1
by reduced mass system formula
wew have
ω=√3k2m
xmax can be found using two equations
momentum conservation
4v=3v'
v'=4v/3
and
energy conservation
kxmax2=2mv'2
now
x=xmaxsinωt
Q1. Two blocks of mass m and 2m respectively are placed on smooth horizontal ground connected by a spring of spring const. K. Initially the spring is unstretched. (natural length L)
m is on the left and 2m is on the right
m is given a velocity 2v towards right and 2m is given a velocity v towards right,
Find:
(i) time period (of individual blocks)
(ii) max and min dist between the blocks
(iii) range of speeds obtained by the block m
(iv) Eqn of motion of the two blocks (if possible)
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9 Answers
1
49hey AKARI...y 4v=3v'???
shudn't it be mv=2mv', so v=2v'??
no no...instead of asking this i should have asked what are v and v'??
106but akari..
why in the equation u get xmax why not xmin ??
As I see it using that eqn #2 .. will (should, may) give xmin
23asish ,
let m move to the right by x1, and 2m by x2 to the right
ten distance between m and 2m is( L+ x2 - x1)
using momentum conservation and conservation of KE , as akari has done , we can find
x1 - x2 = ± (2mv2/3k)1/2
x1 - x2 will be negative when x1 is to the left , i.e negative of wat we assumed earlier ,
and hence it will account for the condition of max distance
dus distance max = L + (2mv2/3k)1/2
distancemin = L - (2mv2/3k)1/2
and w ecan find eqn by the method wich akari gave
106wat abt range of speeds then..
agreed abt the previous questions...
