11
virang1 Jhaveri
·2009-03-21 02:53:37
Which is heavier . That is important for taking decisions
11
virang1 Jhaveri
·2009-03-21 02:57:40
If m1 is stopped the string will be taut the very next second since m1 was being pulled by m2
11
virang1 Jhaveri
·2009-03-21 03:00:15
if we consider m1>m2
Then m1 is pulling m2. Since m1 stops it stops pulling m2. m2 will be having a velocity as it was accelerated by m1 before it stopped. The velocity is upwards. Therefore the string will become taut again after v becomes v(with opposite direction.
1
Rohan Ghosh
·2009-03-21 03:12:46
for m1>m2
well just before stopping they will have their velocities equal=v
when m1 is stopped its velocity goes to zero
then the tension vanishes and regains again when distance between them = length of thread again
after stopping m1 faces g downwards with zero velocity
m2 also faces g downwards with v velocity upwards
so let the string become taut after time t
vt-(1/2)gt2=1/2gt2
In this way we can obtain the time
also if the lighter mass is stopped in the above case
we have (let t again be time of tension regain)
vt+1/2gt2+1/2gt2=0
which gives t=0 which means the string will instantly become taut
1
AARTHI
·2009-03-21 21:21:18
tx..me 2 had da same doubt