Question

A rocket is fired vertically upward. At the instant it reaches an altitude of 1450 m and a speed of 265 m/s, it explodes into three equal fragments. One fragment continues to move upward with a speed of 292 m/s following the explosion. The second fragment has a speed of 396 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

Solution:

The intial speed of the rocket

Mass of the rocket = M

Let the rocket explodes into three equal fragments of mass and speed of the fragments be , and . The third fragment move at angle as show in figure.

  

Given, ,

Since 3 equal fragments are formed, .

According to the law of conservation of momentum,

Momentum before explosion = Momentum after explosion

Thus, we have, from figure in y-direction

  

   .....................,..(1)

In x- direction, we have,

  

...........................(2)

Dividing eq.(2) by eq.(1), we get

  

  

Substituting for in (1)

The third fragment will move with a speed of 640m/s at an angle of 38.2 degrees to west of upward direction.