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All Mechanics DOUBTS HERE!!!

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astarmathsandphysics:
For 2kg 2g-t=2a
For 1 kg t-g=a
Add there two equations g=3a so a=g/3
When string breaks particle falls under gravity so a=g

astarmathsandphysics:
Height=1.5+0.5*g/3*0.5^2 then use this height with s=1/2at^2 to find find t

halosh92:

--- Quote from: astarmathsandphysics on May 04, 2010, 04:23:11 pm ---Height=1.5+0.5*g/3*0.5^2 then use this height with s=1/2at^2 to find find t

--- End quote ---

how did u exactly get this?
0.5*g/3*0.5^2    and for s=1/2at^2  do we use 9.8 for the acceleration?
thx

halosh92:
two particles A and B of masses 0.5kg and 0.7kg respectively are connected by a light inextensible string. particle A rests on a smooth horizontal table 1m from a fixed smooth pulley and B hangs freely 0.75m above the floor. the system is released from rest with the string taut. after 0.5s the string breaks.
find:
a) the distance the particles haved moved when the string broke
b) the velocity of the particles when the string breaks
c) the furthur time that elapses before A reaches the floor, given that the table is 0.9m high.


how to do part c ??

Saladin:
can you please send the exact question paper. I need to see that.

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