## Question

Particles *P* and *Q* of mass 20 *g* and 40*g* respectively are simultaneously projected from points *A* and *B* on the ground. The initial velocities of *P*and *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/s^{2})

### Solution

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 *v*_{Q} = 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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