A fireworks rocket is fired vertically upward. At its maximum height of 80.0 m, it explodes and breaks into two pieces, one with mass 1.40-kg and the other with mass 0.28-kg. In the explosion, 860 J of chemical energy is converted into kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) 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.

The given data is listed below as-

• The chemical energy converted to kinetic energy is, Q = 860 J
• The mass of the one piece of rocket is, $m_A=1.4kg$
• The mass of the second piece of rocket is, $m_B=0.28kg$

The highest point at which rocket exlplode is, h = 80.0 m

The momentum of a particle is the product of the particle’s mass and velocity and is given by-

$\overrightarrow{\mathbf{p}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{v}}$

Where m is the mass of the particle and V is the velocity of the particle.

The total momentum of the system is constant since there is no external horizontal force on the system.

${\overrightarrow{p}}_{total}=cons\tan t$

The total momentum is zero since the rocket is launched vertically and its velocity at the highest point is ${\overrightarrow{p}}_A+{\overrightarrow{p}}_B=0$.

So,

$p_A=-p_B$ ………..(1)

The rocket’s piece moves in opposite direction after the explosion.

Therefore, after the explosion, the one piece moves in +x direction with velocity $v_A$, the second piece moves in -x direction with $v_B$.

Now, ${\overrightarrow{p}}_A=m_A{v}_A$

${\overrightarrow{p}}_B=m_B{v}_B$

Substituting in equation (1)

$m_A{v}_A=m_B{v}_B$

Now substitute values for $m_A$ and $m_B$

For, $m_A=1.4kg$ and $m_B=0.28kg$

$1.4v_A=0.28v_B$

role="math" localid="1665115505110" $v_A=0.200v_B$ …….(1)

The kinetic energy of rocket is zero as the rocket’s velocity just before the explosion was zero. Therefore, the total kinetic energy of rocket’s pieces is given by chemical energy.

role="math" localid="1665115628049" $$\begin{gathered} K_A+K_B=Q \\ \frac{1}{2}{m}_A{v^2}_A+\frac{1}{2}{m}_B{v^2}_B=Q \end{gathered}$$

Substitute $m_A=1.4kg,m_B=0.28kg,Q=860J$and $v_A=0.200v_B$in above equation to obtain the value of$v_B$

$$\begin{gathered} \frac{1}{2}\left(1.40\right){\left(0.200v_B\right)}^2+\frac{1}{2}\left(0.280\right){v^2}_B=860 \\ 0.056{v^2}_B+\left(0.280\right){v^2}_B=1720 \\ {v^2}_B=\sqrt{\left(\frac{1720}{0.056+0.280}\right)} \\ {v^2}_B=72.2m/s \end{gathered}$$

Now, substitute $v_B$in equation (1) to get the value of $v_A$

$$\begin{gathered} v_A=0.200v_B \\ {v}_A=0.200(72.2) \\ {v}_A=14.4m/s \end{gathered}$$

Thus, the speed of each fragment just after the explosion is $v_A=14.4m/s$and$v_B=72.2m/s$

The time taken by a piece to reach the ground is calculated first.

Use the kinematic equation $s=ut+\frac{1}{2}a{t}^2$

Therefore, $y_f-y_i=v_{iy}t+0.5\hspace{0.33em}{a}_y{t}^2$

Substitute ${\mathrm{y}}_{\mathrm{f}}=0,{\mathrm{y}}_{\mathrm{i}}=80\mathrm{m},{\mathrm{v}}_{\mathrm{iy}}=0\mathrm{and}{\mathrm{a}}_{\mathrm{y}}=9.8\mathrm{m}/{\mathrm{s}}^2$

$$\begin{gathered} -80=0-0.5\times 9.8\hspace{0.33em}{\mathrm{t}}^2 \\ \mathrm{t}=\sqrt{\frac{80}{0.5\times 9.8}} \\ \mathrm{t}=4.04\hspace{0.33em}\mathrm{s} \end{gathered}$$

So, piece 1 will travel horizontal distance of

$$\begin{gathered} {\mathrm{x}}_{\mathrm{A}}={\mathrm{v}}_{\mathrm{A}}\cdot \mathrm{t} \\ {\mathrm{x}}_{\mathrm{A}}=14.4\times 4.04 \\ {\mathrm{x}}_{\mathrm{A}}=58.3\hspace{0.33em}\mathrm{m} \end{gathered}$$

And piece 2 will travel horizontal distance of

$$\begin{gathered} {\mathrm{x}}_{\mathrm{B}}={\mathrm{v}}_{\mathrm{B}}\cdot \mathrm{t} \\ {\mathrm{x}}_{\mathrm{B}}=72.2\times 4.04 \\ {\mathrm{x}}_{\mathrm{B}}=292\hspace{0.33em}\mathrm{m} \end{gathered}$$

Now, on the ground the distance between two points where each fragment piece land is calculated as-

$$\begin{gathered} \mathrm{X}={\mathrm{x}}_{\mathrm{A}}+{\mathrm{x}}_{\mathrm{B}} \\ =58.3\times 292 \\ =350.3\hspace{0.33em}\mathrm{m} \end{gathered}$$

Thus, on the ground the distance between two points where each fragment piece land is 350.3 m