A 12.0-kg shell is launched at an angle of 55⁰ above the horizontal with an initial speed of 150 m/s. At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how energy was released in the explosion?

The given data is listed below as-

• The mass of the shell is m =12. kg
• The mass of the lighter fragment is $m_A=3.00\mathrm{kg}$
• The mass of the heavier fragment is $m_B=9.00\mathrm{kg}$
• Initial speed of the shell is v =150 m/s
• Angle of projectile above the horizontal is $\theta =55°$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

There are three stages during the motion of the projectile:

First stage- The motion of the projectile to its highest point

Second stage- the explosion

Third stage-motion of a fragment from explosion to the landing.

The first stage

Now, take +x in the horizontal direction and +y as the upward movement.

The initial velocity is the vertical component of the velocity and the projectile moves with a negative acceleration for a vertical motion.

So, the equation of initial velocity is:

$$\begin{gathered} V_{iy}=150.0\sin 55° \\ =123.\mathrm{m}/\mathrm{s} \end{gathered}$$

The vertical velocity comes to zero at the highest point of the shell.

Now, the kinematic equation of displacement, is v =u+at.

$or,\hspace{0.33em}{v}_{fy}=v_{iy}+a_{y}t$

Substitute $V_{fy}=0$, $a_y=-9.8\hspace{0.33em}m/s^{-2}$$V_{iy}$in the above equation

$0=123-9.8\times t_1$

Therefore, $$\begin{gathered} t_1=\frac{123}{9.8} \\ {t}_1=12.5s \end{gathered}$$

The time taken by the shell to reach its highest point is calculated as 12.5 s.

The horizontal component of initial velocity is equal to the horizontal motion of the projectile.

$$\begin{gathered} V_x=150\hspace{0.33em}cos55° \\ {V}_x=86.0\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

So, $x_1=v\cdot t_1$

$$\begin{gathered} x_1=86.0\times 12.5 \\ {x}_1=1079\hspace{0.33em}\mathrm{m} \end{gathered}$$

The second stage: The velocity of projectile before the explosion is:

$$\begin{gathered} \overrightarrow{v}=u_x\stackrel{⏜}{i}+0\stackrel{⏜}{j} \\ \overrightarrow{v}=86\mathrm{m}/\mathrm{s}\stackrel{⏜}{i} \end{gathered}$$

Equation of momentum is

$p=m\cdot v$

$p=(12.0)(86.0\stackrel{⏜}{i})=(1032\mathrm{kg}\cdot m/s)\stackrel{⏜}{i}$

Now, the momentum of the shell before explosion equals the total momentum of two fragments after explosion.

$\overrightarrow{p}={\overrightarrow{p}}_A+{\overrightarrow{p}}_B$

${\overrightarrow{p}}_A+{\overrightarrow{p}}_B=\left(1032kg.m/s\right)\stackrel{⏜}{i}$

The initial velocity of the shell is same as the final velocity of shell as the heavier segment lands back at the firing point.

$v_B=-v=-86\hspace{0.33em}\mathrm{m}/\mathrm{s}$

$m_A{v}_A+m_B{v}_B=1032$

Now, substitute, $m_A=3.00\mathrm{kg}$, $m_B=9.00\mathrm{kg}$and $v_B=-86\hspace{0.33em}\mathrm{m}/\mathrm{s}$in above equation

$$\begin{gathered} 3v_A+9(-86)=1032 \\ {v}_A=602\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

The third stage

The time taken by fragments to fall is same as the time taken by shell to reach its highest point.

So, $t_2=t_1=12.5\hspace{0.33em}\mathrm{s}$

The horizontal distance travelled by lighter fragment from the explosion point:

$$\begin{gathered} x_2=v_{Ax}.t_2 \\ =602\times 12.5 \\ =7551\mathrm{m} \end{gathered}$$

So, the lighter fragment lands at:

$$\begin{gathered} d=x_1+x_2 \\ =1079+7551 \\ =8630\mathrm{m} \end{gathered}$$

Thus, from where the shell was launched, the lighter fragment lands at 8630 m.

The energy released during explosion is the change in total kinetic energy of the system

$$\begin{gathered} \mathrm{\Delta }K=K_2-K_1 \\ =\left(K_A+K_B\right)-K_1 \\ =\frac{1}{2}\left(3\right)(602{)}^2+\frac{1}{2}(9\left)\right(86{)}^2-\frac{1}{2}\left(12\right){\left(86\right)}^2=5.33\times {10}^5\mathrm{J} \end{gathered}$$

Thus, the kinetic energy released during explosion is $=5.33\times {10}^5\mathrm{J}$