Question

Two balls A and B of equal mass $m$ are connected by an ideal string of length 2.5 m and kept contact with each other on smooth horizontal floor. When ball A is given an upward velocity maximum height attain by the center of mass of the balls is $\frac{\alpha}{\beta}$ where $\alpha$ and $\beta$ are coprime. Find $\alpha$ - $\beta$.

Answer 1

1

Answer 2

We shall show that—with the correct “just–lift‐off” choice of the imparted speed—the maximum height reached by the centre‐of–mass (CM) comes out as   CM height = 5⁄4 m, so that if this is written as α⁄β where α and β are coprime then   α − β = 5 − 4 = 1.

Below is one acceptable concise “JEE–style” solution.

Solution:

• Two equal masses m lie on a smooth floor. They are “connected” by a string of length L = 2.5 m but are kept in contact (the slack in the string means initially the centres are nearly together).

• At t = 0, ball A is given an upward speed v₀ while ball B remains at rest. Thus the initial CM speed is   v_CM(0) = (v₀ + 0)/2 = v₀⁄2. Since only weight acts on them as a system, the CM acceleration is −g so that if no further impulsive effect occurred the maximum additional rise would be   Δh_CM = (v_CM(0))²⁄(2g) = v₀²⁄(8g).

• However the string is slack until A separates from B by L. Choosing the “just–lift–off” (minimum–velocity) case, ball A rises exactly L = 2.5 m before the slack is taken up (B still at the floor). Let time to reach that height be t₁. Then   2.5 = v₀ t₁ − ½g t₁²            (1) and at that instant A’s speed is   v_A = v₀ − g t₁. For the minimum v₀ required so that B is just raised by the impulsive force transmitted when the string becomes taut, one may show that the impulse brings about an “equalization” of the speeds. Conservation of momentum in the impulsive event gives   m v_A + m·0 = 2m·u ⟹ u = v_A⁄2. For the least v₀, the common speed u must be zero (any leftover upward speed would lift the CM even higher). Hence we require   v_A = 0 ⟹ v₀ = g t₁.            (2) Substitute (2) into (1):   2.5 = g t₁² − ½g t₁² = ½g t₁²  ⟹ t₁² = (5/g) ⟹ t₁ = √(5/g). Then   v₀ = g t₁ = √(5g).

• Now, the centre–of–mass in the system before the impulse is at   y_CM = (y_A + y_B)/2 = (2.5 + 0)/2 = 1.25 m. Since at the impulse the common speed becomes u = 0, there is no further kinetic energy available to lift the CM. Thus the maximum height reached by the CM is 1.25 m or   1.25 = 5⁄4 m. Writing this as α⁄β with α and β coprime gives α = 5, β = 4 so that   α − β = 5 − 4 = 1.

Summary (Minimal Core Points):

1. Initial CM speed = v₀⁄2.
2. In the minimum–velocity case the slack is taken up when ball A has risen 2.5 m and v₀ = g t₁.
3. At that moment, impulsive equalization makes both speeds zero so that CM height = (2.5⁄2) = 5⁄4 m.
4. Thus, if 5⁄4 = α⁄β then α − β = 1.