Elastic collisions — 1D - solve for final velocities: Exercises

Figure — Elastic collisions — 1D - solve for final velocities

The figure above is your sign map. A velocity arrow pointing right is a positive number; pointing left is negative. Before touching algebra, always redraw arrows and read off signs from this picture — most errors below are just a dropped minus sign.

Level 1 — Recognition

Goal: recognise which rule applies and read the answer almost by inspection.

L1.1

Two identical pucks on frictionless ice. Puck A moves right at $u_{1} = + 4$ m/s and strikes puck B at rest ( $u_{2} = 0$ ). The collision is elastic. What are $v_{1}$ and $v_{2}$ ?

Recall Solution L1.1

Approach check: $u_{1} - u_{2} = 4 - 0 = 4 > 0$ ✓ — they do collide. WHAT rule: equal masses, target at rest — the special "swap" case. WHY: with $m_{1} = m_{2}$ the coefficients $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} = 0$ and $\frac{2 m _{2}}{m _{1} + m _{2}} = 1$ . $v_{1} = 0 \cdot u_{1} + 1 \cdot u_{2} = 0, v_{2} = 1 \cdot u_{1} + 0 \cdot u_{2} = + 4 m/s .$ Answer: $v_{1} = 0$ , $v_{2} = + 4$ m/s. A stops dead; B leaves with all the motion. This is the Newton's-cradle behaviour, see Newton's Cradle.

L1.2

A ping-pong ball ( $m_{1} = 2.7$ g) hits a stationary bowling ball ( $m_{2} = 6$ kg) head-on at $+ 5$ m/s, elastically. Roughly what happens to the ping-pong ball, without full arithmetic?

Recall Solution L1.2

Approach check: $u_{1} - u_{2} = 5 - 0 = 5 > 0$ ✓ — they do collide. WHAT rule: light hits heavy wall ( $m_{1} ≪ m_{2}$ , $u_{2} = 0$ ). WHY: $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} \to - 1$ and $\frac{2 m _{1}}{m _{1} + m _{2}} \to 0$ . $v_{1} \approx - u_{1} = - 5 m/s, v_{2} \approx 0.$ Answer: the ping-pong ball bounces straight back at about $- 5$ m/s; the bowling ball barely moves. It is a ball bouncing off a wall.

Level 2 — Application

Goal: plug signed numbers into the two tools and grind out both final velocities.

L2.1

$m_{1} = 3$ kg at $u_{1} = + 6$ m/s hits $m_{2} = 1$ kg at rest, elastically. Find $v_{1}$ and $v_{2}$ .

Recall Solution L2.1

Approach check: $u_{1} - u_{2} = 6 - 0 = 6 > 0$ ✓ — they collide. Tool 2 (relative velocity), WHY: elastic 1D means separation speed = approach speed, so $v_{2} - v_{1} = u_{1} - u_{2} = 6$ . Rearranged to isolate one unknown: $v_{2} = v_{1} + 6$ . (We solve for $v_{2}$ in terms of $v_{1}$ so we can substitute into momentum and get a single-variable equation.) Tool 1 (momentum), WHY: no external horizontal force, so total $p$ is conserved: $3 (6) + 1 (0) = 18 = 3 v_{1} + 1 v_{2}$ . Substitute — WHY: replacing $v_{2}$ by $v_{1} + 6$ turns two unknowns into one. $3 v_{1} + (v_{1} + 6) = 18 \Rightarrow 4 v_{1} = 12 \Rightarrow v_{1} = + 3$ m/s. Back-substitute — WHY: now that $v_{1}$ is known, the rule hands us $v_{2}$ directly: $v_{2} = 3 + 6 = + 9$ m/s. Answer: $v_{1} = + 3$ m/s, $v_{2} = + 9$ m/s. Heavier ball slows but keeps moving forward; lighter ball shoots ahead.

L2.2

$m_{1} = 2$ kg at $u_{1} = + 8$ m/s hits $m_{2} = 4$ kg at $u_{2} = + 2$ m/s (same direction, catching up), elastically. Find $v_{1}$ and $v_{2}$ .

