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 xf at which the larger piece lands, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.
Before the explosion, the projectile has only horizontal momentum, which is given by:
P_initial = m1 * V_initial
After the explosion, the smaller piece has returned to the launch point, so its momentum is:
P_m2 = m2 * V_m2
The larger piece lands at a distance xm from the launch point. Since it is moving horizontally, its momentum is given by:
P_m3 = m3 * V_m3
According to the conservation of momentum, the total momentum after the explosion is equal to the total momentum before the explosion:
P_initial = P_m2 + P_m3
Substituting the expressions for momentum, we have:
m1 * V_initial = m2 * V_m2 + m3 * V_m3
Now, we can use the fact that the smaller piece returns to the launch point to find the value of V_m2. Since it returns to the launch point, its vertical velocity component at the time of the explosion must be zero:
V_m2 = 0
Substituting this value into the conservation of momentum equation, we have:
m1 * V_initial = m2 * 0 + m3 * V_m3
m1 * V_initial = m3 * V_m3
Solving for V_m3, we get:
V_m3 = (m1 * V_initial) / m3
Now, we can use the equation of motion for the larger piece to find the distance xf at which it lands:
xf = V_m3 * t
Since air resistance and the curvature of the Earth are neglected, the time of flight t is the same for both the smaller and larger pieces. It is given by the time taken for the projectile to reach the top of its trajectory and then return to the launch point:
t = 2 * (V_initial / g)
Substituting the expression for V_m3 and t into the equation for xf, we have:
xf = ((m1 * V_initial) / m3) * (2 * (V_initial / g))
xf = (2 * m1 * V_initial^2) / (m3 * g)
xf = (2 * m1 * (V_initial^2)) / (m3 * g)
Therefore, the expression for xf, in terms of m1, V_initial, m3, and g, is:
xf = (2 * m1 * (V_initial^2)) / (m3 * g)
Note: This solution assumes that the explosion does not impart any vertical velocity component to the smaller or larger piece. It also neglects any effects due to air resistance and the curvature of the Earth.