Question

Problem 4. 15 pts Superballs. This problem is an amusing generalization of Problem 3. Suppose there are n balls stacked on top of each other, labeled B1, B2, . . . Bn. Their masses are m1 m> mn, so you can make the same sort of approximation as in P1. As before, the bottom of the Bi is initially at a height h above the floor, and the h+ height of ball Bn is at height h +l above the floor. The stack is dropped, and each ball collides elastically with its neighbor (you can think of i as a chain reaction: first Bi and B2 collide, then B2 and B3, etc) B4 B3 B2 B1 Obtain an expression for the velocity of Bn after the chain of collisions. How large must n be in order for it to achieve escape velocity from earth? (Ignoring friction and other realities)

Answer 1

m_1 > > m_2 > > m_3 > > m_4 > > ...... Initially B_1 strike the ground with speed squareroot 2gh and the respects back with same velocity situation before collision for B_1 & B_2 After collision using formula for elastic collision V_2 = u_2 (m_2 - m_1) + 2m_1u_1/m_1 + m_2 = -squareroot 2gh (m_2 - m_1) + 2m_1 squareroot 2gh/m_1 + m_2 = squareroot 2gh m_1 + 2 squareroot 2gh m_1/m_1 [Because m_1 > > m_2] (velocity of B_2) V_2 = 3 squareroot 2gh for collision of B_2 & B_3 before collision B_3 o downarrow squareroot 2gh B_2 o uparrow 3 squareroot 2gh after collision B_3 O uparrow v_3 B_2 O uparrow V_2 V_3 = -squareroot 2gh (m_3 - m_2) + 2m_2 times 3 squareroot 2gh/m_3 + m_2 = squareroot 2gh m_2 + 6 squareroot 2gh m_2/m_2 [Because m_2 > > m_3] \r\n (velocity of B_3) V_3 = 7 squareroot 2gh for collision of B_3 & B_4 situation before collision B_4 O downarrow squareroot 2gh B_3 O uparrow 7 squareroot 2gh After collision B_4 O uparrow V_4 B_3 O uparrow V_3