Elastic vs. Inelastic Collisions - Statics and Dynamics

1. A 4 kg object moving 5 m/s collides with a 2 kg object moving 2 m/s the opposite way. They stick. Find final speed.
   Answer: 2.67 m/s Take direction of the 4 kg object as positive: 4*5 - 2*2 = 16 kg m/s. After they stick, total mass is 6 kg, so final speed is 16/6 = 2.67 m/s. The tempting 4.00 m/s divides the net momentum by only the 4 kg object's mass, but the 2 kg object is also moving with the combined mass.
2. What was the bullet's initial speed $v_0$, and what percentage of the original kinetic energy was lost during the embedding (collision) phase?
   Answer: $v_0 = 182 \ \text{m/s}$; approximately $99\%$ of the original KE is lost, found by applying energy conservation through the swing to get the post-collision velocity, then momentum conservation through the collision to get $v_0$, and finally computing the ratio of lost to original KE. Ballistic pendulum problems test your ability to recognize that two different physical laws govern two different phases of the event — and mixing them up is the most common mistake. Here's the correct approach: During the swing (post-collision), energy is conserved, so $\frac{1}{2}(M+m)V^2 = (M+m)gh$, giving $V = \sqrt{2gh} = \sqrt{2(10)(0.2)} = 2 \ \text{m/s}$. During the collision itself, momentum is conserved (but energy is not): $mv_0 = (M+m)V$, so $v_0 = \frac{(9+0.1)(2)}{0.1} = 182 \ \text{m/s}$. The original KE is $\frac{1}{2}(0.1)(182)^2 \approx 1656 \ \text{J}$, and the post-collision KE is $\frac{1}{2}(9.1)(2)^2 = 18.2 \ \text{J}$. The fraction lost is $\frac{1656 - 18.2}{1656} \approx 99\%$. This confirms B is correct. A is wrong because it applies energy conservation across the entire event — including the inelastic collision — which violates physics. Energy is lost during embedding and cannot be recovered; this gives a drastically underestimated $v_0$. C correctly identifies the formula for energy retention in a perfectly inelastic collision as $\frac{m}{M+m}$, and the numerical answer for $v_0$ is right, but this is a coincidence of framing — the answer in B arrives at the same ~99% figure by actually computing the KE values, which is the methodologically sound approach the exam rewards. D invents a false rule. Perfectly inelastic collisions do not always lose exactly 50% of KE; the fraction depends on the mass ratio. Strategy tip: Always split ballistic pendulum problems into two phases — use energy conservation only for the swing, and momentum conservation only for the collision. Never apply energy conservation across an inelastic collision.
3. In a head-on elastic collision, a 1 kg ball moving 3 m/s hits a 2 kg ball at rest. Find the 1 kg ball's speed after.
   Answer: 1.00 m/s In a head-on elastic collision, the 1 kg ball rebounds because it is lighter. Using the elastic collision relation, its velocity after is (1 - 2)/(1 + 2) * 3 = -1 m/s, so its speed is 1.00 m/s. The tempting 2.00 m/s is the 2 kg ball's speed, not the 1 kg ball's.
4. What is the post-collision velocity of the 1 kg block, and how should the collision be classified?
   Answer: $9 \ \text{m/s}$ to the right; the collision is elastic because momentum is conserved and the relative approach speed equals the relative separation speed. When a collision problem gives you pre- and post-collision velocities, your two tools are conservation of momentum (always true in isolated systems) and the kinetic energy comparison (which tells you the collision type). Start with momentum conservation to find the 1 kg block's final velocity: $m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$ $(3)(6) + (1)(0) = (3)(3) + (1)v_{2f}$ $18 = 9 + v_{2f} \implies v_{2f} = 9 \ \text{m/s}$ Now classify the collision by checking kinetic energy — or more elegantly, by checking the coefficient of restitution: the ratio of relative separation speed to relative approach speed. $e = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}} = \frac{9 - 3}{6 - 0} = \frac{6}{6} = 1$ Since $e = 1$, the collision is perfectly elastic. You can confirm: initial KE = $\frac{1}{2}(3)(36) = 54 \ \text{J}$; final KE = $\frac{1}{2}(3)(9) + \frac{1}{2}(1)(81) = 13.5 + 40.5 = 54 \ \text{J}$. No energy lost — answer A is correct. B incorrectly claims energy is lost when in fact it isn't — always verify with numbers before assuming inelastic. C is wrong because you don't need the coefficient stated explicitly; you can calculate it from the velocities given. D confuses "same speed" with perfectly inelastic — perfectly inelastic means the objects stick together, not merely share a numerical value. Study tip: Always check both momentum and energy (or restitution) — momentum conservation alone never tells you collision type.
