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Physics · Ch 5 — Work, Energy and Power

Collisions in Two Dimensions

5.12

Collisions in Two Dimensions

When the colliding bodies are not confined to a single straight line -- a glancing or oblique collision, such as one billiard ball striking another slightly off-centre -- the outgoing velocities generally point in different, non-collinear directions, and momentum conservation, being a vector statement, must be applied separately along two independent, usually perpendicular, directions (conventionally chosen along the original direction of motion, the xx-axis, and perpendicular to it, the yy-axis).

Setting up the two-dimensional equations. Consider a body of mass m1m_1 moving with initial speed u1u_1 along the xx-axis, striking a second body of mass m2m_2 initially at rest. After the collision, body 11 moves off at speed v1′v_1' at angle θ1\theta_1 above the xx-axis, and body 22 moves off at speed v2′v_2' at angle θ2\theta_2 below the xx-axis (or, more generally, at whatever two angles the actual collision produces). Momentum conservation along each axis separately gives

x-direction: m1u1=m1v1′cos⁡θ1+m2v2′cos⁡θ2x\text{-direction: } \quad m_1u_1 = m_1v_1'\cos\theta_1 + m_2v_2'\cos\theta_2

y-direction: 0=m1v1′sin⁡θ1−m2v2′sin⁡θ2y\text{-direction: } \quad 0 = m_1v_1'\sin\theta_1 - m_2v_2'\sin\theta_2

(the two yy-components must be equal and opposite, since there was no yy-momentum at all before the collision). If, in addition, the collision is elastic, a third equation, kinetic-energy conservation, 12m1u12=12m1v1′2+12m2v2′2\tfrac12 m_1u_1^2 = \tfrac12m_1v_1'^2 + \tfrac12m_2v_2'^2, is also available; together, these equations can be solved for the two unknown outgoing speeds when the two outgoing angles are known (or, conversely, for the angles if the speeds are known), though in general a two-dimensional collision needs more information to be fully determined than a one-dimensional one, since there are more unknowns (two speeds and two angles) than equations.

A well-known special result -- equal masses, elastic, one initially at rest. When m1=m2=mm_1 = m_2 = m and the collision is elastic, a short vector argument gives a striking geometric fact. Momentum conservation as a single vector equation reads u⃗1=v⃗1′+v⃗2′\vec u_1 = \vec v_1' + \vec v_2'; squaring both sides (taking the dot product of each side with itself) gives

u12=v1′2+v2′2+2v⃗1′⋅v⃗2′u_1^2 = v_1'^2 + v_2'^2 + 2\vec v_1'\cdot\vec v_2' …

Figure 1Vector diagram of an oblique (two-dimensional) elastic collision

What this figure shows. A diagram with a horizontal dashed reference line representing the initial direction of motion. On the left, a single solid arrow labelled u⃗1\vec u_1 represents the incoming ball's velocity before collision, drawn along the dashed line, striking a second, stationary ball drawn as a small circle at the origin marked with u⃗2=0\vec u_2 = 0. On the right, after the collision point, two solid arrows radiate outward from the origin at different angles to the dashed reference line: v⃗1′\vec v_1', drawn at an angle θ1\theta_1 above the dashed line, and v⃗2′\vec v_2', drawn at an angle θ2\theta_2 below the dashed line, with a small square-corner mark placed at the origin between the two outgoing arrows to indicate that, for this equal-mass elastic case, θ1+θ2=90∘\theta_1 + \theta_2 = 90^\circ. Faint dashed construction lines drop perpendiculars from the tips of v⃗1′\vec v_1' and v⃗2′\vec v_2' onto the original dashed line, illustrating how the xx- and yy-components of …

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