Two particles of masses m_(1) andm_(2) in projectile motion have velocities vecv_(1) and vecv_(2) respectively at time t= 0. They collide at time t_(0). Their velocities become vecv_(1) and vecv_(1) at time 2t_(0) while still moving in air. The value of |(m_(1)vecv_(1)^(1)+m_(2)vecv_(2)^(1))-(m_(1)vecv_(1)+m_(2)vecv_(2))| is

Two particles of masses m_(1) andm_(2) in projectile motion have veloc

1.	Two particles of masses m_(1) andm_(2) in projectile motion have velocities vecv_(1) and vecv_(2) respectively at time t= 0. They collide at time t_(0). Their velocities become vecv_(1) and vecv_(1) at time 2t_(0) while still moving in air. The value of \|(m_(1)vecv_(1)^(1)+m_(2)vecv_(2)^(1))-(m_(1)vecv_(1)+m_(2)vecv_(2))\| is
Answer» ZERO `(m_(1)+m_(2))g t_(0)` `2(m_(1)+m_(2))g t_(0)` `(1)/(2)(m_(1)+m_(2))g t_(0)` Solution :`mv^(1)=m(u+at)` `m_(1)v_(1)^(1)+m_(2)v_(2)^(1)=m[v_(1)+2g t_(0)]+m[v_(2)+2g t_(0)]`