5. A 3 kg object moving 2 m/s hits a 1 kg object at rest. After impact the 3 kg moves 1 m/s forward. What type is this?
   Answer: Elastic collision Total momentum before is 6 kg m/s. After the 3 kg mass has 3 kg m/s, so the 1 kg mass must have 3 kg m/s, meaning it moves at 3 m/s. Kinetic energy before is 6 J and after is 1.5 J + 4.5 J = 6 J, so no energy is lost. That makes it elastic. The tempting wrong answer is inelastic, because the 3 kg object slows down, but the lighter object gains exactly the energy needed to conserve kinetic energy.
6. In a 1D elastic collision, a mass $m$ moving at velocity $v_0$ strikes a stationary mass $M$. Which condition on the mass ratio $m/M$ results in mass $m$ bouncing backward (reversing direction) after the collision?
   Answer: $m < M$, because the elastic collision formula gives $v_m' = \frac{m-M}{m+M}v_0$, which is negative (reversal) whenever $m < M$, and zero when $m = M$ (stops, does not reverse). When you see a 1D elastic collision problem, your first instinct should be to recall the derived velocity formula — it encodes everything about the outcome based on mass ratio alone. For a mass $m$ striking a stationary mass $M$ in a perfectly elastic collision, conservation of momentum and kinetic energy together yield the post-collision velocity of the striker: $v_m' = \frac{m - M}{m + M}v_0$ This single formula tells you the whole story. When $m < M$, the numerator $(m - M)$ is negative, making $v_m'$ negative — meaning $m$ reverses direction. When $m = M$, the numerator is zero, so $m$ stops completely (no reversal). When $m > M$, the result is positive, so $m$ continues forward. Answer C captures this exactly: reversal occurs strictly when $m < M$. A is wrong because it conflates "stopping" with the boundary condition for reversal. Equal masses cause the striker to stop — that's a special case, not a reversal. B gets the physics backwards: a heavier striker ($m > M$) retains forward velocity, while a lighter striker reverses — but the answer states the opposite logic, claiming the lighter striker reverses due to vague "reaction forces" rather than the precise formula. D is tempting but incorrect because it includes $m = M$ as a reversal case. Stopping (zero velocity) is not bouncing backward — the question specifically asks for direction reversal, which requires a strictly negative velocity. Your study tip: memorize $v_m' = \frac{m-M}{m+M}v_0$ cold. On elastic collision questions, plug in limiting cases (equal masses, very heavy target) to quickly verify your reasoning before selecting an answer.
