Sol29
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(1) For the balls in the semicircle to deflect by an angle of π/N in each collision, the ratio of the mass of each ball (μ) to the total mass (m) cannot be too small. The maximum deflection angle is given by sinθ = μ/m, so π/N ≤ μ/m, or π ≤ M/mN. (2) After the first bounce, the speed of the ball (m) is approximately its initial speed (V) multiplied by 1 - μ/m. Each subsequent bounce reduces the speed by this same factor of 1 - μ/m. (3) For the minimum mass ratio found in part
Sol29
- 1. Solution Week 29 (3/31/03) Balls in a semicircle (a) Let µ ≡ M/N be the mass of each ball in the semicircle. We need the deflection angle in each collision to be θ = π/N. However, if the ratio µ/m is too small, then this angle of deflection is not possible. From Problem of the Week #25, the maximal angle of deflection in each collision is given by sin θ = µ/m. (We’ll just invoke this result here.) Since we want θ = π/N here, this sin θ ≤ µ/m condition becomes (using sin θ ≈ θ, for the small angle θ) θ ≤ µ m =⇒ π N ≤ M/N m =⇒ π ≤ M m . (1) (b) Referring back to the solution to Problem #25, we see that m’s speed after the first bounce is obtained from the following figure. mV m+µ µV m+µ ____ ____V θmax f Looking at the right triangle, we see that the speed after the bounce is Vf = V m2 − µ2 m + µ . (2) To first order in the small quantity µ/m, this equals Vf ≈ mV m + µ ≈ V 1 − µ m . (3) The same reasoning holds for each successive bounce, so the speed decreases by a factor of (1 − µ/m) after each bounce. In the minimum M/m case found in part (a), we have µ m = M/N m = M/m N = π N . (4) Therefore, the ratio of m’s final speed to initial speed equals Vfinal Vinitial ≈ 1 − π N N ≈ e−π . (5) That’s a nice result, if there ever was one! Since e−π is roughly equal to 1/23, only about 4% of the initial speed remains. 1