Collisions in Two Dimensions — Question 2

Question 2

Why equal-mass elastic departures are perpendicular

A puck strikes an identical stationary puck elastically on a frictionless plane. Initially $\vec u=(10,0)\,\mathrm{m/s}$. Both pucks move after collision, and the incident puck leaves at $30^\circ$ above $+x$.

1. Prove that their final velocities are perpendicular.

2. Find both final speeds and the target's direction.

Question 2 - Solution

Why equal-mass elastic departures are perpendicular

(a) Derivation and calculation

For equal masses, momentum conservation gives $\vec u=\vec v_1+\vec v_2$. Square this relation and compare with elastic energy conservation: $$u^2=v_1^2+v_2^2+2\vec v_1\cdot\vec v_2,\qquad u^2=v_1^2+v_2^2.$$ $$\boxed{\vec v_1\cdot\vec v_2=0}.$$ Both velocities are nonzero, so their departure directions are perpendicular.

(b) Result and interpretation

The velocity triangle has hypotenuse $u=10$ m/s: $$\boxed{v_1=u\cos30^\circ=5\sqrt3\ \mathrm{m/s}},\qquad\boxed{v_2=u\sin30^\circ=5\ \mathrm{m/s}}.$$ Vertical momentum cancellation selects the target direction below the axis: $$\boxed{\theta_2=30^\circ-90^\circ=-60^\circ}.$$

Physical check

The squared final speeds sum to $75+25=100$ m$^2$/s$^2$.