A block of mass 10 kg is at rest on a frictionless table at the origin of the co

Question

A block of mass 10 kg is at rest on a frictionless table at the origin of the coordinate system. A 5 kg block is moving along the negative x axis towards the origin with a velocity of 5 m/s. The blocks collide, stick together, and move off along the positive x axis.
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

Navigate

• Browse (All)
• Browse A
• Subjects

---