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 solve this problem, we can use the principle of conservation of momentum.
First, let's consider the initial momentum of the projectile before the explosion. Since there is no horizontal component, the initial momentum is only in the vertical direction and is given by:
p_initial = m1 * v1
where v1 is the vertical velocity of the projectile.
Next, let's consider the final momentum of the system after the explosion. The smaller piece, m2, returns to the launching station which means its final velocity is zero. The larger piece, m3, moves horizontally and lands at a distance xf from the launching point. The final momentum of the system is then:
p_final = m2 * 0 + m3 * v3
where v3 is the horizontal velocity of the larger piece.
According to the principle of conservation of momentum, the initial and final momenta must be equal. So we have:
m1 * v1 = m3 * v3
Now, let's consider the horizontal motion of the larger piece. Since there is no horizontal force acting on it (neglecting air resistance), the horizontal component of its velocity remains constant throughout its motion. Therefore, we can use the equation of motion for horizontal motion:
x = v * t
where x is the distance, v is the velocity, and t is the time of flight.
In this case, the time of flight is the same for both the smaller and larger pieces since they are part of the same projectile. Let's denote it as t_flight.
For the larger piece, the horizontal distance xf is given by:
xf = v3 * t_flight
Finally, we need to find an expression for t_flight in terms of the given variables.
To do this, we can consider the vertical motion of the smaller piece. Since it returns to the launching station, its maximum height is reached at the explosion point. Using the equations of motion for vertical motion, and considering the time it takes for the smaller piece to reach its maximum height, we can write:
xm = (1/2) * g * t_max^2
where g is the acceleration due to gravity.
From this equation, we can solve for t_max:
t_max = sqrt(2 * xm / g)
Since the time of flight is the same as the time to reach the maximum height, we have:
t_flight = 2 * t_max = 2 * sqrt(2 * xm / g)
Now, substituting this expression for t_flight into the equation for xf, we get:
xf = v3 * 2 * sqrt(2 * xm / g)
Therefore, the distance xf from the original launching point where the larger piece lands is equal to v3 multiplied by 2 times the square root of 2 times xm divided by g.