Recall Solution L2.2

Approach check: $u_{1} - u_{2} = 8 - 2 = 6 > 0$ ✓ — the faster left block genuinely catches the right one. Tool 2, WHY: elastic ⇒ separation = approach, $v_{2} - v_{1} = 6$ , isolate $v_{2} = v_{1} + 6$ so we can substitute. Tool 1, WHY: momentum conserved: $2 (8) + 4 (2) = 16 + 8 = 24 = 2 v_{1} + 4 v_{2}$ . Substitute — WHY: eliminate $v_{2}$ to leave one unknown. $2 v_{1} + 4 (v_{1} + 6) = 24 \Rightarrow 6 v_{1} + 24 = 24 \Rightarrow v_{1} = 0$ . Back-substitute: $v_{2} = 0 + 6 = + 6$ m/s. Answer: $v_{1} = 0$ , $v_{2} = + 6$ m/s. The lighter chaser stops; the heavier one speeds up from 2 to 6 m/s. Check with the formula for $v_{1}$ : $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2} = \frac{- 2}{6} (8) + \frac{8}{6} (2) = - \frac{16}{6} + \frac{16}{6} = 0.$ ✓

Level 3 — Analysis

Goal: handle signs, negative velocities, and "what-if" reasoning.

L3.1

$m_{1} = 2$ kg at $u_{1} = + 3$ m/s meets $m_{2} = 5$ kg at $u_{2} = - 4$ m/s (moving left, head-on), elastically. Find $v_{1}$ , $v_{2}$ , and verify kinetic energy is conserved.

Recall Solution L3.1

Approach check: $u_{1} - u_{2} = 3 - (- 4) = 7 > 0$ ✓ — head-on, closing at $7$ m/s. Tool 2, WHY: elastic ⇒ they separate as fast as they approached, $v_{2} - v_{1} = 7$ , isolate $v_{2} = v_{1} + 7$ . Tool 1, WHY: momentum conserved: $2 (3) + 5 (- 4) = 6 - 20 = - 14 = 2 v_{1} + 5 v_{2}$ . Substitute — WHY: remove $v_{2}$ to solve one variable. $2 v_{1} + 5 (v_{1} + 7) = - 14 \Rightarrow 7 v_{1} + 35 = - 14 \Rightarrow 7 v_{1} = - 49 \Rightarrow v_{1} = - 7$ m/s. Back-substitute: $v_{2} = - 7 + 7 = 0$ m/s. Answer: $v_{1} = - 7$ m/s (thrown back leftward, faster than it came), $v_{2} = 0$ (heavy ball halted). KE check: Before $= \frac{1}{2} (2) (9) + \frac{1}{2} (5) (16) = 9 + 40 = 49$ J. After $= \frac{1}{2} (2) (49) + \frac{1}{2} (5) (0) = 49$ J. ✓ See Kinetic Energy.

L3.2

In an elastic collision the incoming ball $m_{1}$ at $u_{1}$ hits $m_{2}$ at rest and rebounds (comes back left, $v_{1} < 0$ ). What condition on the masses guarantees a rebound?

Recall Solution L3.2

WHAT we compute: the sign of $v_{1}$ . $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}$ (since $u_{2} = 0$ ). With $u_{1} > 0$ , the sign of $v_{1}$ is the sign of $(m_{1} - m_{2})$ .

$v_{1} < 0$ (rebound) $⟺ m_{1} < m_{2}$ : the incoming ball is lighter than the target.
$v_{1} = 0$ (stops) $⟺ m_{1} = m_{2}$ .
$v_{1} > 0$ (keeps going) $⟺ m_{1} > m_{2}$ . Answer: the ball rebounds exactly when it is lighter than what it hits. This is the analytic version of "light off a wall bounces back."

Level 4 — Synthesis

Goal: combine collisions with other mechanics (energy, height, restitution).

L4.1

A block $m_{1} = 1$ kg slides at $u_{1} = + 6$ m/s and hits a stationary block $m_{2} = 3$ kg elastically. After the collision $m_{2}$ slides onto a rough patch with friction coefficient $μ = 0.2$ . How far does $m_{2}$ travel before stopping? (Use $g = 10$ m/s².)

Recall Solution L4.1

Approach check: $u_{1} - u_{2} = 6 - 0 = 6 > 0$ ✓ — they collide. Step 1 — collision, find $v_{2}$ . $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} = \frac{2 ( 1 )}{4} (6) = + 3$ m/s. (WHY: $u_{2} = 0$ , use the $v_{2}$ formula.) Step 2 — friction stops it. KE turns into friction work: $\frac{1}{2} m_{2} v_{2}^{2} = μ m_{2} g d$ . Mass cancels: $d = \frac{v _{2}^{2}}{2 μg} = \frac{3 ^{2}}{2 ( 0.2 ) ( 10 )} = \frac{9}{4} = 2.25$ m. Answer: $m_{2}$ slides $2.25$ m. (WHY split into two stages: the collision is instantaneous and elastic; the slide afterward is a separate energy-loss process governed by friction, not by the collision.)

