In the figure below, two particles are launched from the origin of the coordinat
ID: 2187481 • 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 15.0 m/s. Particle 2 of mass m2 = 3.00 g is shot with a velocity of magnitude 30.0 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? 1 m (b) What is the magnitude and direction of the velocity of the center of mass when the center of mass reaches this height? 2 m/s 3Explanation / Answer
The horizontal component of partical 2's initial velocity has to be equal to partical 1's velocity. That should allow you to get the angle of partical 2's initial velocity. a. You can calculate the maximum height of parftical 2 using the kinematic formula Vf^2 = Vi^2 + 2*a*y where Vf is the vertical component of partical 2's velocity at the time of max height (hint - if it's at the highest point, is it still going up? So what's the vertical component?), Vi = the vertical component of partical 2's initial velocity, a = -g. Solve for y. Look up the method of finding center of mass. b. Partical 1 only goes along x and what did you decide about the vertical component ov #2's velocity? c. This is a bit more difficult. Some points to consider. #1 moves with a constant speed -- so no acceleration. #2 has been under the influence of gravity the whole time. That's why the upward motion stopped. If the 2 particles were the same mass, the acceleration of the center of mass would be 1/2 that of #2. I'm leaving quite a bit of this for you to figure out. Mostly I've pointed out things that should make sense to you. With that understanding of what's going on, I hope you can make it the rest of the way.