Question
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
Answers
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
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