Two particles of masses $$m1$$ and $$m2$$ in projectile motion have velocities v→$$1$$ and v→$$2$$ respectively at time t $$=$$ $$0$$. They collide at time $$t0$$. Their velocities become v→$$1$$′ and v→$$2$$' at time $$2t0$$ while still moving in air. The value of $$m1v$$→$$1$$′$$+m2v2$$→′−$$m1v1$$→$$+m2v2$$→ is

Question

Two particles of masses $$m1$$ and $$m2$$ in projectile motion have velocities v→$$1$$ and v→$$2$$ respectively at time t $$=$$ $$0$$. They collide at time $$t0$$. Their velocities become v→$$1$$′ and v→$$2$$' at time $$2t0$$ while still moving in air. The value of $$m1v$$→$$1$$′$$+m2v2$$→′−$$m1v1$$→$$+m2v2$$→ is

Infinity Learn · Accepted Answer

The momentum of the $$two-particle$$ system, at t $$=$$ $$0$$ isP→$$i=m1v$$→$$1+m2v$$→$$2Collision$$ between the two does not affect the total momentum of the system.A constant external force $$m1+m2g$$ acts on the system. The impulse given by this force, in time t $$=$$ $$0$$  to $$t=2t0$$ is $$m1+m2g$$×$$2t0$$∴ |Change in momentum in this interval $$=m1v$$→$$1+m2v$$→$$2$$−$$m1v$$→$$1+m2v2$$→∣$$=2m1+m2gt0$$

Two particles of masses m1 and m2 in projectile motion have velocities v→1 and v→2 respectively at time t = 0. They collide at time t0. Their velocities become v→1′ and v→2' at time 2t0 while still moving in air. The value of m1v→1′+m2v2→′−m1v1→+m2v2→ is

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