Particles P and Q of mass 20g and 40g respectively are simu ltaneously proejected from points A and B on the ground. The initial velocities of P and Q make `45^(@)` and `135^(@)` angles respectivley with the horizontal AB as shown in the Fig. 5.44 Each particle has an initial speed of 49m/s . the separation AB is 249m. both particles travel in the same vertical plane and undergo a collision. After collision P retraces its path. Determine the position of q when it hits the grou.d How much time after the collision does the particle Q take to reach the ground? (Take g`=9.8m//s^(2)`)

Explanation

As the horizontal speed of two particles towards each other is same `(u cos 45^(@))` ,they will meet at the middle of AB, i.e., at distance (245/2)=122.5 from A towards B.
Now As `R=(u^(2)sin 2theta)/(g)=(49xx49xx1)/(9.8)=245m`
So AB is the range and as the collision takes place at the middle of AB, so it is at the highest point of the trajectory.
Now applying conservation of linear momentum at the highest point along horizontal direction keeping in mind, `v_(P)=-u_(P)cos 45^(@)`
`20xx10^(-3)cos 45^(@)-40xx10^(-3)cos45^(@)=-20xx10^(-3)cos 45^(@)+v_(Q)`
This given `v_(Q)=0`. i.e., after collision the velcoity of Q at highest point is zero. so Q will fall freely under gravity and will hit the ground in the middle of AB, i.e., 122.5 m from A towards B.
Now as `H=(u^(2)sin^(2)theta)/(2g)=(49xx49xx1)/(2xx9.8xx2)=(490)/(8)=61.25m`
So, time taken by Q to reach ground.
`t=sqrt((2H)/(g))=sqrt(((2xx490)/(8xx9.8)))=(5)/(sqrt(2))s=3.536s`