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=
To find the distance that the larger piece, m3, lands from the original launching point, we need to consider the conservation of momentum.
Before the explosion, the total momentum of the system is zero since there is no horizontal motion. Therefore, the momentum after the explosion must also be zero due to the absence of external forces.
Let's analyze the situation after the explosion:
The smaller piece, m2, returns to Earth at the launching station, which means it has a change in momentum equal to zero. Since the momentum is given by the product of mass and velocity, we can express this as:
m2 * v2 = 0,
where v2 is the velocity of the smaller piece.
However, the larger piece, m3, also has a momentum that needs to be canceled out to achieve a total momentum of zero. Since the smaller piece has returned to the launching station, we can assume the larger piece moves forward and lands at a distance xf.
To determine the velocity of the larger piece, we can use the principle of conservation of momentum:
(m3 * v3) - (m1 * v1) = 0,
where v3 is the velocity of the larger piece and v1 is the velocity of the original projectile before the explosion.
Given that m3 = 3 * m2, we can rewrite the equation as:
(3 * m2 * v3) - (m1 * v1) = 0.
Now, we need to relate the velocities of the pieces to the distance xm. Since we are neglecting air resistance and the effects of the Earth's curvature, we can assume that the entire trajectory is symmetric.
The time taken for the smaller piece to reach the maximum height is equal to the time taken for it to fall back down to the launching station. This gives us:
t_total = 2 * t_max_height.
The total time, t_total, can be expressed as:
t_total = (2 * xm) / v2,
where xm is the horizontal distance.
The time taken to reach the maximum height, t_max_height, can be expressed in terms of the velocity of the original projectile before the explosion, v1, as:
t_max_height = (v1) / g,
where g is the acceleration due to gravity.
Combining these equations, we can now express v2 in terms of v1 and g:
(2 * xm) / v2 = (v1) / g.
Simplifying, we find:
v2 = (2 * xm * g) / v1.
Finally, substituting this expression for v2 into the earlier conservation of momentum equation, we get:
(3 * m2 * v3) - (m1 * v1) = 0,
(3 * m2 * v3) - (m1 * v1) = 0,
(3 * m2 * v3) = (m1 * v1),
(3 * m2 * v3) = (m1 * v1).
Since the smaller piece, m2, has returned to the launching station, its velocity, v2, is zero. Therefore, we have:
(3 * m2 * v3) = (m1 * v1),
(3 * m2 * v3) = (m1 * v1),
(3 * m2 * v3) = (m1 * v1).
Substituting the expression for v2 and rearranging the equation, we find:
3 * (2 * xm * g) = v1 * (m1 + 3 * m2),
6 * xm * g = v1 * (m1 + 3 * m2),
xf can be expressed as:
xf = v3 * t_max_height,
Substituting the expression for v3 from the conservation of momentum equation, we find:
xf = (m1 * v1) / (3 * m2) * (v1) / g,
xf = (m1 * v1^2) / (3 * m2 * g),
xf = (m1 * v1^2) / (3 * m2 * g),
xf = (m1 * v1^2) / (3 * m2 * g),
Therefore,
xf = (m1 * v1^2) / (3 * m2 * g).