the two particles experience the same force (the tension in the string)
since A has twice the mass of B, its acceleration will be half of B's
so A's final velocity will be half of B's
(1/2 * 2kg * v^2) + (1/2 * 1kg * (2v)^2) = 12J
2 v^2 + 4 v^2 = 24
v^2 = 4 ___ v = 2 ___ 2v = 4
because B's acceleration and final velocity are twice that of A, B travels twice the distance of A
A + 2A = .9m
A = .3m , B = .6m
a particle a of mass 2kg and particle b of mass 1kg are connected by alight elastic string c, and initially held at rest 0.9m apart on a smooth horizontal table with the string in tension. they are then simultaneously released . the string releases 12j of energy as it contracts to its natural length. calculate the velocity acquired by each of the particles. Where do the particles collide
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1 answer