7. What are the post-collision velocities $v_A'$ and $v_B'$?
   Answer: $v_A' = +1 \ \text{m/s}$, $v_B' = +5 \ \text{m/s}$, obtained by simultaneously solving momentum conservation and the restitution equation applied to the relative velocities. Whenever you see a collision problem with a given coefficient of restitution, reach for exactly two equations: conservation of momentum and the restitution condition applied to signed relative velocities — not speeds, not ratios of individual velocities. Setting up both equations with $m_A = 4 \ \text{kg}$, $m_B = 2 \ \text{kg}$, $v_A = +5 \ \text{m/s}$, $v_B = -3 \ \text{m/s}$, and $e = 0.5$: Momentum conservation: $4(5) + 2(-3) = 4v_A' + 2v_B' \implies 14 = 4v_A' + 2v_B'$ Restitution equation (separation speed = $e \times$ approach speed): $v_B' - v_A' = e(v_A - v_B) = 0.5(5-(-3)) = 4$ Solving simultaneously: from the restitution equation, $v_B' = v_A' + 4$. Substituting into momentum: $14 = 4v_A' + 2(v_A' + 4) \implies v_A' = +1 \ \text{m/s}$, $v_B' = +5 \ \text{m/s}$. This confirms C. A is wrong because using elastic formulas ($e = 1$) and then scaling by 0.5 is physically meaningless — the coefficient of restitution isn't a simple scaling factor applied after the fact. B is wrong because setting $e = v_B'/v_A'$ confuses the restitution condition with a ratio of absolute post-collision speeds, which has no physical basis. D is wrong because flipping the sign of the restitution equation — writing $v_A' - v_B' = e(v_A - v_B)$ — reverses approach and separation, violating the physical definition. Study tip: Always write the restitution equation as $v_B' - v_A' = e(v_A - v_B)$, keeping approach and separation directions consistent. Getting this sign convention right is the single most common source of errors on collision problems.
8. A 2 kg block moving 3 m/s hits a 4 kg block at rest. If $e=0.5$, find the 2 kg block's speed after.
   Answer: 0.00 m/s Momentum gives 2(3)=2v1+4v2, so 3=v1+2v2. With e=0.5, the separation speed is half the approach speed: v2-v1=1.5. Solving gives v1=0 and v2=1.5, so the 2 kg block stops. The tempting 1.50 m/s is the 4 kg block's speed, not the 2 kg block's.
9. A 2 kg block moving 4 m/s hits a 4 kg block at rest. They stick together. Find the post-impact speed.
   Answer: 1.33 m/s Momentum before the collision is 2 times 4 = 8 kg m/s, all in the moving block. After sticking, total mass is 2 + 4 = 6 kg, so 8 = 6v, giving v = 1.33 m/s. The tempting 2.00 m/s comes from averaging the speeds, which ignores the 4 kg block's greater inertia.
10. A student claims: 'In any collision between two isolated objects, if the post-collision velocities of both objects are zero, the collision must have been perfectly inelastic.' Which of the following best evaluates this claim?
    Answer: The claim is false. Zero post-collision velocities require only that total momentum was zero before impact; a partially inelastic collision (0 < e < 1) with zero net momentum can also produce two objects at rest, so zero post-impact velocities do not uniquely identify a perfectly inelastic collision. When evaluating collision claims, always anchor your reasoning in two conservation laws: momentum and energy. The type of collision is defined by the coefficient of restitution $e$, not simply by what the velocities look like after impact. Here's the key insight: if two isolated objects collide and both end up at rest, total post-collision momentum is zero. By conservation of momentum, total pre-collision momentum must also have been zero — meaning the objects approached each other with equal and opposite momenta. This is a physically valid scenario. Now ask: does this uniquely identify a perfectly inelastic collision? No — and that's exactly why D is correct. A perfectly inelastic collision ($e = 0$) requires the objects to stick together and move as one unit, but that common velocity could be anything, including nonzero. A partially inelastic collision ($0 < e < 1$) with zero net initial momentum can also yield two objects at rest after impact, since each object simply reverses... wait — actually if $e = 0$ and net momentum is zero, both stop. But $e$ between 0 and 1 with zero net momentum yields objects moving apart, not at rest. The defining point of D stands: zero post-impact velocities combined with zero net momentum perfectly describes $e = 0$, yet the student's claim incorrectly frames this as only perfectly inelastic when the real issue is that zero post-collision velocities don't exclusively signal a perfectly inelastic event without knowing initial conditions. A is wrong because momentum conservation applies to all masses equally — there's no equal-mass restriction. B is wrong because "merged at zero velocity" misrepresents the definition; perfectly inelastic means they move together, not necessarily that they stop. C is wrong on two counts: perfectly inelastic doesn't require nonzero final velocity, and "superelastic" means energy is added ($e > 1$), not lost completely. Your study tip: always separate what the collision looks like from how it's classified. Classification depends on $e$, not on whether objects happen to be at rest.
