Particles `P` and `Q` of masses `20g` and `40g`, respectively, are projected from positions `A` and `B` on the ground. The initial velocities of `P` and `Q` make angles of `45^(circ)` and `135^(circ)`, respectively with the horizontal as shown in the fig. Each particle has an initial speed of `49m//s`. The separation `AB` is `245m`. Both particles travel in the same vertical plane and undergo a collision. After the collision `P` retraces its path. The separation of `Q` from its initial position when it hits the ground is

Options

A. `245m`
B. `(245)/(3)m`
C. `(245)/(2)m`
D. `(245)/(sqrt(2))m`

Answer

C

Explanation

Both the particles will collide at the highest point of their path. Momentum of the particle `P` just before collision
`m_(A)V_(A) cos 45^("circ") = 20 xx 10^(-3) xx 49 xx (1)/(sqrt(2))`
`(980)/(sqrt(2)) xx 10^(-3)kg m//s`
Momentum of particle `Q` just before collision is
`m_(B)V_(B) cos 135^("circ") = -40 xx 10^(-3) xx 49 xx (1)/(sqrt(2)) = (1960)/(sqrt(2)) xx 10^(-3) kg m//s`
After collision, let velocity of `Q` be `V` and velocity of `P` be `49 cos 45^("circ")`. Therefore, momentum after collision is
`40 xx 10^(-3)V - (49 xx 20 xx 10^(3))/(sqrt(2))`
Applying law of conservation of momentum
`(980)/(sqrt(2)) xx 10^(-3) - 1960 xx (10^(-3))/(sqrt(2)) = 40 xx 10^(-3)V-(980 xx 10^(-3))/(sqrt(2))`
or `v = 0`
Thus, particle `Q` will fall freely after collision. So, the distance travelled by it will be `245//2m`.
i.e., `122.5m`