Question

A 1500 kg weather rocket accelerates upward at 10.0 m/s . It explodes 2.00 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion

Answer 1

20 m/s is the speed of the heavier fragment just after the explosion.

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

F = ma                   where, F = Force (Newton)

                                          m= mass

                                          a = acceleration

Given:

mass of rocket = M = 1500 kg

acceleration of rocket = a = 10 m/s²

elapsed time = t = 2.00 s

mass of lighter fragment = m₁ = m = 500 kg

mass of heavier fragment = m₂ = 2m = 1000 kg

maximum height of lighter fragment = h = 530 m

Let's calculate the final speed of the rocket just before the explosion:

v = u + at

v = 0 + 10(2)

v = 20 m/s

Then, we will calculate the height of the rocket just before the explosion:

$h' = ut + \frac{1}{2}at^{2}$

$h' = 0 + \frac{1}{2} (10)(2.00)^{2}$

$h' =20m$

The initial speed of lighter fragment just after the explosion:

$v^{2}$₁ $= u^{2}$₁ - $2g$Δ$h$

$v^{2}$₁ $= u^{2}$₁ - $2g$ $(h - h')$

$0^{2} = u^{2}$₁$- 2(9.8) (530-20)$

$u^{2}$₁$=9996$

$u$₁ $=\sqrt{9996} m/s$

Using Conservation of Momentum Law :

$M v=m$₁ $u$₁ $+ m$₂$u$₂

$1500 (20)= 500(\sqrt{9996} ) + 1000u$₂

$u$₂ ≅ $- 20 m/s$

Learn more about Newton's law of motion here:

brainly.com/question/10454047

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