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)`)

Answer

3.53

Explanation

The horizontal velocities of both the particle are the same and since both are projected simultaneously, these particle will meet exactly in the middle of AB (horizontally). Since
`R = (u^2 sin 2 theta)/(g) = 245 m`

`implies`The collision takes place at the maximum height where the velocities of both the particles will be in the horizontal direction.
`H = (u^2 sin^2 theta)/(2g) = ((49)^2 xx sin^2 45)/(2 xx 9.8) = 61.25 m`
Applying conservation of linear momentum in the horizontal direction with the information that P retraces its path, therefore its momentum will be the same in magnitude but different in direction.
Linear momentum of system before collision = Linear momentum of system after collision.
`m_pvecv_p + m_Q vecv_Q = m_pvecv_p + m_Q vecv_Q`
Here we have `m_p = 20 gm = m ("say") and m_Q = 40 gm = 2m ("say")`,
`m u_x + 2m -(u_x) = m(-u_x) + 2m vecv_Q`
Which gives `vecv_Q = 0`
Since `v_Q = 0`, therefore the particle Q falls down vertically so it falls down on the mid-point of AB. Now considering vertical motion of Q
Using `S_y - u_y t + 1/2 a_yt^2`
Here `u_y = 0,s_y = –61.25 m and a_y = – 9.8 m//s^2" " implies " " -61.25 = 0.t - 1/2 9.8 t^2`.
Which gives t = 3.53 sec.