Particles P and Q of mass 20 g and 40g respectively are simultaneously projected from points A and B on the ground. The initial velocities of Pand Q make 45° and 135° angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49 m/s. The separation AB is 245 m. Both particles travel in the same vertical and plane and undergo a collision. After collision P retraces its path. Determine the position of Q when it hits the ground. How much time after the collision does the particle Q take to reach the ground? (Take g = 9.8 m/s2)
As the horizontal speed of two particles towards each other is same they will meet at the middle of AB, i.e., at distance (245/2) = 122.5 from A towards B.
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,
This gives vQ = 0, i.e., after collision, the velocity 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.
So time taken by Q to reach ground,
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