InterviewSolution
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InterviewSolution
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A proton moves from a large distance with a speed u m/s directly towards a free proton originally at rest. Find the distance of closest of closest approach for the two protons in terms of mass of proton m and its charge e. |
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Answer» Since the particle rest is free to move, when one particle approaches to other, due to electrostatic replusion, the other particle will also start moving. So the velocity of the first particle will decrease while that of the other particle will increase, and at the closest approach both will move with the same velocity. So, if v is the common velocity of each particle at the closest approach, as no external forces are acting on them (the system of particles), the linear momentum of the system of particles will be conserved. By conservation of momentum, we get `m u = mv + mv`, `i.e., v = (1)/(2) u` And by conservation of energy, we have `(1)/(2) m u^2 = (1)/(2) m v^2 + (1)/(2) m V^2 + (1)/(4 pi epsilon _0) (e^2)/( r)` So, `r = (e^2)/(pi epsilon_0 m u^2)` `(as v = u//2)`. |
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