In the figure below, two particles are launched from the origin of the coordinat
ID: 2146147 • 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 13.0 m/s. Particle 2 of mass m2 = 3.60 g is shot with a velocity of magnitude 26.0 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?
____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
horizontal component of velocity of m2 = 26cos
26cos = 13
= 60 degrees
Max height reached by m2 = v^2 sin^2 /2g
= 25.87 m
Center of mass is at its maximum height when m2 is at its maximum height.
(a)Height of center of mass = 25.87*3.6/(5+3.6) = 10.83 m
(b) At maximum height, m2 has a horizontal velocity of 13 m/s
So, velocity of center of mass = 13 m/s, 0 degrees to the horizontal.
(c) Acceleration of m1 = 0
acceleration of m2 = 9.8 m/s downward.
Acceleration of center of mass = m2g/(m1+m2)
= 4.1 m/s^2 , 270 degrees counterclockwise from the positive x axis