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A two-sphere impact with unequal masses

Example

Take $m_{1} = 2$ , $m_{2} = 3$ , any radii $r_{1}, r_{2} > 0$ , centers with $q_{1} - q_{2} = (r_{1} + r_{2}) n$ for $n = (1, 0)$ , pre-impact velocities $v_{1}^{-} = - n$ , $v_{2}^{-} = n$ , and restitution $e = \frac{1}{2}$ . The incoming relative normal speed is $- 2$ .

Facts & Assumptions

Given: The two-sphere impact data in the Example.

[F1]

Impulse, momentum, and kinetic-energy laws for an isolated frictionless sphere impact: for $v_{n} < 0$ , $J = - (1 + e) v_{n} / (1 / m_{1} + 1 / m_{2})$ , the post-impact velocities are updated by equal and opposite normal impulses, and the kinetic-energy change is $- (1 - e^{2}) v_{n}^{2} / [2 (1 / m_{1} + 1 / m_{2})]$ .

Verification

1.1F1givenalgebra

The inverse reduced mass is $1 / m_{1} + 1 / m_{2} = 1 / 2 + 1 / 3 = 5 / 6$ , and $v_{n} = ⟨ - n - n, n ⟩ = - 2$ . Thus $J = - (3 / 2) (- 2) / (5 / 6) = 18 / 5$ . The post-impact velocities are $v_{1}^{+} = - n + (9 / 5) n = (4 / 5) n$ and $v_{2}^{+} = n - (6 / 5) n = (- 1 / 5) n$ .

2.1F1step 1.1givenalgebra∎

Total momentum is $2 (- n) + 3 n = n$ before impact and $2 (4 n / 5) + 3 (- n / 5) = n$ after. The relative normal speed after impact is $4 / 5 - (- 1 / 5) = 1 = - e v_{n}$ . The kinetic energies are $K^{-} = 1 + 3 / 2 = 5 / 2$ and $K^{+} = 16 / 25 + 3 / 50 = 7 / 10$ , so $K^{+} - K^{-} = - 9 / 5$ , matching $- (3 / 4) \cdot 4 / (2 \cdot 5 / 6) = - 9 / 5$ from [F1].

Depends on

Impulse, momentum, and kinetic-energy laws for an isolated frictionless sphere impact

Used by

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Dependency tree · two levels

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Sources

David E. Stewart, Dynamics with Inequalities: Impacts and Hard Constraints (standard reference, not scraped)