Maximum deflection angle!

Let $V'$ and $v'$ be the final speeds, and let $φ$ and $γ$ be the scattering angles of $M$ and $m$ , respectively. Then conservation of $p_x, p_y$ and $E$ gives

$MV = MV' \cos φ + mv' \cos γ ........ (1)$

$0 = MV' \sin φ - mv' \sin γ ....... (2)$

$\frac{MV^ 2}{2} =\frac{M{V'}^ 2}{2} +\frac{m{v'}^ 2}{2} ........ (3)$

On simplification, we get

$(M + m){V'}^2 - (2MV \cos φ)V' + (M - m)V^2 = 0$

A solution to this quadratic equation in $V'$ exists if and only if the discriminant is non-negative. Therefore, we must have

$(2MV \cos φ)^2 - 4(M + m)(M - m)V^2 ≥ 0$

$⇒ m^2 ≥ M^2(1 - \cos^2 φ)$

$⇒ m^2 ≥ M^2 \sin^2 φ$

$⇒\frac{m}{M} ≥ \sin φ$