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We have two particles on frictionless rails:

S1: with Mass $m_1$ and Velocity $v_1=0\frac{m}{s}$

S2: with Mass $m_2$ and Velocity $v_2$

$`m_1 >= m_2`$

So particle S2 is moving toward particle S1 which is initially at rest with a constant speed $V_2$

S1 and S2 collide: So the question is how can we choose a ratio $\frac{m_1}{m_2}$ in order to get a maximum distance L covered on the railenter image description here

Knowing that: $L= \frac{4m_{2}^{2}V_{2}^{2}}{g(m_{1}+m_{2})^2}$

The answer is 1 but i just don't understand why?

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  • $\begingroup$ If the equation is correct , then $m_1 = 0 $ results in the highest $L$, since that gives the lowest denominator. $\endgroup$
    – JAlex
    Dec 27, 2023 at 19:43

3 Answers 3

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In the expression for $L$, divide numerator and denominator by $m_2^2$. Then choose a value for $\frac{m_1}{m_2}$ which makes the denominator as small as possible, subject to the condition $m_1 \ge m_2$.

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Consider what happens in the 2 extreme cases where S1 is very heavy and where it is very light.

  1. If S1 is light it will not have much kinetic energy to impart to S2, it will bounce off.
  2. If S1 is heavy it will not impart much of its kinetic energy to S2, it will retain most of it. Because this scenario is a situation that is actually the the same as 1) with the roles of S1 and S2 reveresed. If the frame of reference is centered on the heavy particle, S1 in this case. (In actuality the frame of reference for resolving collisions is focused on the center of mass of both particles, but in the extreme cases this is very similar to focus on the heavy particle)
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For maximum distance up the incline you need maximum energy imparted to mass 1. This is exactly the case here when both masses are equal because then mass 1 takes over the full energy of mass 2 (which comes to rest after the collision).

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