Arocket is fired vertically upward. at the instant it reaches an altitude of 2700 m and a speed of 274 m/s, it explodes into three equal fragments. one fragment continues to move upward with a speed of 235 m/s following the explosion. the second fragment has a speed of 484 m/s and is moving east right after the explosion. what is the magnitude of the velocity of the third fragment? answer in units of m/s.

Explanation:

Given

initial velocity of particle u=274 m/s

one Particle moves up with velocity of v=235 m/s

and other moves u=484 m/s towards east

let and be the velocity of third Particle is Y and x direction

conserving momentum in y direction  

Now conserving momentum in  x direction

Net Velocity of third Particle

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//