In the figure below, two particles are launched from the origin of the coordinat
ID: 2061628 • 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 12.0 m/s. Particle 2 of mass m2 = 3.60 g is shot with a velocity of magnitude 24.0 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?Explanation / Answer
We are given the following information m1 = 5.00 g v1x = 12.0 m/s v1y = 0.0 m/s m2 = 3.60 g v2 = 24.0 m/s We are also told that mass m2 has to remain above mass m1 at all times during the motion which leads to the x velocity v2x relation for mass m2 v2x = v1x = 12.0 m/s From this we can also find the y velocity v2y of mass m2 v2^2=v2x^2 + v2y^2 v2y^2=v2^2 - v2x^2 v2y=v(v2^2 - v2x^2) v2y=v(24.0^2-12.0^2) v2y=v(507) v2y=20.784 m/s We are also told that mass m1 moves across the floor at constant speed while mass m2 is in projectile motion so that we have a1x = 0 m/s^2 a1y = 0 m/s^2 a2x = 0 m/s^2 a2y = -9.8 m/s^2 From this information we can now calculate the center of mass velocity and acceleration Center of Mass Velocity vx=S(mnvnx)/S(mn) vx=(m1 v1x + m2 v2x)/(m1 + m2) vx=(5*12+3.6*12)/(5+3.6) vx=104/8 vx=12 m/s vy=S(mnvny)/S(mn) vy=(m1 v1y + m2 v2y)/(m1 + m2) vy=(5*0+3.6*20.784)/(5+3.6) vy=67.551/8.6 vy=8.7003 m/s Center of Mass Acceleration ax=S(mnanx)/S(mn) ax=(m1 a1x + m2 a2x)/(m1 + m2) ax=(5*0+3*0)/(5+3) ax=0/8 ax=0 m/s^2 ay=S(mnany)/S(mn) ay=(m1 a1y + m2 a2y)/(m1 + m2) ay=(5*0+3.6*-9.8)/(5+3.6) ay=-29.4/8 ay=-4.1023 m/s^2 The center of mass motion can now be solved with the two dimensional kinematics equations. Since the questions all pertain to the maximum height, we have the condition that the final y velocity will be equal to zero. Collectively we have x direction ax=0 m/s^2 vx initial=13 m/s xinitial=0 m y direction ay=-4.1023 m/s^2 vy initial=8.7003 m/s vy final=0 m/s yinitial=0 m Since we have more information in the y direction, we can start there with the y direction time implicit kinematics formula vy^2=vy initial^2+2ay(y-yinitial) 0^2=8.7003^2 + 2*(-4.1023)*(y-0) y=9.226m