In the figure below, two particles are launched from the origin of the coordinat
ID: 2004114 • Letter: I
Question
In the figure below, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m1 = 5.00 g is shot directly along an x axis (on a frictionless floor), where it moves with a constant speed of 14.0 m/s. Particle 2 of mass m2 = 3.20 g is shot with a velocity of magnitude 28.0 m/s, at an upward angle such that it always stays directly above particle 1 during its flight.(a) What is the maximum height Hmax reached by the center of mass of the two-particle system?
(b) What is the magnitude and direction of the velocity of the center of mass when the center of mass reaches this height?
m/s
° (counterclockwise from the positive x axis)
(c) What is the magnitude and direction of the acceleration of the center of mass when the center of mass reaches this height?
m/s2
° (counterclockwise from the positive x axis)
Explanation / Answer
(a) Since #2 is always directly above #1, we know that #2 must have a horizontal speed of 14 m/s. So it has an initial vertical speed of (using pythag theorem) sqrt (28^2 - 14^2) = 24.25 m/s the max height obtained by #2 must be max height = v^2 / 2g = 24.25^2 / 2 * 9.8 = 30.0 meters So the max height of the center of mass is the weighted average of the two particles: max ht of cm = (1 / total mass) ( mass 1 * height 1 + mass 2 * height 2) = = (1/8.20) (5.00 * 0 + 3.20 * 30.0) = = 11.71 meters (b) When #2 is at max height, its only velocity is the horizontal speed of 14.0 m/s. Since both particles are moving this speed at that instant, the center of mass (i.e. the average of the two particles) is also this speed. So. velocity of center of mass = 14.0 m/s in the x direction (i.e. zero degrees) (c) acceleration of cm = (1 / total mass) ( mass 1 * acc 1 + mass 2 * acc 2) = = (1 / 8.20) (5.00*0 + 3.20 * -9.80) = -3.82 m per sec squared in the y direction This is 3.82 m/s^2 at an angle of 270 deg counterclockwise from x axis