Collisions in Two Dimensions — Question 1

Question 1

Perpendicular approaches followed by sticking

A $3\,\mathrm{kg}$ puck moving east at $4\,\mathrm{m/s}$ sticks to a $4\,\mathrm{kg}$ puck moving north at $3\,\mathrm{m/s}$. They move on a frictionless horizontal plane and are treated as particles, with no final rotation.

1. Find the combined velocity vector, speed and direction.

2. Find the kinetic-energy loss and explain why adding the two speeds would fail.

Question 1 - Solution

Perpendicular approaches followed by sticking

(a) Derivation and calculation

Resolve the incident momenta into east and north components: $$\vec P_i=3(4,0)+4(0,3)=(12,12)\ \mathrm{kg\,m/s}.$$ $$\boxed{\vec V=\frac{\vec P_i}{7}=(12/7,12/7)\ \mathrm{m/s}}.$$ $$\boxed{|\vec V|=12\sqrt2/7=2.424\ \mathrm{m/s}},\qquad\boxed{\theta=\tan^{-1}(12/12)=45^\circ}.$$

(b) Result and interpretation

The energies are $$K_i=\tfrac12(3)(4)^2+\tfrac12(4)(3)^2=42\ \mathrm J,\qquad K_f=\tfrac12(7)\frac{288}{49}=\frac{144}{7}\ \mathrm J.$$ $$\boxed{L=42-144/7=150/7=21.43\ \mathrm J}.$$ Perpendicular momentum components combine by vector addition; adding scalar speeds discards their directions and mass weights.

Physical check

The two equal momentum components produce a 45-degree direction despite unequal masses and speeds.