I don\'t know how to solve this problem. The figure below, two particles are lau

Question

I don't know how to solve this problem.

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 18.0 m/s. Particle 2 of mass m2 = 2.40 g is shot with a velocity of magnitude 36.0 m/s, at an upward angle such that it always stays directly above particle 1 during its flight. What is the maximum height Hmax reached by the center of mass of the two-particle system? m What is the magnitude and direction of the velocity of the center of mass when the center of mass reaches this height? m/s degree (counterclockwise from the positive x axis) What is the magnitude and direction of the acceleration of the center of mass when the center of mass reaches this height? m/s2 degree (counterclockwise from the positive x axis)

Explanation / Answer

a) stay upward means 36cosA = 18 (horizontak compo equal)

A = 60 degrees .

max. height = v^2sin^2(A) / 2g = 36^2 x sin^2(60) /2g

       = 49.60 m

HC.m = 5 x0 + 49.60 x 2.40 / (5 + 2.40) = 16.08 m

b) vertical compo of partical 2 will be zero at max. height .

so velocity will be along x- axis.

Vc.m. = 5 x 18 + 2.4 x 36cos60) / 5+2.4 = 18 m/s

angle = 0

c) a = 5 x 0 + 2.4 x 9.8 / (5 + 2.4) = 3.18 m/s2

downwards

angle = 270 degrees

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