Question

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A fireworks rocket is fired vertically up. at its maximum height of 60.0m, it explodes and breaks into two pieces, one with mass 1.20kg and the other with mass 0.30kg, both with horizontal initial velocities. in the explosion, 0.960kJ of chemical energy is converted to the kinetic energy of the two fragments.
What is the speed of each fragment just after the explosion? it is observed that the two fragments hit the ground at the same time. what is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.
Explain the difference in the distances travelled by the fragments if any.

Answer 1

Answer on Question #56681-Physics-Mechanics-Relativity

A fireworks rocket is fired vertically up. at its maximum height of 60.0m, it explodes and breaks into two pieces, one with mass 1.20kg and the other with mass 0.30kg, both with horizontal initial velocities. in the explosion, 0.960kJ of chemical energy is converted to the kinetic energy of the two fragments.

What is the speed of each fragment just after the explosion? It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land?

Solution

What is the speed of each fragment just after the explosion?

This can be answered by the law of conservation of momentum.

At its maximum height, the velocity, and hence momentum, of the rocket is 0.

Right after the rocket explodes and breaks into 2 fragments, the total momentum of these fragments is 0 and their total kinetic energy is 960 J. So,

$$1.2 v_A + 0.3 v_B = 0$$ $$1.2 v_A^2 + 0.3 v_B^2 = 960$$

Solving for $v_A$ and $v_B$:

$$v_B = \left(- \frac{12}{3}\right) v_A$$ $$1.2 v_A^2 + 0.3 \left(- \frac{12}{3}\right)^2 v_A^2 = 960$$ $$v_A^2 = \frac{960}{1.2 + 0.3 \cdot 16} = 160$$

Supposing $v_A$ is in the positive direction,

$$v_A = +12.6 \frac{m}{s}$$ $$v_B = \left(- \frac{12}{3}\right) 12.6 = -50.6 \frac{m}{s}$$

What is the distance between the points on the ground where they land?

The time it takes for the fragments to fall from the 60-meter height to the ground:

$$t = \sqrt{\frac{2h}{g}} = \sqrt{2 \cdot \frac{60}{9.81}} = 3.50 \, s$$

Position of each fragment when hitting the ground:

$$x_A = v_A t = 12.6 \cdot 3.50 = 44.1 \, m$$ $$x_B = v_B t = -50.6 \cdot 3.50 = -177.1 \, m.$$

The distance between these positions:

$$d = x _ {A} - x _ {B} = 44.1 + 177.1 = 221 \text{ m}.$$

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