Classical-Mechanics · Unit 12 · Video 6 · Interactive Practice

The Explosion Gravity Never Noticed: Finding the Larger Fragment's Landing Point

IKey Formulas

Formula	Name	What you need
$\vec{F}^{\,\text{ext}} = m_{\text{sys}}\vec{A}_{cm} \;\Rightarrow\; \vec{A}_{cm} = \vec{g}$	Center-of-mass acceleration	Gravity is the only external force
$\vec{R}_{cm} = \dfrac{m_2\vec{r}_2 + m_3\vec{r}_3}{m_2 + m_3}$	Two-particle center of mass	Both masses and both positions
$\vec{R}_{cm}(t_f) = 2x_i\,\hat{i}$	Center of mass at landing	Parabola symmetry, equal fall times
$m_1v_1 = m_2v_2 + m_3v_3$	Momentum across the blast	An instantaneous explosion

Key Insight: Internal forces cancel in pairs, so the center of mass keeps flying the undisturbed parabola. With $m_2 = \tfrac{1}{4}m_1$ back at the origin, $2x_i = \tfrac{3}{4}x_{3,f}$ gives $x_{3,f} = \tfrac{8}{3}x_i$ — with no peak height, no launch angle, and no $g$ .

IIThe Center of Mass Never Notices

The blast scatters the pieces, yet the mass-weighted average of their positions stays on the original parabola.

t / t₁ 0.00

IIIWhere the Larger Piece Must Land

With $m_2$ home at the pad, exactly one landing point for $m_3$ puts the center of mass at $2x_i$ .

Mass split x₃ 5.00

💡 Only the ratio matters — $m_1$ divides out, so a 2 kg probe and a 2000 kg probe drop their large fragments on exactly the same spot.

IVMethod 2 — Momentum Across the Blast

The same answer without the center of mass: the small piece's return trip fixes every velocity.

Step 1

💡 Method 1 never asked how fast anything was moving — but had the question been the large piece's impact speed, only this route could have answered it.

VQuiz Questions

Problem 1 · Splitting the Mass

Given: a projectile of mass $m_1$ breaks into exactly two pieces with $m_3 = 3m_2$ — find each fragment's mass in terms of $m_1$ .

✅ Correct! Four equal shares: the small piece takes one of them, the large piece the other three.

❌ That is the split for $m_3 = 2m_2$ . Here the large piece is three times the small one, so the total is

m_2 + 3m_2 = 4m_2

, not

3m_2

❌ Not quite. Mass is conserved in an explosion:

m_2 + m_3 = m_1

. Substitute

m_3 = 3m_2

into that equation before solving.

Show solution

Nothing leaves the system, so mass is conserved:

m_2 + m_3 = m_1

Substitute the given ratio $m_3 = 3m_2$ :

m_2 + 3m_2 = 4m_2 = m_1 \quad\Rightarrow\quad m_2 = \tfrac{1}{4}m_1

and therefore

m_3 = 3m_2 = \tfrac{3}{4}m_1

These two fractions are the only thing the mass information contributes — the actual value of $m_1$ cancels out of the final answer.

Problem 2 · Where the Center of Mass Lands

Given: the explosion happens at the peak, a horizontal distance $x_i$ downrange, and both fragments reach the ground at the same instant $t_f = 2t_1$ — find the center of mass position at that instant.

✅ Correct!

\vec{A}_{cm} = \vec{g}

right through the blast, so the center of mass finishes the undisturbed parabola — and a parabola peaking at

x_i

returns to the ground at

2x_i

❌ That is where $m_3$ lands, not the center of mass. The center of mass sits between the two pieces; only the heavy piece travels the full

\tfrac{8}{3}x_i

❌ Not quite. The explosion is driven by internal forces, which cannot change

\vec{A}_{cm} = \vec{g}

— the center of mass keeps moving along the original parabola instead of stalling at the peak.

Show solution

Step 1 — the only external force is gravity. The chemical forces of the blast come in third-law pairs and cancel, so

m_1\vec{A}_{cm} = m_1\vec{g} \quad\Rightarrow\quad \vec{A}_{cm} = \vec{g}

before, during, and after the explosion. The center of mass therefore flies the very same parabola the intact projectile would have flown.

Step 2 — use the symmetry of that parabola. Its peak lies a horizontal distance $x_i$ from the launch pad, and the descending half mirrors the ascending half, so it meets the ground at

\vec{R}_{cm}(t_f) = 2x_i\,\hat{i}

Why the timing matters: at the peak the velocity is purely horizontal and the pieces separate horizontally, so neither fragment gains vertical velocity. Both fall from the same height with $v_y = 0$ , both take the time $t_1$ , and both are on the ground at $t_f = 2t_1$ — the one instant at which we know where both pieces are.