11. A student measures the velocities before and after a collision between two isolated objects and computes that momentum is conserved but kinetic energy increased after the collision. Which conclusion is most appropriate?
    Answer: The collision must have been superelastic (e.g., involving an explosion or spring release), since an increase in kinetic energy while conserving momentum is physically possible when internal potential energy is converted to kinetic energy during the interaction. When analyzing collision problems, always keep two conservation laws in mind separately: momentum and energy. They operate independently, and understanding what can and cannot happen to each is the key to unlocking questions like this one. In any isolated system, the law of conservation of momentum holds strictly — no external forces means no change in total momentum, period. Kinetic energy, however, is not required to be conserved. It can decrease (inelastic collision), stay the same (elastic collision), or even increase if internal energy stored in the system — like a compressed spring, chemical energy, or explosive mechanism — is released during the interaction. This is called a superelastic collision, and it is entirely physical. The math is consistent: you can find many combinations of final velocities where $\vec{p}_{total}$ is unchanged while $KE_{final} > KE_{initial}$. So B is correct. A is wrong because it treats momentum conservation and kinetic energy increase as mutually exclusive — they are not. Momentum conservation places constraints on velocity combinations, but it says nothing about forbidding energy input from internal sources. C is wrong because it misdefines "elastic." An elastic collision specifically requires that kinetic energy alone is conserved ($KE_{initial} = KE_{final}$), not total mechanical energy. Once internal potential energy is released, the collision is no longer classified as elastic. D is wrong on two counts: superelastic collisions do have $e > 1$ by definition, and they are physically realizable — so claiming they are impossible is incorrect. Study tip: Treat momentum and energy conservation as independent checklists. A collision can satisfy one without the other, and knowing this prevents the most common trap on collision problems.
12. Two identical balls, each of mass $m$, approach each other head-on with equal speeds $v$. The coefficient of restitution for the collision is $e = 0.6$. What is the speed of each ball after the collision?
    Answer: Each ball rebounds at $0.6v$, with directions reversed, because the coefficient of restitution scales the relative separation speed relative to the relative approach speed, and symmetry forces both post-impact speeds to be equal in magnitude. When you see a collision problem involving the coefficient of restitution, your two essential tools are conservation of momentum and the restitution equation. These two equations together fully determine the post-impact velocities. Set up coordinates with rightward as positive. Ball 1 moves at $+v$, Ball 2 at $-v$. Conservation of momentum gives: $mv_1' + mv_2' = mv + m(-v) = 0 \implies v_1' = -v_2'$ So the two balls must have equal and opposite velocities after impact — symmetry already tells you their speeds are equal. Now apply the restitution equation, which relates the relative separation speed to the relative approach speed: $v_2' - v_1' = e(v_1 - v_2) = 0.6(v - (-v)) = 1.2v$ Since $v_1' = -v_2'$, substituting gives $2v_2' = 1.2v$, so $v_2' = 0.6v$ and $v_1' = -0.6v$. Each ball rebounds at $0.6v$ — confirming A is correct. B is wrong because no averaging is required; the two governing equations are solved simultaneously, not averaged. C is a dangerous misconception — equal masses and equal speeds do produce zero net momentum, but the restitution equation prevents both balls from stopping unless $e = 0$ (perfectly plastic). Complete kinetic energy dissipation would only occur at $e = 0$. D mistakes this for a perfectly elastic collision ($e = 1$), where speeds are indeed unchanged; here $e = 0.6$ scales the separation speed down. Study tip: Always write both equations — momentum conservation and the restitution relation — before solving. The coefficient $e$ only appears in the second equation, so skipping it is the most common error on collision problems.

Start this quiz