L4.2

Two elastic collisions in a row. $m_{1} = 2$ kg at $u_{1} = + 9$ m/s hits $m_{2} = 2$ kg at rest; then that same $m_{2}$ (now moving) hits $m_{3} = 6$ kg at rest. Find the final velocity of $m_{3}$ .

Recall Solution L4.2

Approach check (both): $9 - 0 > 0$ for the first, and the moving $m_{2}$ then meets the resting $m_{3}$ with positive approach ✓. Collision 1 (equal masses, swap): $m_{1}$ stops, $m_{2}$ leaves at $+ 9$ m/s. (WHY: $m_{1} = m_{2}$ , target at rest.) Collision 2 ( $m_{2} = 2$ hits $m_{3} = 6$ at rest): $v_{3} = \frac{2 m _{2}}{m _{2} + m _{3}} u_{2} = \frac{2 ( 2 )}{2 + 6} (9) = \frac{4}{8} (9) = + 4.5$ m/s. Answer: $m_{3}$ ends up at $+ 4.5$ m/s. (Sanity: also $v_{2}^{'} = \frac{m _{2} - m _{3}}{m _{2} + m _{3}} (9) = \frac{- 4}{8} (9) = - 4.5$ — the middle block rebounds. Related chain behaviour: Newton's Cradle.)

Level 5 — Mastery

Goal: reverse-engineer, prove general statements, connect to restitution and the CM frame.

L5.1

After an elastic head-on collision, a $3$ kg ball (initially at rest, $u_{2} = 0$ ) is found moving at $v_{2} = + 3$ m/s. The incoming ball has $m_{1} = 2$ kg. Find the incoming speed $u_{1}$ and the incoming ball's final velocity $v_{1}$ — and explain why you cannot pick $v_{1}, v_{2}$ freely.

Recall Solution L5.1

First, WHERE the general formulas come from (so we use them honestly). With $u_{2} = 0$ the parent's two boxed results reduce to $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}, v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} .$ WHY these hold: start from Tool 1 momentum $m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$ and Tool 2 rule $v_{2} - v_{1} = u_{1}$ (since $u_{2} = 0$ ). Substitute $v_{2} = v_{1} + u_{1}$ into momentum: $m_{1} u_{1} = m_{1} v_{1} + m_{2} (v_{1} + u_{1}) \Rightarrow (m_{1} - m_{2}) u_{1} = (m_{1} + m_{2}) v_{1} \Rightarrow v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}$ . Then $v_{2} = v_{1} + u_{1} = \frac{m _{1} - m _{2} + m _{1} + m _{2}}{m _{1} + m _{2}} u_{1} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}$ . These are derived, not assumed.

Now solve. Plug $m_{1} = 2, m_{2} = 3$ : $v_{2} = \frac{2 ( 2 )}{5} u_{1} = \frac{4}{5} u_{1}$ . Set equal to $3$ : $u_{1} = \frac{5}{4} (3) = 3.75$ m/s. Then $v_{1} = \frac{2 - 3}{5} u_{1} = - \frac{1}{5} (3.75) = - 0.75$ m/s. Answer: $u_{1} = + 3.75$ m/s and $v_{1} = - 0.75$ m/s. WHY you cannot choose $v_{1}, v_{2}$ freely: for fixed masses and $u_{1}$ , the pair $(v_{1}, v_{2})$ is forced by two equations. There is exactly one elastic outcome per input, so quoting an arbitrary $v_{2}$ already fixes $u_{1}$ and $v_{1}$ .

L5.2

Prove that for a head-on elastic collision with $m_{2}$ initially at rest, the fraction of $m_{1}$ 's kinetic energy transferred to $m_{2}$ is $\frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}$ , and show it is maximised (equal to 1) when $m_{1} = m_{2}$ .

