1We can use the concept of work energy theorm .
Work doen by gravity and work done by tensions = change in K.E
initial velocity must be zero, final we have to calculate.
when released ,the mass 2m falls a distance x before the limp string becomes taut.thereafter both the mass on the left rise with the same speed.what is the final speed.
i have applied impulse momentum theoram.but not gettin te final answer
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6 Answers
1
3g/3 = accn until it becomes taut ...
v^2 = 2(g/3)x
v(i)=√(2gx/3)
now a sudden impulsive force [tnsion] acts .. say T for time dt
for block just rising
Tdt = mv [for lowermost block][neglecting mg and all dat ]
(T1-T)dt = mv - mv(i)
-T1dt = 2mv - 2mv(i)
add all three
4mv -3mv(i) = 0
v =(3/4)v(i) ..... but v(i) value from before
3assumin the right block is not on the floor when the left string becomes taut ........
[and for clarification T1 is the implusive tension developed in the second string due to the first]