202409051606 Pastime Exercise 012

About the figure below, tell some stories as probable as probable can be.

Roughwork.

Ignoring gravitational effect, denote the initial and the final velocity before and after impact by $\mathbf{u}=(u_x,u_y)$ and $\mathbf{v}=(v_x,v_y)$ . Taking the positive an up and a right, then $u_x,u_y,v_x<0$ and $v_y>0$ . Under conservation of momentum,

$\begin{aligned} m|\mathbf{u}|+M|\mathbf{U}| & =m|\mathbf{v}|+M|\mathbf{V}| \\ m|(u_x\,\hat{\mathbf{i}}+u_y\,\hat{\mathbf{j}})| + M|(\mathbf{0})| & = m|(v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}})| + M|(\mathbf{0})| \\ |u_x| & = |v_x| \\ |u_y| & = |v_y| \\ \textrm{( } |u_i| & = |v_i| \textrm{ )}\\ \end{aligned}$

and conservation of energy

$\begin{aligned} \frac{1}{2}mu^2 + \frac{1}{2}MU^2 & = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 \\ m(u_x^2+u_y^2) + M(0)^2 & = m(v_x^2+v_y^2) + M(0)^2 \\ u_x^2+u_y^2 & = v_x^2 + v_y^2 \\ \sum u_i^2 & = \sum v_i^2 \\ \end{aligned}$

Therefore

$\begin{aligned} \frac{|u_x|}{|u_y|} & = \frac{|v_x|}{|v_y|} \\ \tan\theta_i & = \tan\theta_r \\ \theta_i & = \theta_r \\ \end{aligned}$

(to be continued)

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