Question 2

Let the initial velocity of the heavier particle ($2m$) be $\vec{u}$, and its final velocity be $\vec{v}_1$ at an angle $\theta$ with the initial direction. Let the final velocity of the lighter particle ($m$) be $\vec{v}_2$.

Step 1: Conservation of Linear Momentum Since there are no external forces, the initial momentum equals the final momentum:

$\vec{P}_{initial} = \vec{P}_{final}$

$2m\vec{u} = 2m\vec{v}_1 + m\vec{v}_2$

Dividing the entire equation by $m$: $2\vec{u} = 2\vec{v}_1 + \vec{v}_2 \implies \vec{v}_2 = 2(\vec{u} - \vec{v}_1)$

To find the magnitude squared of $\vec{v}_2$, we take the dot product of $\vec{v}_2$ with itself:

$v_2^2 = \vec{v}_2 \cdot \vec{v}_2 = 4(\vec{u} - \vec{v}_1) \cdot (\vec{u} - \vec{v}_1)$

$v_2^2 = 4(u^2 + v_1^2 - 2uv_1 \cos\theta)$ --- (Equation 1)

Step 2: Conservation of Kinetic Energy Because the collision is perfectly elastic, kinetic energy is conserved: $K_{initial} = K_{final}$ $\frac{1}{2}(2m)u^2 = \frac{1}{2}(2m)v_1^2 + \frac{1}{2}mv_2^2$

Multiplying the equation by $\frac{2}{m}$ to simplify: $2u^2 = 2v_1^2 + v_2^2 \implies v_2^2 = 2(u^2 - v_1^2)$ --- (Equation 2)

Step 3: Equating and forming a Quadratic Equation Equating the expressions for $v_2^2$ from Equation 1 and Equation 2:

$4(u^2 + v_1^2 - 2uv_1 \cos\theta) = 2(u^2 - v_1^2)$

Dividing by 2 and expanding: $2u^2 + 2v_1^2 - 4uv_1 \cos\theta = u^2 - v_1^2$

$3v_1^2 - (4u\cos\theta)v_1 + u^2 = 0$

Step 4: Condition for Real Roots (Maximum Angle) This is a quadratic equation in terms of $v_1$. For a real physical collision to occur, the final velocity $v_1$ must be a real number. Therefore, the discriminant ($D$) of this quadratic equation must be greater than or equal to zero ($D \ge 0$).

$b^2 - 4ac \ge 0$

$(-4u\cos\theta)^2 - 4(3)(u^2) \ge 0$

$16u^2\cos^2\theta - 12u^2 \ge 0$

$4u^2(4\cos^2\theta - 3) \ge 0$

Since $4u^2$ is always positive, we get:

$4\cos^2\theta - 3 \ge 0 \implies \cos^2\theta \ge \frac{3}{4}$

Taking the square root (considering $\theta$ is an acute angle in forward scattering): $\cos\theta \ge \frac{\sqrt{3}}{2}$

For the maximum scattering angle $\theta_{max}$, $\cos\theta$ must be at its minimum allowable value:

$\cos\theta_{max} = \frac{\sqrt{3}}{2}$

$\theta_{max} = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \text{ radians}$

Final Answer: The maximum angular deviation $\theta$ is $\frac{\pi}{6}$ radians.