In the figure below, two particles are launched from the origin of the coordinat
ID: 2175902 • 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.(b) What is the magnitude and direction of the velocity of the center of mass when the center of mass reaches this height?
(c) What is the magnitude and direction of the acceleration of the center of mass when the center of mass reaches this height?
Explanation / Answer
Let me try to help you but don't rely on the answer too much since I have not done this calculation for a long time. let a be the angle at which the second particle (P2) was shot. The velocity of P2 can be split into 2 vectors - vertical and horizontal The horizontal vector has the same velocity magnitude as that of particle 1 (P1) Therefore 17cos(a) = 11.5 m/s. This gives cos(a) = 11.5/17 = 0.6765. The angle a = acos(0.6765) = 47.4315 The vertical vector velocity of P2 = 17 sin(a) = 12.52 m/s. This is the initial magnitude of the vertical vector. This vertical velocity decelerate over time by a factor of 9.8 m/s/s (by gravity) and is = 12.52 - 9.8t At max height, the velocity = 0, therefore t = 12.52/9.8 = 1.2775 sec. The max height = 12.52 x 1.2775 - 9,8/2 x square of 1.2775 = 8 m The answer: (a) 8 m (b) 0 (c) 9.8 m/s/s