An instrument-carrying projectile of mass m1 accidentally explodes at the top of its trajectory. The horizontal distance between launch point and the explosion is xm. The projectile breaks into two pieces which fly apart horizontally. The larger piece, m3, has three times the mass of the smaller piece, m2. To the surprise of the scientist in charge, the smaller piece returns to earth at the launching station. Neglect air resistance and effects due to the earth's curvature.

How far away, xf, from the original launching point does the larger piece land? Express your answer in terms of some or all of the given variables m1, xm, and g (enter m_1 for m1 and x_m for xm).

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To solve this problem, let's break down the given information and determine the necessary steps to find the value of xf.

Given variables:
- Mass of the instrument-carrying projectile: m1
- Horizontal distance between launch point and explosion: xm
- Mass of the larger piece: m3 (3 times the mass of the smaller piece)
- Mass of the smaller piece: m2
- Acceleration due to gravity: g

To find the distance, xf, where the larger piece lands, we can consider the horizontal motion of the two pieces separately.

1. Horizontal motion of the smaller piece (m2):
Since the smaller piece returns to the launching station, we know that its horizontal distance traveled equals the horizontal distance between the launch point and explosion (xm).

Therefore, the horizontal distance traveled by the smaller piece is xf = xm.

2. Horizontal motion of the larger piece (m3):
The larger piece will continue to move horizontally after the explosion. Since no external horizontal forces are acting on the system, there will be no horizontal acceleration.

Therefore, the horizontal velocity of the larger piece will remain constant throughout its motion.

Now, let's find the initial horizontal velocity (v_x) of the larger piece:
- The initial momentum in the horizontal direction is conserved.
- Before the explosion, both projectiles had the same horizontal velocity and momentum.
- After the explosion, the smaller piece moves backward to the launch point, while the larger piece moves forward.
- Since the larger piece has three times the mass of the smaller piece, the velocity of the larger piece must be one-third that of the smaller piece.

Therefore, the initial horizontal velocity of the larger piece is v_x = (1/3)v_x (where v_x represents the initial horizontal velocity of the smaller piece).

The time of flight (t) for both pieces can be calculated using the vertical motion equations:
- For the smaller piece, its time of flight is the time taken to reach the maximum height and return to the launching station.
- For the larger piece, its time of flight is the time taken to reach the ground (since it does not return to the launching station).

To find the time of flight for each piece, we can consider their vertical motions separately:
- The smaller piece:
- The vertical distance traveled by the smaller piece is the same as the maximum height reached by the entire projectile.
- At the maximum height, the final vertical velocity (v_y) is 0 m/s.
- We can use the equation v_y^2 = v_0y^2 + 2gΔy, where v_0y is the initial vertical velocity and Δy is the vertical displacement.
- Since the smaller piece returns to the launching station, its vertical displacement equals 2 times the maximum height.
- Thus, Δy = 2(maximum height).

- The larger piece:
- The vertical distance traveled by the larger piece is the same as the maximum height reached by the entire projectile.
- At the maximum height, the final vertical velocity (v_y) is 0 m/s.
- We can use the same equation as above to find the time of flight for the larger piece, using Δy = maximum height.

Now, let's calculate the time of flight:
1. For the smaller piece:
- v_y^2 = v_0y^2 + 2gΔy
- Since the initial vertical velocity, v_0y, is unknown, we need to find the initial speed (v_0) first.
- The initial speed can be calculated using the equation v_0^2 = v_x^2 + v_y^2, where v_x represents the initial horizontal velocity of the smaller piece (which is equal to xm/t) and v_y represents the initial vertical velocity.
- Once we find the initial speed, we can calculate the initial vertical velocity, v_0y, using the equation v_0y = v_0sinθ, where θ is the launch angle.
- Finally, substitute the known values (v_0y, g, and Δy) into the equation to find the time of flight, t.

2. For the larger piece:
- Since the vertical motion of the larger piece is symmetrical to the smaller piece (reaching the same maximum height and falling back down), the time of flight, t, will be the same as that of the smaller piece.

Once we have the value of t, we can find the horizontal distance traveled by the larger piece using the equation xf = v_x * t.

In summary, the steps to find xf are as follows:
1. Calculate the initial horizontal velocity, v_x (which is 1/3 times the initial velocity of the smaller piece).
2. Use the equations of horizontal and vertical motions to find the time of flight, t (for both pieces separately).
3. Substitute the values of v_x and t into the equation xf = v_x * t to find the horizontal distance traveled by the larger piece.