Question

A 1500 kg weather rocket accelerates upward at 10m/s2. It explodes 2.0 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 was the speed of the heavier fragment just after the explosion?

Answer 1

The speed of the heavier fragment just after the explosion was about 20 m/s

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Further explanation

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.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of rocket = M = 1500 kg

acceleration of rocket = a = 10 m/s²

elapsed time = t = 2.0 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

Asked:

the speed of the heavier fragment = ?

Solution:

Firstly , we will calculate the final speed of rocket just before explosion:

$v = u + at$

$v = 0 + 10(2)$

$\boxed{v = 20 \texttt{ m/s}}$

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

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

$h' = 0 + \frac{1}{2}(10)(2.0)^2$

$\boxed{h' = 20 \texttt{ m}}$

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We could calculate the initial speed of lighter fragment just after explosion:

$v_1^2 = u_1^2 - 2g\Delta h$

$v_1^2 = u_1^2 - 2g(h - h')$

$0^2 = u_1^2 - 2(9.8)(530 - 20)$

$u_1^2 = 9996$

$\boxed{u_1 = \sqrt{9996} \texttt{ m/s}}$

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Finally , we will use Conservation of Momentum Law as follows:

$Mv = m_1u_1 + m_2u_2$

$1500(20) = 500(\sqrt{9996}) + 1000(u_2)$

$30000 = 500(\sqrt{9996}) + 1000(u_2)$

$\boxed{u_2 \approx -20 \texttt{ m/s}}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer 2

Answer:

Approximately $\rm 19.8\; m\cdot s^{-1}$ (downwards.)

Assumptions:

• the rocket started from rest;
• the gravitational acceleration is constantly $\rm -9.8\; m \cdot s^{-2}$;
• there's no air resistance on the rocket and the two fragments.
• Both fragments traveled without horizontal velocity.

Explanation:

The upward speed of the rocket increases by $\rm 10\; m \cdot s^{-1}$. If the rocket started from rest, the vertical speed of the rocket should be equal to $\rm 20\; m \cdot s^{-1}$.

The mass of the rocket (before it exploded) is 1500 kilograms. At 20 m/s, its momentum will be equal to $\rm 20 \times 1500 = 30,000\; kg \cdot m\cdot s^{-1}$.

What's the initial upward velocity, $u$, of the lighter fragment?

The upward velocity of the lighter fragment is equal to $v = 0$ once it reached its maximum height of $x = \rm 530\; m$.

$v^2 - u^2 = 2g \cdot x$.

$\begin{aligned}u &= \sqrt{v^2 - 2g\cdot x} \\ &= \sqrt{-2 (-9.8) \times 530}\\ &\approx \rm 101.922\; m \cdot s^{-1}\end{aligned}$.

Mass of the two fragments:

• Lighter fragments: $\displaystyle \frac{1}{1 + 2} \times 1500 =\rm 500\; kg$.
• Heavier fragment: $\displaystyle \frac{2}{1 + 2} \times 1500 =\rm 1000\; kg$.

Initial momentum of the lighter fragment:

$m \cdot v = \rm 10192.2\; kg \cdot m \cdot s^{-1}$.

If there's no air resistance, momentum shall conserve. The momentum of the lighter fragment, plus that of the heavier fragment, should be equal to that of the rocket before it exploded.

The initial momentum of the heavier fragment should thus be equal to the momentum of the two pieces, combined, minus the initial momentum of the lighter fragment.

$\rm 30000 - 10192.2 = 19807.8\;kg \cdot m \cdot s^{-1}$.

Velocity of the heavier fragment:

$\displaystyle \rm \frac{19807.8\;kg \cdot m \cdot s^{-1}}{1000\; kg} \approx 19.8\; m \cdot s^{-1}$.