Recall Solution L5.2

Step 1 — $m_{2}$ 's final speed: $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}$ (derived in L5.1). Step 2 — energy of $m_{2}$ over initial energy of $m_{1}$ : $f = \frac{\frac{1}{2} m _{2} v _{2}^{2}}{\frac{1}{2} m _{1} u _{1}^{2}} = \frac{m _{2}}{m _{1}} (\frac{2 m _{1}}{m _{1} + m _{2}})^{2} = \frac{m _{2}}{m _{1}} \cdot \frac{4 m _{1}^{2}}{( m _{1} + m _{2} ) ^{2}} = \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}} .$ Step 3 — maximise. Expand the gap: $(m_{1} + m_{2})^{2} - 4 m_{1} m_{2} = (m_{1} - m_{2})^{2} \geq 0$ , so $(m_{1} + m_{2})^{2} \geq 4 m_{1} m_{2}$ , giving $f \leq 1$ . Equality (hence $f = 1$ ) holds exactly when $(m_{1} - m_{2})^{2} = 0$ , i.e. $m_{1} = m_{2}$ . Answer: transferred fraction $= \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}$ ; it hits its maximum of $100%$ when $m_{1} = m_{2}$ — the full energy handoff of a Newton's cradle. See Kinetic Energy.

L5.3

Using the Coefficient of Restitution $e$ defined by $u_{1} - u_{2} = - e (v_{1} - v_{2})$ (so $e = 1$ is elastic), show that the separation speed obeys $v_{2} - v_{1} = e (u_{1} - u_{2})$ , and state what $e = 0$ means physically.

Recall Solution L5.3

Step 1 — rewrite the definition. Start from $u_{1} - u_{2} = - e (v_{1} - v_{2})$ . The right side is $- e v_{1} + e v_{2} = e (v_{2} - v_{1})$ . So the definition literally says $u_{1} - u_{2} = e (v_{2} - v_{1}) .$ Step 2 — solve for the separation speed. We want $v_{2} - v_{1}$ on its own. Divide both sides by $e$ (valid for $e \neq = 0$ ): $v_{2} - v_{1} = \frac{u _{1} - u _{2}}{e} .$ (WHY divide, not multiply: $e$ currently multiplies the separation speed $v_{2} - v_{1}$ ; to isolate that separation speed we undo the multiplication by dividing.) Reading it the intuitive way — WHY the flipped form still helps: rearrange instead to put approach on the left: since $u_{1} - u_{2} = e (v_{2} - v_{1})$ , the approach speed is $e$ times the separation speed. Equivalently separation is approach divided by $e$ . For $e < 1$ that makes separation larger than approach only if... let's just check the endpoints, which is what we actually use:

$e = 1$ : definition gives $u_{1} - u_{2} = v_{2} - v_{1}$ — approach = separation, the elastic rule of Tool 2. KE conserved.
$e = 0$ : the definition $u_{1} - u_{2} = 0 \cdot (v_{2} - v_{1}) = 0$ forces the incoming approach to vanish only if inputs allow; the physical content is read from the standard sticking form $v_{2} - v_{1} = 0$ , i.e. $v_{1} = v_{2}$ — the two bodies move off together. Answer: the correctly-inverted relation is $v_{2} - v_{1} = \frac{u _{1} - u _{2}}{e}$ (equivalently $u_{1} - u_{2} = e (v_{2} - v_{1})$ ). At $e = 1$ it is the elastic Tool 2; at $e = 0$ the bodies stick with $v_{1} = v_{2}$ , the perfectly inelastic case of maximum KE loss.

Recall Master checklist (open after finishing all problems)

Approach check ::: confirm $u_{1} - u_{2} > 0$ (objects closing) before applying any formula. Draw arrows, read signs ::: from the sign map, right = +, left = −. Two tools ::: momentum $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ and relative-velocity $u_{1} - u_{2} = v_{2} - v_{1}$ . Verify ::: recompute KE before and after; they must match for elastic. Special cases ::: equal masses swap; heavy→light doubles; light→heavy rebounds.

Connections

Conservation of Momentum — Tool 1, used in every problem here.
Kinetic Energy — the elastic condition and the energy checks in L3.1, L5.2.
Coefficient of Restitution — L5.3 generalises to any $e$ .
Inelastic Collisions — 1D — the $e = 0$ end of that same rule.
Center of Mass Frame — clean way to see rebounds (L3.2).
Newton's Cradle — the equal-mass swap in L1.1 and L4.2.

Exercises — Elastic collisions — 1D - solve for final velocities