Question

question 6.2b:
during a fireworks display, a 0.290 - kg bottle rocket approaches the top of its trajectory and explodes into two pieces as in the figure. the first piece (piece 1) has mass 0.190 kg and the second piece (piece 2) has mass 0.100 kg. immediately after the explosion, the first piece is observed traveling to the left at a speed of 4.40 m/s and the second piece is observed traveling at an angle \\( \theta=32.0^{\circ} \\) above the horizon at a speed of 16.0 m/s.

determine the components of the bottle rockets velocity immediately before the explosion.
\\( v_{x i}=\\) \\( \mathrm{m} / \mathrm{s} \\)
\\( v_{y i}=\\) \\( \mathrm{m} / \mathrm{s} \\)
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Explanation:

Step 1: Apply conservation of momentum in the x - direction

The law of conservation of momentum states that \(P_{initial}=P_{final}\). In the \(x\) - direction, \(m_{total}v_{xi}=m_1v_{1x}+m_2v_{2x}\).
We know that \(m_{total}=0.290\space kg\), \(m_1 = 0.190\space kg\), \(v_{1x}=- 4.40\space m/s\), \(m_2 = 0.100\space kg\), and \(v_{2x}=16.0\cos(32.0^{\circ})\space m/s\).
Substitute the values into the equation:

$$
LATEXBLOCK0
$$

Solve for \(v_{xi}\): \(v_{xi}=\frac{0.517}{0.290}\approx1.78\space m/s\)

Step 2: Apply conservation of momentum in the y - direction

In the \(y\) - direction, \(m_{total}v_{yi}=m_1v_{1y}+m_2v_{2y}\). Since before the explosion (at the top of the trajectory), the vertical component of the velocity of the rocket is \(v_{yi}\) (because at the top of the trajectory for the original rocket \(v_{yi}\) is non - zero if there is an explosion with a vertical component of momentum). The first piece has \(v_{1y} = 0\) (moving horizontally to the left), and \(v_{2y}=16.0\sin(32.0^{\circ})\space m/s\)
Substitute into \(m_{total}v_{yi}=m_1v_{1y}+m_2v_{2y}\):

$$
LATEXBLOCK1
$$

Answer:

\(v_{xi}\approx1.78\space m/s\), \(v_{yi}\approx2.92\space m/s\)