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=
http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W05D1-2.pdf
To find the distance xf at which the larger piece lands, we can use the principle of conservation of momentum along the horizontal direction.
The momentum of an object is defined as the product of its mass and velocity. In this case, since we are neglecting air resistance and the effects of the Earth's curvature, the only force acting on the system is gravity in the vertical direction, which does not affect the horizontal motion.
Before the explosion, the instrument-carrying projectile of mass m1 had a horizontal momentum of m1 * 0 = 0, since it was initially at rest horizontally. After the explosion, the momentum is still conserved, but now it is divided between the smaller piece m2 and the larger piece m3.
Since the smaller piece m2 returns to Earth at the launching station, its final horizontal velocity component is 0. Therefore, the momentum it carries is also 0.
The momentum of the larger piece m3 is given by m3 * vf, where vf is its final horizontal velocity.
Since the explosion occurred at the top of the trajectory, the larger piece m3 had already reached the highest point of its trajectory. At this point, the vertical component of its velocity is 0. Therefore, we can use the equation for projectile motion to relate the horizontal distance xm between the launch point and the explosion to the initial velocity and time of flight.
The time of flight is given by the equation t = sqrt((2h)/g), where h is the maximum height reached by the projectile, and g is the acceleration due to gravity.
At the highest point, the final vertical velocity is 0, and the initial vertical velocity is given by Vy = gt. Therefore, the time of flight is t = 2Vy/g.
Combining the two equations, we have xm = Vx * t = Vx * (2Vy/g), where Vx is the initial horizontal velocity component.
To find Vx in terms of the given variables, we can use the fact that the horizontal motion is uniform (no horizontal acceleration). Therefore, Vx does not change during the flight of the projectile.
The horizontal velocity is given by Vx = (m1 * V1x) / m1, where V1x is the initial horizontal component of the velocity of the projectile before the explosion.
Let's denote the final horizontal velocity of the larger piece m3 as Vfx.
Applying the law of conservation of momentum, we have:
0 = m2 * 0 + m3 * Vfx
Solving for Vfx, we find Vfx = 0.
This means that the larger piece m3 lands with a final horizontal velocity of 0, at the same point where the smaller piece m2 landed.
Therefore, the distance xf at which the larger piece lands is equal to the distance xm from the launch point to the explosion.
Thus, we can conclude that xf = xm.
Therefore, the answer is xf = xm.