Collisions in Two Dimensions — Question 10

Section 9-10 · Physics pages 1350–1351

Question 10 Crossing paths is a separate requirement from vector conservation Two small pucks move freely on a horizontal plane. Puck A starts at (−2, 0) m with velocity (2, 0) m/s; puck B starts at (0, −3) m with velocity (0, 2) m/s. Treat them as points. x y A B Initial geometry; both speeds are 2 m/s in part (a). (a) Determine whether they collide and find their minimum separation. (b) What speed must B have along +y for collision at the crossing, and why must this be checked before applying collision equations?
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Question 10 - Solution Crossing paths is a separate requirement from vector conservation x y A B (a) Derivation and calculation The trajectories are⃗rA = (−2 + 2t, 0) and⃗rB = (0, −3 + 2t). Their origin-crossing times are 1 s and 1.5 s, so they miss. The separation satisfies d2 = (2 −2t)2 + (−3 + 2t)2 = 8(t −1.25)2 + 0.50. tmin = 1.25 s, dmin = 1/ √ 2 m . (b) Result and interpretation For B to reach the origin simultaneously with A, −3 + vB(1) = 0 ⇒vB = 3 m/s . A collision requires the same position at the same time. Conservation equations for an interaction apply only once this contact condition is satisfied. Physical check Intersecting geometric paths alone do not guarantee a collision.

Question 10

Crossing paths is a separate requirement from vector conservation

Two small pucks move freely on a horizontal plane. Puck A starts at (−2,0)m(-2,0)\,\mathrm m with velocity (2,0)m/s(2,0)\,\mathrm{m/s}; puck B starts at (0,−3)m(0,-3)\,\mathrm m with velocity (0,2)m/s(0,2)\,\mathrm{m/s}. Treat them as points.

Diagram shown in the original SVG worksheet.

  1. Determine whether they collide and find their minimum separation.

  2. What speed must B have along +y+y for collision at the crossing, and why must this be checked before applying collision equations?

Question 10 - Solution

Crossing paths is a separate requirement from vector conservation

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The trajectories are r→A=(−2+2t,0)\vec r_A=(-2+2t,0) and r→B=(0,−3+2t)\vec r_B=(0,-3+2t). Their origin-crossing times are 1 s and 1.5 s, so they miss. The separation satisfies d2=(2−2t)2+(−3+2t)2=8(t−1.25)2+0.50.d^2=(2-2t)^2+(-3+2t)^2=8(t-1.25)^2+0.50. tmin=1.25s,dmin=1/2m.\boxed{t_{\min}=1.25\ \mathrm s,\qquad d_{\min}=1/\sqrt2\ \mathrm m}.

(b) Result and interpretation

For B to reach the origin simultaneously with A, −3+vB(1)=0⇒vB=3m/s.-3+v_B(1)=0\Rightarrow\boxed{v_B=3\ \mathrm{m/s}}. A collision requires the same position at the same time. Conservation equations for an interaction apply only once this contact condition is satisfied.

Physical check

Intersecting geometric paths alone do not guarantee a collision.