A block of mass 10 kg is at rest on a frictionless table at the origin of the co
ID: 2259967 • Letter: A
Question
When the 5kg block is at x=-10 meters, where is the center of mass of the system?
Find the momentum of the 5kg mass, the 10kg mass and the center of mass before the collision What is the final velocity of the combined system. I know the correct answers, but again not what I'm doing wrong. Please explain.
Explanation / Answer
m1 = 10 kg, u1 = 0
m2 = 5 kg, u2 = 5 m/s
xcm = (m1*x1+m2*x2)/(m1+m2)
here x1 = 0m, x2 = -10m
Xcm = -3.333 m
p1 = m1*u1 = 0
p2 = m2*u2 = 5*5 = 25 kg.m/s
pcm = (m1*p1 + m2*p2)/(m1+m2) = 8.333 kg.m/s
m1*u1 + m2*u2 = (m1+m2)*v
v = (m1*u1+m2*u2)/(m1+m2) = 1.67 m/s
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.