Problem 3 · A Different Split

Given: the same setup — explosion at the peak, smaller piece returns to the pad — but now $m_3 = 2m_2$ . Find the larger piece's mass and its landing distance $x_{3,f}$ .

What is the larger piece's mass?

How far from the pad does it land?

✅ Correct! A lighter large piece must travel farther to hold the center of mass at

2x_i

\tfrac{2}{3}x_{3,f} = 2x_i

gives

3x_i

❌ Check the mass split. With

m_3 = 2m_2

the total is

m_2 + 2m_2 = 3m_2 = m_1

, so the pieces are thirds, not quarters.

❌ Check the balance.

m_2

contributes nothing (it sits at the origin), so

m_3x_{3,f} = m_1\cdot 2x_i

— solve that for

x_{3,f}

Show solution

Step 1 — the masses. With $m_2 + m_3 = m_1$ and $m_3 = 2m_2$ :

3m_2 = m_1 \quad\Rightarrow\quad m_2 = \tfrac{1}{3}m_1, \qquad m_3 = \tfrac{2}{3}m_1

Step 2 — the center of mass at landing. Nothing about the geometry changed, so $\vec{R}_{cm}(t_f) = 2x_i\,\hat{i}$ still holds. With $\vec{r}_2 = \vec{0}$ the first term vanishes:

2x_i = \frac{m_2(0) + m_3\,x_{3,f}}{m_1} = \frac{\tfrac{2}{3}m_1\,x_{3,f}}{m_1} = \tfrac{2}{3}x_{3,f}

Step 3 — solve.

x_{3,f} = \tfrac{3}{2}\cdot 2x_i = 3x_i

Sanity check: this piece is lighter than the $\tfrac{3}{4}m_1$ of the video, so it must reach farther out to keep the average at $2x_i$ — and $3x_i > \tfrac{8}{3}x_i$ ✓

Problem 4 · When the Small Piece Just Drops

Given: the original masses $m_2 = \tfrac{1}{4}m_1$ and $m_3 = \tfrac{3}{4}m_1$ , but this time the blast leaves the small piece with zero horizontal velocity, so it falls straight down from the peak. The intact projectile had speed $v_1$ at the peak and $x_i = v_1t_1$ . Find $v_3$ and the landing distance $x_{3,f}$ .

What is the larger piece's velocity just after the blast?

How far from the pad does the larger piece land?

✅ Excellent! Nothing is thrown backward now, so the large piece needs less forward speed —

\tfrac{4}{3}v_1

instead of

\tfrac{5}{3}v_1

— and lands short of the old answer, at

\tfrac{7}{3}x_i

❌ Redo the momentum balance. The small piece now carries no horizontal momentum:

m_1v_1 = \tfrac{1}{4}m_1(0) + \tfrac{3}{4}m_1v_3

, so

v_3

must be

\tfrac{4}{3}v_1

, not the video's

\tfrac{5}{3}v_1

❌ Add the two stretches.

\Delta x_3 = v_3t_1

measures travel after the blast; the explosion point was already

x_i

downrange, so

x_{3,f} = x_i + \Delta x_3

Show solution

Step 1 — momentum across the blast. The explosion is instantaneous, so gravity's impulse is negligible and the horizontal momentum is unchanged. With $v_2 = 0$ :

m_1v_1 = \tfrac{1}{4}m_1(0) + \tfrac{3}{4}m_1v_3

v_1 = \tfrac{3}{4}v_3 \quad\Rightarrow\quad v_3 = \tfrac{4}{3}v_1

Step 2 — kinematics to the ground. The small piece has no horizontal velocity but still $v_y = 0$ , so both pieces fall for the same time $t_1$ . The large piece travels

\Delta x_3 = v_3t_1 = \tfrac{4}{3}v_1t_1 = \tfrac{4}{3}x_i

Step 3 — add the pre-explosion stretch.

x_{3,f} = x_i + \tfrac{4}{3}x_i = \tfrac{7}{3}x_i

Check with method 1. The small piece now lands directly below the peak, at $x_i$ , so

R_{cm}(t_f) = \tfrac{1}{4}x_i + \tfrac{3}{4}\left(\tfrac{7}{3}x_i\right) = \tfrac{1}{4}x_i + \tfrac{7}{4}x_i = 2x_i \ \checkmark

Both routes agree, exactly as they did in the video.

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