The balls, having linear momenta `vecp_1=vecpi and vecp_2_2=-vecpi`, undergo a collision in free space. There is no external force acting on the balls. Let `vecp'_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`.

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_(1)hat(k)`
D. `vec(p_(1)^('))=a_(1)hat(i)+b_(1)hat(j), vec(p_(2)^('))=a_(2)hat(k)+b_(1)hat(j)`

Answer

A, D

Explanation

Initail linear momentum `=p hat(i) (-p hat(i))=0`
According to law of conservation of momentum, the final momentum should be zero. ltbr. For optio `(a)`
`vec(p_(1)^('))+vec(p_(2)^('))=(a_(1)+a_(2))hat(i)+(b_(1)+b_(2))hat(j) + c_(1) hat(k)`
The term `c_(1) hat(k)` cannot be zero
For option (d) `vec(p_(1)^('))+vec(p_(2)^('))=(a_(1)+a_(2))hat(i)+2b_(1)hat(j)`
Here, term `2 b_(1) hat(j)` cannot be zero .
Therefore , choices `(a)` and `(d)` are not allowed.