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Two particles A (mass 2m) and B (mass m) hang from the ends of a light inextensible string of total length 4a that runs over a small smooth peg fixed 3a above an inelastic table. Released from rest with both particles a height a above the table, find: (i) B's speed at the instant A reaches the table, (ii) the time taken for A to first reach the table, and (iii) how long A remains on the table after this before the string jerks it up again.

1. (i) sqrt(2ga/3), (ii) sqrt(6a/g), (iii) 2*sqrt(2a/(3g))
2. (i) sqrt(2ga/3), (ii) sqrt(3a/g), (iii) sqrt(2a/(3g))
3. (i) sqrt(ga/3), (ii) sqrt(6a/g), (iii) 4*sqrt(2a/(3g))
4. (i) sqrt(4ga/3), (ii) 2*sqrt(6a/g), (iii) 2*sqrt(a/(3g))

Correct answer: (i) sqrt(2ga/3), (ii) sqrt(6a/g), (iii) 2*sqrt(2a/(3g))

Solution

While both hang, acceleration a_sys = (2m-m)g/(3m) = g/3 over distance a, giving v = sqrt(2(g/3)a). After A lands B rises freely then the string re-tightens, jerking A up. The on-table time is the free-flight time of B until the string is taut again.

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