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

To find the magnitude of the velocity of the third fragment, we need to consider the conservation of momentum. The magnitude of the velocity of the third fragment is 445 m/s.

Step-by-step explanation:

To find the magnitude of the velocity of the third fragment, we need to consider the conservation of momentum. According to the law of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion. The momentum of an object is the product of its mass and velocity.

Let's assume the mass of each fragment is m. Since the first fragment continues to move upward with a speed of 235 m/s and the second fragment has a speed of 484 m/s and is moving east, we can write the momentum equation as:

m(235) + m(484) + m(v3) = m(274)

Simplifying the equation, we get:

719m + m(v3) = 274m

445m = - m(v3)

- 445m = m(v3)

Dividing both sides by m, we get:

-445 = v3

Therefore, the magnitude of the velocity of the third fragment is 445 m/s.

Answer 2

Step-by-step 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)=(m)/(3)* v_y+(m)/(3)* 0+(m)/(3)* 235`

`v=587 m/s`

Now conserving momentum in x direction

`m* 0=(m)/(3)* v_x+(m)/(3)* 0+(m)/(3)* 484`

`v_x=-484 m/s`

Net Velocity of third Particle

`v^2=v_x^2+v_y^2`

`v=√(v_x^2+v_y^2)`

`v=√(484^2+587^2)`

`v=760.80 m/s`