Two particles of masses 100 gm and 300 gm have at a given time, position 2hati + 5hatj + 13hatkand -6hati + 4hatj -2hatkrespectively and velocities 10hati - 7hatj - 3hatk and -7hati - 9hatj - 6hatk m/s respectively. Deduce the instantaneous position and velocity of the Centre of mass.

Two particles of masses 100 gm and 300 gm have at a given time, positi

1.	Two particles of masses 100 gm and 300 gm have at a given time, position 2hati + 5hatj + 13hatkand -6hati + 4hatj -2hatkrespectively and velocities 10hati - 7hatj - 3hatk and -7hati - 9hatj - 6hatk m/s respectively. Deduce the instantaneous position and velocity of the Centre of mass.
Answer» Solution :Let the position vector of CM. be `vecR` `m_(1) = 100 gm = 0.1 kg` `m_(2) = 300 gm = 0.3 kg` `vecr_(1) = 2hati + 5hatj + 13 hatk` `vecr_(2) = -6hati + 4hatj - 2HATK` `vecR `=? `vecR = (m_(1)vecr_(1) + m_(2)vecr_(2))/(m+m_(2))` `=(0.1(2hati + 5hatj + 13hatk) + 0.3 (-6hati + 4hatj - 2hatk))/(0.1 + 0.3)` `=(-16hati + 17hatj + 7 hatk)/5 m` `vecr_(1) = 10 hati -7hatj - 3hatk` `vecr_(2)=7hati - 9hatj - 6hatk` The VELOCITY of C.M. `vecv = (m_(1)vecv_(1) + m_(2)vecv_(2))/(m + m_(2))` `=(0.1(10 hati -7hatj - 3hatk) + 0.3(7hati - 9hatj + 6hatk))/(0.1 + 0.4)` `=(3HATI -2hatj + 9hatk)/4 = ms^(-1)`