Respuesta :
Answer:[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Explanation:
Given
mass of first particle is [tex]m_1=3.12\times 10^{-3}\ kg[/tex]
mass of second particle is [tex]m_2=7.1\times 10^{-3}\ kg[/tex]
Charge on both the particle [tex]q=8.8\times 10^{-6}\ C[/tex]
Now final speed of first particle is [tex]v_1=131\ m/s[/tex]
Final separation between particles is [tex]r=0.15\ m[/tex]
As there is no external force therefore linear momentum is conserved
[tex]0+0=m_1v_1+m_2v_2[/tex]
[tex]0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2[/tex]
[tex]v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131[/tex]
[tex]v_2=-57.56\ m/s[/tex]
Conserving total energy
Initial Kinetic energy +Initial Potential energy=Final Kinetic energy +Final Potential energy
[tex]\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}[/tex]
[tex]\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=43.177[/tex]
[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Using the law of conservation of linear momentum and conservation of total energy, the obtained initial separation between the particles is 0.01612 m.
Conservation of Linear Momentum
Given that the masses of the two particles are;
[tex]m_1 = 3.12\times 10^{-3}\,kg\\m_2 = 7.1\times 10^{-3}\,kg[/tex]
Also, the charges of both the particles are equal.
[tex]q_1 = q_2 = +8.8\times 10^{-6} \,C[/tex]
The final separation between the particles is;
[tex]r_f = 0.15\,m[/tex]
Also, the final speed of the first particle is;
[tex](v_1)_{final} = 131\,m/s[/tex]
There is no external force applied here; so by the law of conservation of linear momentum, we have;
[tex](m_1v_1)_{initial} +(m_2v_2)_{initial}=(m_1v_1)_{final} +(m_2v_2)_{final}[/tex]
But initially, the particles are at rest, so the initial velocities are zero.
[tex]0+0=(3.12\times 10^{-3}\,kg \, \times 131\,m/s ) +(7.1\times 10^{-3}\,kg\,)\,(v_2)_{final}\\[/tex]
[tex]\implies (v_2)_{final}=-\frac{3.12\times 10^{-3}\,kg \, \times 131\,m/s}{7.1\times 10^{-3}\,kg} =-57.57\,m/s[/tex]
Conservation of Energy
Now, applying the law of conservation of total energy, we get;
[tex](KE)_{\,initial}\, + \,(PE)_{\,initial} = (KE)_{\,final}\, + \,(PE)_{\,final}[/tex]
But initially, the particles are at rest; so they have no initial kinetic energy.
They have electrostatic potential alone initially.
[tex]0+k\frac{q^2}{r_i} = \frac{1}{2}(m_1 v_1)_{initial} + \frac{1}{2}(m_2 v_2)_{final} + k\frac{q^2}{r_f}[/tex]
Substituting the known values, we get;
[tex](9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_i} =[ \frac{1}{2}(3.12\times 10^{-3}\,kg )\times (131\,m/s)^2] + [\frac{1}{2}(7.1\times 10^{-3}\,kg) \times (-57.57\,m/s)^2 + (9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_f}[/tex]
[tex]\implies\frac{0.696\,Nm^2}{r_i} =26.77\,J+11.76\,J+ 4.64\,J[/tex]
[tex]\implies r_i =\frac{0.696\,Nm^2}{43.17\,J}=0.01612\,m[/tex]
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