Two 100 g blocks hang at the ends of a light flexible cord passing over a small frictionless pulley. A 40 g block rests on the block an right and removed after 2 sec.
(a) How far will each block move in the first second after the 40 g block is removed?
(b) What was the tension in the cord before the 40 g block was removed?
(c) What was the tension in the cord supporting the pulley after the 40 g block was removed?

Explanation

When two blocks of masses `m_(1)` and `m_(2)` are hanging from the ends of a string passing over a fixed pulley their acceleration is:
`a=((m_(2)-m_(1))g)/((m_(2)+m_(1))`
and the tension is: `T=(2m_(1)m_(2)g)/(m_(1)+m_(2))`
From `t=0 " to "t=2 (M=100g,m=40g)`
`a=([(M+m)-M]g)/([M+m+M])=g/6`
`v=u+at=0+(g//6)(2)=980//3 " cm"//s`
Hence, at `t=2` s left block is moving up with velocity `980//3 " cm"//s` and right block is moving down with velocity `980//3 " cm//s` Tension in the string during this interval is:
`T=(2M(m+M)g)/(m+2M)`
`=1.143 N`
From `t=2 " to" t=3` when m is removed, blocks move with a constant velocity
The distance covered `=s=(980//3)xx1`
`=326.6 cm=4.226 m`
Tension in the string = weight of each block
`T=Mg=0.98N`
`and T' = 2T=1.96N`