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).
xf=
Problem 3 http://mikebloxham.com/H7A/HW7soln.pdf
To solve this problem, we need to use the principles of projectile motion. Since the smaller piece returns to the launching station, we know that its horizontal distance travelled is equal to the horizontal distance of the explosion, xm.
Let's analyze the motion of the smaller piece first. Since there is no air resistance, the only force affecting its motion is gravity. Therefore, its vertical motion can be treated as free fall.
We know that the time it takes for an object to fall from a certain height h under free fall can be calculated using the equation:
t = sqrt(2h / g)
where g is the acceleration due to gravity.
In this case, the height h is the maximum height reached by the projectile before it exploded. We know that at the top of its trajectory, the projectile momentarily comes to a stop before starting to fall. Hence, the height h is the same as the maximum height reached by the projectile. We can find this by using the kinematic equation for vertical motion:
v^2 = u^2 + 2as
where v is the final vertical velocity (0 m/s at the top), u is the initial vertical velocity (which we can find from the initial velocity of the projectile), a is the acceleration due to gravity (-9.8 m/s^2), and s is the vertical displacement (the maximum height reached).
Since we neglect air resistance, the initial vertical velocity of the projectile is zero, and thus we have:
v^2 = 0 + 2as
0 = 2as
s = 0
This means that the maximum height reached by the projectile is zero. Therefore, the height h is also zero, and the time it takes for the smaller piece to fall back to the launching station is zero as well.
Since the time is zero, the horizontal distance travelled by the smaller piece is also zero. Therefore, xf = 0.
Hence, the larger piece, with three times the mass of the smaller piece, would land at the horizontal distance xm from the original launching point.
Therefore, we have:
xf = xm
So, the answer is xf = xm.