At one instant, the center of mass of a system of two particles is located on the x-axis at x = 2.0 m and has a velocity of One of the particles is at the origin. The other particle has a mass of 0.10 kg and is at rest on the x-axis at x = 8.0 m. (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

Solution 56E Step 1 : It is given Center of velocity v = 5.0 m/s c Position of center of mass x =c2.0 m Velocity of particle 1 v =1 Mass of particle 1 m =1 Velocity of particle 2 v = 0 2 Mass of particle m =20.10 kg By the law of conservation of momentum , momentum at the center of mass has to be equal to the sum of momentum of each particle That is m c c m v + m v 1 1 2 2 But m =cm + m1and v2= 5m/s c Substituting values we get (m 1 m ) 2 5m/s = m v + 1 110 kg × 0) (m + 0.10 kg) × 5m/s = m v 1 1 1 (5m 1 0.50 kg m/s) = m v 1 1 Consider particles m × (x x ) = m × (x x ) c 1 c 1 2 2 Substituting values we get m 1 (2.0 m 0 ) = 0.10 kg × (8.0 m 2.0 m ) m 1 (2.0 m 0 ) = 0.10 kg × (6.0 m ) m = 0.60 /2.0 m 1 m 1 0.3 kg Hence the particle 1 is found to be 0.3 kg