Solution .
m1 = 3 * 10 4 kg
m2 = ½ of mass of A = 1.5 * 10 4 kg u1 = 0.6 m / s u2 = -0.4 m / s Before collision : m1u1 + m2u2 = 3 * 10 4 * 0.6 + 1.5 * 10 4 * ( -0.4)
= 1.2 * 10 4 kg m / s After collision : ( m1 + m2 ) v = 4.5 * 10 4 v kg m / s As per conservation of momentum , 1.2 * 10 4 = 4.5 * 10 4 v V = 1.2 / 4.5 = 0.27 m / s
Therefore , the common velocity is 0.27m / s
Equilibrium of a Particle o
A particle is said to be in equilibrium when the net external force on it is zero