Collisions in Two Dimensions — Question 4

Question 4

Perpendicular departures do not prove elasticity for unequal masses

A $2\,\mathrm{kg}$ projectile moving at $6\,\mathrm{m/s}$ along $+x$ hits a stationary $1\,\mathrm{kg}$ target. The projectile leaves at $30^\circ$ and the target at $-60^\circ$. External impulse is negligible.

1. Use the two momentum components to find both final speeds.

2. Test whether the claimed event can be elastic or passive inelastic.

Question 4 - Solution

Perpendicular departures do not prove elasticity for unequal masses

(a) Derivation and calculation

The two momentum equations are $$12=2v_1\cos30^\circ+v_2\cos60^\circ,\qquad0=2v_1\sin30^\circ-v_2\sin60^\circ.$$ The second gives $v_2=2v_1/\sqrt3$. Substitution into the first gives $$12=\frac4{\sqrt3}v_1\Rightarrow\boxed{v_1=3\sqrt3,\quad v_2=6\ \mathrm{m/s}}.$$

(b) Result and interpretation

Test the energy rather than assuming the right-angle rule: $$K_i=\tfrac12(2)(6)^2=36\ \mathrm J,\qquad K_f=\tfrac12(2)(27)+\tfrac12(1)(36)=45\ \mathrm J.$$ $$\boxed{\Delta K=+9\ \mathrm J}.$$ The event requires internal energy release. It is neither elastic nor passive inelastic; the usual perpendicular-departure result assumes equal masses.

Physical check

The momentum constraints permit the directions, but energy identifies the required active release.