Sol19
0 likes170 views
1) The document analyzes the collision between a bouncing ball and a block. It shows that the quantity (v-V), where v is the ball's speed and V is the block's speed after each collision, remains constant. 2) It then derives an expression for the minimum distance Lmin the block reaches from the wall in terms of the mass ratio m/M of the ball and block. 3) An alternative solution is presented that uses conservation of momentum to derive expressions for the speeds of the ball and block after each collision in terms of the number of bounces N. This allows calculating the mass ratio m/M needed for the block to come to rest after a given number of bounces N
Sol19
- 1. Solution Week 19 (1/20/03) Block and bouncing ball (a) Consider one of the collisions. Let it occur at a distance from the wall, and let v and V be the speeds of the ball and block, respectively, after the collision. We claim that the quantity (v − V ) is invariant. That is, it is the same for each collision. This can be seen as follows. The time to the next collision is given by V t + vt = 2 (because the sum of the distances traveled by the two objects is 2 ). Therefore, the next collision occurs at a distance from the wall, where = − V t = − 2 V V + v = (v − V ) v + V . (1) Therefore, (v + V ) = (v − V ). (2) We now make use of the fact that in an elastic collision, the relative speed before the collision equals the relative speed after the collision. (This is most easily seen by working in the center of mass frame, where this scenario clearly satisfies conservation of E and p.) Therefore, if v and V are the speeds after the next collision, then v + V = v − V . (3) Using this in eq. (2) gives (v − V ) = (v − V ), (4) as we wanted to show. What is the value of this invariant? After the first collision, the block continues to move at speed V0, up to corrections of order m/M. And the ball acquires a speed of 2V0, up to corrections of order m/M. (This can be seen by working in the frame of the heavy block, or equivalently by using eq. (3) with V ≈ V = V0 and v = 0.) Therefore, the invariant (v − V ) is essentially equal to L(2V0 − V0) = LV0. Let Lmin be the closest distance to the wall. When the block reaches this closest point, its speed is (essentially) zero. Hence, all of the initial kinetic energy of the block now belongs to the ball. Therefore, v = V0 M/m, and our invariant tells us that LV0 = Lmin(V0 M/m − 0). Thus, Lmin = L m M . (5) (b) (This solution is due to Slava Zhukov) With the same notation as in part (a), conservation of momentum in a given collision gives MV − mv = MV + mv . (6) 1
- 2. This equation, along with eq. (3),1 allows us to solve for V and v in terms of V and v. The result, in matrix form, is V v = M−m M+m −2m M+m 2M M+m M−m M+m V v . (7) The eigenvectors and eigenvalues of this transformation are A1 = 1 −i M m , λ1 = (M − m) + 2i √ Mm M + m ≡ eiθ , A2 = 1 i M m , λ2 = (M − m) − 2i √ Mm M + m ≡ e−iθ , (8) where θ ≡ arctan 2 √ Mm M − m ≈ 2 m M . (9) The initial conditions are V v = V0 0 = V0 2 (A1 + A2). (10) Therefore, the speeds after the nth bounce are given by Vn vn = V0 2 (λn 1 A1 + λn 2 A2) = V0 2 einθ 1 −i M m + e−inθ 1 i M m = V0 cos nθ M m sin nθ . (11) Let the block reach its closest approach to the wall at the Nth bounce. Then VN = 0, and so eq. (11) tells us that Nθ = π/2. Using the definition of θ from eq. (9), we have N = π/2 arctan 2 √ Mm M−m ≈ π 4 M m . (12) Remark: This solution is exact, up to the second line in eq. (12), where we finally use M m. We can use the first line of eq. (12) to determine the relation between m and M for which the Nth bounce leaves the block exactly at rest at its closest 1 Alternatively, you could use conservation of energy, but this is a quadratic statement in the velocities, which makes things messy. Conservation of energy is built into the linear eq. (3). 2
- 3. approach to the wall. For this to happen, we need the N in eq. (12) to be an integer. Letting m/M ≡ r, we can rewrite eq. (12) as 2 √ r 1 − r = tan π 2N ≡ 1 − cos β 1 + cos β , (13) where we have used the tan half-angle formula, with β ≡ π/N. Squaring both sides and solving the resulting quadratic equation for r gives r = 3 + cos β − 2 √ 2 + 2 cos β 1 − cos β . (14) If we want the block to come to rest after N = 1 bounce, then β = π gives r = 1, which is correct. If we want N = 2, then β = π/2 gives r = 3 − 2 √ 2 ≈ 0.172. If we want N = 3, then β = π/3 gives r = 7 − 4 √ 3 ≈ 0.072. For general N, eq. (14) must be computed numerically. For large N, the second line in eq. (12) shows that r goes like 1/N2 . More precisely, r ≈ π2 /(16N2 ). 3