Explain the motion of blocks connected by a string in (i) vertical motion (ii) horizontal motion .

Explanation

Case 1: vertical motion consinder two blocks of masses `m_(1) and m_(2) (m_(1)gtm_(2))` connected by a light and inextensible string that passes over a pilley as shown in figure .
Let the tension in the string be T and acceleration a when the system is released both the blocks start moving `m_(2)` vertically upward and `m_(1)` downward with same acceleration a The gravitaional force `m_(1)g` on mass ` m_(1)` is used in lifting the mass `m_(2)`.
The upward direction is choosen as y direction The free body diageams of both masses are shown in figure
Appying newton"s second law for mass `m_(2)`
`ThatJ-m_(2)ghatj=m_(2)ahatj`
The left hand side above equation is the total force that acts on `m_(2)` and the right hand side is the product of mass and acceleration on the both sides, we get .
`Thatj-m_(2)ghatj=m_(2)ahatj`
Si,ilarly applying newtons.s decond law for mass `m_(1)`
`T-m_(1)g=-m_(1)a`
`m_(1)g--T=m_(1)a`
the shows that if the masses are equal there is no acceleratiion and the system as a whole will be at rest .
To find the tension acting on the string , substitute from the equation (4) into the equation (1) .
by taking `m_(2)g` commn in the RHS of equation ..
`T=m_(2)g(1+(m_(1)-m_(2))/(m_(1)+m_(2)))`
`T=m_(2)g(1+(m_(1)+m_(2)+m_(1)-m_(2))/(m_(1)+m_(2)))`
`T=((2m_(1)m_(2))/(m_(1)+m_(2)))g`
For mass `m_(1)` the acceleration vector is given by `veca=-((m_(1)-m_(2))/(m_(1)+m_(2)))ghatj`
For mass `m_(2)` the acceleration vector is given by `veca=((m_(1)-m_(2))/(m_(1)+m_(2)))ghatj`.