In the figure below, two particles are launched from the origin of the coordinat
ID: 2175985 • 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 20.0 m/s. Particle 2 of mass m2 = 3.00 g is shot with a velocity of magnitude 40.0 m/s, at an upward angle such that it always stays directly above particle 1 during its flight.
(a) What is the maximum heightHmaxreached 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 positivexaxis))
(c) What is the magnitude and direction of the acceleration of the center of mass when the center of mass reaches this height?(m/s)(? (counterclockwise from the positivexaxis))
Explanation / Answer
wrong ans because i used different numbers kindly use nubers in the question 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