Question

A person with mass m₁ = 63 kg stands at the left end of a uniform beam with mass m₂ = 91 kg and a length L = 2.7 m. Another person with mass m₃ = 59 kg stands on the far right end of the beam and holds a medicine ball with mass m₄ = 15 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.

+X X=0

1) What is the location of the center of mass of the system?

2) The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now?

3) What is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?)

4) To return the medicine ball to the other person, both people walk to the center of the beam. At what x-position do they end up?

Answer 1

Cm m,tmt my 636)( ) + (59 X2 .ว ) +C15) ( 2-7) 63 +91+59 +IS (2) The cent er飞man does not change became thou is ne fichen bchweer beam and our C1.4i) 228 )They au mest at the certer dmoss tho