Irodov 1.171: Sharing the Bonus Energy Unevenly – Amit & Deepali Sinha Foundation

Problem Irodov 1.171

A shell of mass $M$ , flying with velocity $v=500$ m/s, bursts into three fragments of equal mass $M/3$ . The internal (explosive) energy released is such that the total kinetic energy of the three fragments immediately after the burst is $\eta=1.5$ times the shell’s kinetic energy just before it. Find the largest possible speed $v_{\max}$ that any one fragment can come away with.

Before the burst the shell moves as one body of mass M with velocity v. After it, three fragments of mass M/3 fly off with velocities v_1, v_2, v_3; the directions drawn are illustrative. — Before the burst the shell moves as one body of mass $M$ with velocity $\vect v$ . After it, three fragments of mass $M/3$ fly off with velocities $\vect v_1$ , $\vect v_2$ , $\vect v_3$ ; the directions drawn are illustrative.

Given

$v=500$ m/s
three identical fragments, mass $M/3$ each
KE increases by factor $\eta=1.5$

Find

v_{\max}

Solution

Step 1Why one shared velocity is the cheapest way to carry a given momentum

Take any two bodies moving along one line, of masses $m_a$ and $m_b$ with velocities $v_a$ and $v_b$ . Three combinations of those four numbers carry the whole argument: the pair’s total mass, its total momentum, and the velocity of either body relative to the other.

\begin{align*}m &= m_a+m_b \\ p &= m_av_a+m_bv_b \\ \Delta v &= v_a-v_b\end{align*}

In those variables the pair’s kinetic energy splits into exactly two terms, and the split is an identity, true for every pair of velocities.

\frac12 m_av_a^2+\frac12 m_bv_b^2 = \frac{p^2}{2m}+\frac12\,\frac{m_am_b}{m}\,\Delta v^2 \tag{1}

For a fixed total momentum, then, the pair’s energy is smallest exactly when $\Delta v=0$ , and the whole of the shortfall is the second term.

Two fragments of mass $M/3$ apiece show the size of that penalty. Sharing a total momentum of $200M$ in SI units, they can move together at 300 m/s, or they can take 200 m/s and 400 m/s, which is the same total momentum differently divided.

\begin{align*}K_{\text{together}} &= \frac{M}{6}\bigl[(300)^2+(300)^2\bigr] = 3.00\times10^{4}M \\ K_{\text{split}} &= \frac{M}{6}\bigl[(200)^2+(400)^2\bigr] = 3.33\times10^{4}M\end{align*}

The divided pair carries $3.3\times10^{3}M$ more for no extra momentum, which is the second term of (1) evaluated at $\Delta v=200$ m/s. In this problem that surplus is energy the burst could have spent on the third fragment instead, which is why the extremum has the other two moving together.

Step 2Reduce to two effective bodies

The extremal arrangement: the two slower fragments merged into one body of mass 2M/3 at the common velocity u, the maximized fragment of mass M/3 at v_, both on the original line of flight. The vertical offset is for legibility only. — The extremal arrangement: the two slower fragments merged into one body of mass $2M/3$ at the common velocity $u$ , the maximized fragment of mass $M/3$ at $v_{\max}$ , both on the original line of flight. The vertical offset is for legibility only.

Call the fragment being maximized fragment 1, with velocity $v_{\max}$ along the shell’s original line of flight (any transverse component would only add speed for no momentum gain, so the extremal fragment moves along that line). By the argument above, the other two fragments, at the extremum, share a single common velocity $u$ ; momentum conservation forces $u$ to lie along that same line too, since fragment 1’s momentum is entirely axial and the shell’s total momentum $Mv$ is axial as well, so the remaining momentum $(2M/3)u$ carried by the merged pair has no transverse part left to supply. This reduces the problem to two effective bodies: mass $M/3$ at $v_{\max}$ , and mass $2M/3$ at $u$ .

Step 3Momentum conservation for the two effective bodies

With the total mass $M$ and original velocity $v$ ,

\frac{M}{3}v_{\max}+\frac{2M}{3}u = Mv.

Dividing through by $M/3$ and solving for $u$ ,

v_{\max}+2u = 3v \;\Longrightarrow\; u = \frac{3v-v_{\max}}{2}. \tag{2}

Step 4Energy conservation for the two effective bodies

The burst raises the total kinetic energy to $\eta$ times the shell’s original kinetic energy,

\frac{1}{2}\cdot\frac{M}{3}v_{\max}^2+\frac{1}{2}\cdot\frac{2M}{3}u^2 = \eta\cdot\frac{1}{2}Mv^2.

Multiplying by $3/M$ ,

v_{\max}^2+2u^2 = 3\eta v^2.

Substituting (2) and expanding,

\begin{align*}v_{\max}^2+2\left(\frac{3v-v_{\max}}{2}\right)^2 &= 3\eta v^2 \\ v_{\max}^2+\frac{1}{2}\left(9v^2-6vv_{\max}+v_{\max}^2\right) &= 3\eta v^2 \\ \frac{3}{2}v_{\max}^2-3vv_{\max}+\frac{9}{2}v^2 &= 3\eta v^2.\end{align*}

Dividing by $3/2$ and collecting terms,

v_{\max}^2-2vv_{\max}+(3-2\eta)v^2 = 0.

Step 5Solve the quadratic and take the larger root

The quadratic formula gives

v_{\max} = v\pm v\sqrt{1-(3-2\eta)} = v\left(1\pm\sqrt{2(\eta-1)}\right).

The larger root is the maximum speed asked for; taking the $+$ sign and substituting $v=500$ m/s, $\eta=1.5$ into it,

\begin{align*}v_{\max} &= 500\left(1+\sqrt{2(0.5)}\right)\text{ m/s} \\ &= 500(1+1)\text{ m/s}\end{align*}

v_{\max} = v\left(1+\sqrt{2(\eta-1)}\right) = 1.0\text{ km/s}

(3)

Check

At $\eta=1$ (no energy released, no burst at all) the formula gives $v_{\max}=v$ , correctly returning the unburst shell’s own speed. The other root, $v\left(1-\sqrt{2(\eta-1)}\right)$ , is the smallest speed a fragment can have (it can even go negative, meaning that fragment reverses direction for large enough $\eta$ ), consistent with the same reduction argument applied to minimize rather than maximize.

Remark

The identity (1) is worth more than this one problem: it is the statement that the kinetic energy of any collection splits into a centre-of-mass part, which conservation nails down, and an internal part, which nothing constrains from below except zero. Every extremal-energy question about a closed system is that same split read one way or the other, and the lumping trick works here only because the quantity being maximized is a speed. It settles “what is the smallest speed a fragment can have” by minimizing instead of maximizing, but it says nothing about a question of a different kind, such as the largest possible angle between two fragments’ paths, where the internal term is not what is being pushed to its limit.