A 3:25 g particle is moving at 1:51 m/s toward a stationary 7:89 g particle.?

Question

A 3:25 g particle is moving at 1:51 m/s toward a stationary 7:89 g particle. (a)With what speed does the heavier parti-
cle approach the center of mass of the two particles? Answer in units of m/s.
(b)What is the magnitude of the momentum of the lighter particle, relative to the center of mass? Answer in units of Ns.

Answer 1

Vcm=[ m1v1+m2v2] / (m1+m2)=0.4405 m/s

(a) speed with which the heavier particle approaches the center of mass is - 0.4405 m/s
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(b) the magnitude of the momentum of the lighter particle= 3.4758 kgm/s

Answer 2

A. CM moves at vcm = m1v1/(m1+m2). The stationary m2 approaches the CM at -vcm.
B. Relative to the moving CM, m1 moves at vrel = v1-vcm. Momentum = m1*vrel.