Question

A rocket 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.

Answer 1

Answer:

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 $v_y$ and $v_x$ be the velocity of third Particle is Y and x direction

conserving momentum in y direction  

$m(274)=\frac{m}{3}\times v_y+\frac{m}{3}\times 0+\frac{m}{3}\times 235$

$v=587 m/s$

Now conserving momentum in  x direction

$m\times 0=\frac{m}{3}\times v_x+\frac{m}{3}\times 0+\frac{m}{3}\times 484$

$v_x=-484 m/s$

Net Velocity of third Particle

$v^2=v_x^2+v_y^2$

$v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{484^2+587^2}$

$v=760.80 m/s$