Problems

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Solution:

Since the coefficient of restitution is \(e\) and the speed of approach is \(3u\), \(v_B = v_A + 3eu\), \begin{align*} \text{COM}: && 2m\cdot2u + m \cdot (-u) &= 2m v_A + m(v_A + 3eu) \\ \Rightarrow && 3u &= 3v_A + 3eu \\ \Rightarrow && v_A &= (1-e)u \\ \Rightarrow && v_B &= (1+2e)u \end{align*} After rebounding from the wall, the velocity of \(B\) will be \(-fv_B\). So for the second collision (between the particles) we will have: \begin{align*} \text{NEL}: && w_B - w_A &= e((1-e)u+(1+2e)fu) \\ \Rightarrow && w_B - w_A &= (1-e+f+2ef)eu \tag{1} \\ \text{COM}: && 2m w_A + w_B &= 2m(1-e)u -m(1+2e)fu \\ \Rightarrow && 2w_A + w_B &= (2-2e -f-2ef)u \tag{2} \\ (2) + 2\times(1): && 3w_B &= (2-2e -f-2ef)u +2(1-e+f+2ef)eu \\ &&&= (2-2e-f-2ef)u+(2e-2e^2+2ef+4e^2f)u \\ &&&= (2-2e^2-f+4e^2f)u \\ &&&= 2(1-e^2)-f(1-4e^2)u \\ \Rightarrow && w_B &= \frac23 (1-e^2)u-\frac13(1-4e^2)fu \end{align*} Since we've always taken towards the wall as positive, the question is whether or not this is positive for all values of \(e\) and \(f\). The first term is clearly positive, so in order to have a chance of being negative, we must have that \(1-4e^2 > 0\) and \(f\) is as large as possible, so wlog \(f = 1\). \begin{align*} 2-2e^2-1+4e^2 = 1+2e^2 > 0 \end{align*} \end{align*}