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).
To find the distance xf where the larger piece lands, we first need to analyze the motion of the smaller piece (m2) after the explosion.
Since the smaller piece returns to Earth, its motion is governed by projectile motion. We know that the projectile was launched with an initial velocity in the vertical direction. When the smaller piece returns to Earth, it has the same vertical velocity component as the larger piece (m3).
Using the equation for the vertical motion of a projectile, we can relate the vertical displacement (Δy), initial vertical velocity (v0y), and time of flight (t) as follows:
Δy = v0y * t - (1/2) * g * t^2
Where g is the acceleration due to gravity and is approximately 9.8 m/s^2.
Since the smaller piece is returning to the same height as the launch point, the vertical displacement is zero. We can solve for the time of flight as follows:
0 = v0y * t - (1/2) * g * t^2
Rearranging the equation, we get:
(1/2) * g * t^2 = v0y * t
Simplifying, we find:
(1/2) * g * t = v0y
Solving for t, we get:
t = (2 * v0y) / g
Now, we can find the horizontal distance xf where the larger piece lands. Since the explosion occurs at the top of the projectile's trajectory, the time of flight for the larger piece is exactly twice the time of flight for the smaller piece.
Therefore, the time of flight for the larger piece is:
t_larger = 2 * t = 2 * (2 * v0y / g) = 4 * v0y / g
The horizontal distance is then given by:
xf = xm + v0x * t_larger
Since the horizontal velocity component (v0x) remains constant throughout the motion, we can use the initial horizontal velocity component from the launch point.
Therefore, the final expression for xf becomes:
xf = xm + v0x * (4 * v0y / g)
Given that m3 has three times the mass of m2, we can relate the initial velocities of the two pieces as follows:
m2 * v0y = m3 * v0y
Since the vertical velocities cancel out, we can conclude that the initial horizontal velocities are also equal:
v0x = v0x
Therefore, substituting v0y = (m3 / m2) * v0y into xf, we get:
xf = xm + (v0x * 4 * (m3 / m2) * v0y) / g
Simplifying further:
xf = xm + (4 * m3 * v0x * v0y) / (m2 * g)
Thus, the distance xf where the larger piece lands is xm plus (4 * m3 * v0x * v0y) divided by (m2 * g).