Two balls , having linear momenta `vec(p)_(1) = p hat(i)` and `vec(p)_(2) = - p hat(i)` , undergo a collision in free space. There is no external force acting on the balls. Let `vec(p)_(1)` and `vec(p)_(2)`, be their final momenta. The following option(s) is (are) NOT ALLOWED for any non -zero value of `p , a_(1) , a_(2) , b_(1) , b_(2) , c_(1)` and `c_(2)`
(i) `vec(p)_(1) = a_(1) hat(i) + b_(1) hat(j) + c_(1) hat(k) , vec(p)_(2) = a_(2) hat(i) + b_(2) hat(j)`
(ii) `vec(p)_(1) = c_(1) vec(k) , vec(p)_(2) = c_(2) hat(k)`
(iii) `vec(p)_(1) = a_(1) hat(i) + b_(1) hat(j) + c_(1) hat(k) ,vec(p)_(2) = a_(2) hat(i) + b_(2) hat(j) - c_(1) hat(k)`
(iv) `vec(p)_(1) = a_(1) hat(i) + b_(1) hat(j) , vec(p)_(2) = a_(2) hat(i) + b_(1) hat(j)`

Options

A. `vec(p)_(1) = a_(1) hat(i) + b_(1) hat(j) + c_(1) hat(k)`
`vec(p)_(2) = a_(2)hat(i) + b_(2) hat(j)`
B. `vec(p)_(1) = c_(1) hat(k)`
`vec(p)_(2) = c_(2)hat(k)`
C. `vec(p)_(1) = a_(1) hat(i) - b_(1) hat(j) + c_(1)hat(k)`
`vec(p)_(2) = a_(2)hat(i) + b_(2) hat(j) - c_(2) hat(k)`
D. `vec(p)_(1) = a_(1)hat(i) + b_(1) hat(j)`
`vec(p)_(2) = a_(2) hat(i) + b_(1) hat(j)`

Answer

A, D

Explanation

As `vec(p)_(1) + vec(p)_(2) = vec(0)` so `vec(p)_(1') + vec(p)_(2') = 0`
For (A) `vec(p)_(1') + vec(p)_(2') = (a_(1) + a_(2)) hat(i) + (b_(1) + b_(2)) hat(j) + c_(1) hat(k)`
For (B) `vec(p)_(1') + vec(p)_(2') = (a_(1) + a_(2)) hat(i) + (b_(1) + b_(2)) hat(j)`
For (C ) `vec(p)_(1') + vec(p)_(2') = (c_(1) + c_(2)) hat(k)`
For (D) `vec(p)_(1') + vec(p)_(2') = (a_(1) + a_(2)) hat(i) + 2b_(1) hat(j)`
But `a_(1'), b_(1'), c_(1'), a_(2), b_(2), c_(2) != 0`
Therefore (A) & (D) is not possible to get
`vec(p)_(1') + vec(p)_(2') = vec(0)`