7D+%5C%5C++%5C%5C++%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5Cdfrac%7B1%7D%7B2%7D+m+%5Ccdot+i%5E%7B2%7D%3D%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5Cdfrac%7B1%7D%7B2%7D+m+%5Ccdot++f_%7Ba%7D+%5E%7B2%7D%2B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5Cdfrac%7B3%7D%7B2%7D+m+%5Ccdot++f_%7Bb%7D+%5E%7B2%7D+%5C%5C++%5C%5C++i%5E%7B2%7D%3D+f_%7Ba%7D+%5E%7B2%7D%2B+3+f_%7Bb%7D%5E%7B2%7D+++%5C++%5C++%5C+%282%29)
Reemplazamos la ecuación (1) en la (2):
![i^{2}=(i-3f_{b})^{2}+3 f_{b}^{2} \\ \\ \diagup\!\!\!\!\!i^{2}= \diagup\!\!\!\!\!i^{2}-6i f_{b}+12f_{b} ^{2} \\ \\12f_{b} ^{2}-6i f_{b}=0 \\ \\ 6f_{b} ^{2}-i f_{b}=0 \\f_{b}(6f_{b}-i)=0~\Longrightarrow \ \left\langle\begin{array}{c} f_{b} =0 \\ \\ 6 f_{b} -i=0\Longrightarrow \ \boxed{f_b=i/6}\end{array}\right \\ \\ i^{2}=(i-3f_{b})^{2}+3 f_{b}^{2} \\ \\ \diagup\!\!\!\!\!i^{2}= \diagup\!\!\!\!\!i^{2}-6i f_{b}+12f_{b} ^{2} \\ \\12f_{b} ^{2}-6i f_{b}=0 \\ \\ 6f_{b} ^{2}-i f_{b}=0 \\f_{b}(6f_{b}-i)=0~\Longrightarrow \ \left\langle\begin{array}{c} f_{b} =0 \\ \\ 6 f_{b} -i=0\Longrightarrow \ \boxed{f_b=i/6}\end{array}\right \\ \\](https://tex.z-dn.net/?f=+i%5E%7B2%7D%3D%28i-3f_%7Bb%7D%29%5E%7B2%7D%2B3+f_%7Bb%7D%5E%7B2%7D+%5C%5C++%5C%5C++%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21i%5E%7B2%7D%3D+%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21i%5E%7B2%7D-6i+f_%7Bb%7D%2B12f_%7Bb%7D+%5E%7B2%7D+%5C%5C++%5C%5C12f_%7Bb%7D+%5E%7B2%7D-6i+f_%7Bb%7D%3D0+%5C%5C++%5C%5C+6f_%7Bb%7D+%5E%7B2%7D-i+f_%7Bb%7D%3D0+%5C%5Cf_%7Bb%7D%286f_%7Bb%7D-i%29%3D0%7E%5CLongrightarrow+%5C+%5Cleft%5Clangle%5Cbegin%7Barray%7D%7Bc%7D++f_%7Bb%7D+%3D0+%5C%5C++%5C%5C+6+f_%7Bb%7D+-i%3D0%5CLongrightarrow++%5C+%5Cboxed%7Bf_b%3Di%2F6%7D%5Cend%7Barray%7D%5Cright+%5C%5C++%5C%5C++++++++++)
Reemplazamos este hecho en la ecuación (1):
![f_{a}=i-3\left(\ \dfrac{i}{6}\right) \\ \\ f_{a}=i- \dfrac{i}{2} \\ \\ \therefore \boxed{f_{a}= i/2} f_{a}=i-3\left(\ \dfrac{i}{6}\right) \\ \\ f_{a}=i- \dfrac{i}{2} \\ \\ \therefore \boxed{f_{a}= i/2}](https://tex.z-dn.net/?f=+f_%7Ba%7D%3Di-3%5Cleft%28%5C+%5Cdfrac%7Bi%7D%7B6%7D%5Cright%29+%5C%5C++%5C%5C++f_%7Ba%7D%3Di-+%5Cdfrac%7Bi%7D%7B2%7D+%5C%5C++%5C%5C++%5Ctherefore++%5Cboxed%7Bf_%7Ba%7D%3D+i%2F2%7D++++)
Espero haberte ayudado, un saludo :)
Llamaremos a las velocidades finales de cada cuerpo:
Sabemos que en la colisión se conserva el vector cantidad de movimiento lineal:
Esa será la ecuación (1). Por otro lado, si el choque es elástico significa que la energía cinética del sistema se conserva:
Reemplazamos la ecuación (1) en la (2):
Reemplazamos este hecho en la ecuación (1):
Espero haberte ayudado, un saludo :)
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