Two particles, one of mass m and the other of mass 2m, are projected horizontally towards each other from the same level above the ground with velocities 10 m/s and 5 m/s, respectively.  They collide in air and stick to each other.  The distance of the combined mass where they fall from point A is

To solve this problem, we need to use the principles of conservation of momentum and the equations of motion.

Step 1: Conservation of Momentum Before the collision, the momentum of the system is the sum of the momenta of the two particles. The momentum of a particle is given by its mass times its velocity. So, the momentum of the first particle is m10 and the momentum of the second particle is 2m(-5) (negative because it's moving in the opposite direction). The total initial momentum is m10 - 2m5 = 0.

After the collision, the two particles stick together and move as one object. The total mass is m + 2m = 3m. Let's say the velocity of the combined object is v. So, the total final momentum is 3m*v.

According to the conservation of momentum, the total initial momentum should equal the total final momentum. So, we have 0 = 3m*v, which implies that v = 0. This means that the combined object doesn't move horizontally after the collision.

Step 2: Distance from Point A Since the combined object doesn't move horizontally after the collision, it falls vertically down from the point of collision. The distance from point A to the point of collision is the same as the distance from point A to the point where the combined object falls. So, the distance is 0.

The problem involves the principles of conservation of momentum and the physics of projectile motion. Here are the steps to solve it:

Step 1: Determine the initial momentum of the system. The initial momentum (P_initial) of the system is the sum of the momentum of each particle. Momentum is mass times velocity, so for the first particle it's m10 and for the second particle it's 2m5. Therefore, P_initial = m10 + 2m5 = 20m.

Step 2: Determine the final momentum of the system. After the collision, the particles stick together and move as one object. The final momentum (P_final) is therefore the total mass times the final velocity (v_final). The total mass is m + 2m = 3m, so P_final = 3m*v_final.

Step 3: Apply the conservation of momentum. The total momentum of the system is conserved, so P_initial = P_final. This gives us 20m = 3m*v_final. Solving for v_final, we get v_final = 20m/3m = 20/3 m/s.

Step 4: Determine the time of flight. The particles are projected horizontally, so the time of flight depends only on the vertical motion. Assuming the initial vertical velocity is 0 and the acceleration due to gravity is g, the time of flight (t) can be found using the equation of motion h = 0.5gt^2, where h is the height above the ground. Solving for t, we get t = sqrt(2h/g).

Step 5: Determine the horizontal distance travelled. The horizontal distance (d) travelled by the particles is the final velocity times the time of flight, so d = v_final*t. Substituting the values we found for v_final and t, we get d = (20/3 m/s)*sqrt(2h/g).

This is the distance from point A where the combined mass falls.