In the figure below, two particles are launched from the origin of the coordinat
ID: 2008412 • 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 16.0 m/s. Particle 2 of mass m2 = 3.80 g is shot with a velocity of magnitude 32.0 m/s, at an upward angle such that it always stays directly above particle 1 during its flight.http://www.webassign.net/hrw/W0174-N.jpg
(a) What is the maximum height Hmax reached by the center of mass of the two-particle system?
m
(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
m1 = 5.00 g, v1 = 16.0 m/s. m2 = 3.80 g, v2 = 32.0 m/s,
note " m2 always stays directly above m1 during its flight", so v2x = v1 = 16.0 m/s, v2y = (v2^2 - v2x^2) = 768
(a) What is the maximum height Hmax reached by the center of mass of the two-particle system?
at the maximum height, vertical velocity = 0, H2max = v2y^2/(2g) = 768/(2*9.81) = 39.1 m
so Hmax = m2*H2max/(m1 + m2) = 16.9 m
(b) What is the magnitude and direction of the velocity of the center of mass when the center of mass reaches this height?
v = (m1*v1x + m2*v2x)/(m1 + m2) = 16.0 m/s
0° (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?
a = (m1*0 + m2*9.81)/(m1 + m2) = 4.24 m/s^2
270° (counterclockwise from the positive x axis)