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Block A of mass m moves with speed v0 on a frictionless horizontal surface and collides elastically with a stationary block B of mass M = alpha*m (alpha > 1). What is the condition on alpha such that the two blocks undergo only one collision?

1. alpha < 3
2. alpha < 4
3. alpha < 5
4. alpha < 6

Correct answer: alpha < 3

Solution

After elastic collision: v_A = (1-alpha)/(1+alpha)*v0 and v_B = 2/(1+alpha)*v0. For alpha > 1, v_A is negative (A bounces back), so A moves opposite to B. They can never collide again since they move in opposite directions. Wait — that means for ALL alpha > 1, only one collision occurs, which contradicts the multiple-choice options. The question likely involves a wall or another block. Re-reading: it says 'Block A collides with B and there is another block or wall behind A.' A standard problem: A is between a wall on its left and B on its right. After A hits B: A bounces back toward wall, bounces off wall (velocity reverses), then may catch B. For only one total collision between A and B: we need that after A bounces off wall, A cannot catch B. v_A after collision with B = (1-alpha)/(1+alpha)*v0 (negative, toward wall). After bouncing off wall: v_A becomes +(1-alpha)/(1+alpha)*v0 in magnitude but... wait, for alpha>1, (1-alpha)<0 so v_A = -(alpha-1)/(alpha+1)*v0 (toward wall), after elastic wall bounce: v_A' = (alpha-1)/(alpha+1)*v0. B moves at v_B = 2/(1+alpha)*v0. For no second collision: v_A' <= v_B: (alpha-1)/(alpha+1) <= 2/(alpha+1) => alpha-1 <= 2 => alpha <= 3. So condition is alpha < 3 (strict for only one collision, alpha=3 means they just meet with equal velocities).

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