Three blocks arranged with pullwy and spring as shown in fig. If the string connecting blocks `m_(2)` and `m_(3)` is cut at point A, find the acceleration of masses `m_(1), m_(2)` and `m_(3)`. Just after the string is cut at point A.

Explanation

Hint : (a) Put `m_(1)=m_(2) and u_(2)=0` in equation (x) & (xi)
(b) Put `m_(1)=0` in equations (x) & (xi)
(c ) Put `m_(2)=0` in equations (x) & (xi)
`u_(2)=0" (given)"`
From equation (x) of. Example 22
`v_(1)=((m_(1)-m_(2))u_(1))/(m_(1)+m_(2))" …(1)"`
From equation (11),
`v_(2)=(2m_(1)u_(1))/(m_(1)+m_(2))" …(2)"`
(a) Put `m_(1)=m_(2)=m` (say) in equations (1) & (2)
`v_(1)=((m-m)u_(1))/(m+m)=0`
`v_(2)=(2m u_(1))/(m+m)=u_(1)`
i.e., body A comes to rest and body B starts moving with the initial velocity of A. `100%` KE of A is transferred to the body B.
(b) `m_(2) gt gt m_(1)`
`m_(1)` can be ignored
Put m = 0 in equations (1) & (2)
`v_(1)=(-m_(2))/(0+m_(2))u_(1)=-u_(1)`
`v_(2)=(2xx0xxu_(1))/(0+m_(2))=0`
When a light body A collides against a heavy body B at rest. A rebounds with its own velocity and B continuous to be at rest e.g., a ball rebounds with same speed (direction of velocity is opposite) on striking a floor.
(c) (When larget body B at rest has negligible mass) `m_(2) ltltm_(1), m_(2)` can be ignored
Put `m_(2)=0` in equations (1) and (2)
`v_(1)=((m_(1)-0))/(m_(1)+0)u_(1)=u_(1)`
`v_(2)=(2m_(1)u_(1))/(m_(1)+0)=2u_(1)`
When a heavy body A undergoes as elastic collision with a light body at rest. A keeps on moving with the same velocity of its own and B starts moving with double the initial velocity of A.