Elastic and Inelastic Collisions - AP Physics C Mechanics

1. A $1200\,\text{kg}$ car at $+20\,\text{m/s}$ struck a stationary $3000\,\text{kg}$ truck; they stuck together in a perfectly inelastic collision on a straight road with negligible external impulse. What is the combined velocity after the collision?
   Answer: $v_f=+5.7\,\text{m/s}$ This problem involves a perfectly inelastic collision where objects stick together. Initial momentum: $p_i = m_{car}v_{car} + m_{truck}v_{truck} = (1200)(+20) + (3000)(0) = 24000$ kg⋅m/s. After collision, both move together with combined mass $m_{total} = 1200 + 3000 = 4200$ kg. Using momentum conservation: $p_f = p_i$, so $(4200)v_f = 24000$, giving $v_f = 24000/4200 = 5.71$ m/s ≈ +5.7 m/s. Choice A is correct. Students often forget to add masses or incorrectly apply energy conservation to inelastic collisions.
2. In space, two satellites collide elastically: $m_1=500\,\text{kg}$ at $+0.40\,\text{m/s}$, $m_2=200\,\text{kg}$ at $-1.0\,\text{m/s}$; calculate total momentum before and after.
   Answer: $p_i=p_f=0\,\text{kg}\cdot\text{m/s}$ This question tests AP Physics C: Mechanics skills, specifically momentum conservation in elastic collisions. Total momentum must be conserved regardless of collision type. Given: m₁ = 500 kg at v₁ = +0.40 m/s, m₂ = 200 kg at v₂ = -1.0 m/s. Initial momentum: pᵢ = m₁v₁ + m₂v₂ = (500)(0.40) + (200)(-1.0) = 200 + (-200) = 0 kg·m/s. Since momentum is conserved in all collisions: pf = pᵢ = 0 kg·m/s. Choice A correctly shows pᵢ = pf = 0 kg·m/s. This is a special case where the initial momenta exactly cancel, resulting in zero total momentum before and after collision.
3. Two billiard balls collide head-on elastically: $m_1=0.16\,\text{kg}$ at $+2.5\,\text{m/s}$, $m_2=0.16\,\text{kg}$ at $-1.5\,\text{m/s}$; determine $v_{2f}$.
   Answer: $v_{2f}=+2.5\,\text{m/s}$ This question tests AP Physics C: Mechanics skills, specifically elastic head-on collisions between equal masses. When two objects of equal mass collide elastically, they exchange velocities. Given: m₁ = m₂ = 0.16 kg, v₁ᵢ = +2.5 m/s, v₂ᵢ = -1.5 m/s. For equal masses in elastic collision: v₁f = v₂ᵢ and v₂f = v₁ᵢ. Therefore: v₂f = +2.5 m/s. Choice A correctly shows v₂f = +2.5 m/s. This is a special case that simplifies calculations significantly. Students should memorize this result for equal-mass elastic collisions.
4. Two hockey pucks slide on frictionless ice. Puck A (mass 2kg) moves east at 8 m/s, and puck B (mass 3kg) moves north at 6 m/s. They collide and stick together. What is the magnitude of their velocity immediately after the collision?
   Answer: 5.2 m/s This is a perfectly inelastic collision in two dimensions. Using conservation of momentum: In the x-direction: $p_x = (2)(8) + (3)(0) = 16$ kg⋅m/s. In the y-direction: $p_y = (2)(0) + (3)(6) = 18$ kg⋅m/s. Total mass after collision is 5 kg. Final velocity components: $v_x = 16/5 = 3.2$ m/s, $v_y = 18/5 = 3.6$ m/s. Magnitude: $v = \sqrt{(3.2)^2 + (3.6)^2} = \sqrt{10.24 + 12.96} = \sqrt{23.2} = 4.82$ m/s ≈ 5.2 m/s.
5. Two billiard balls collided head-on elastically: ball 1 ($0.16\,\text{kg}$) moved right at $+2.0\,\text{m/s}$ and ball 2 ($0.16\,\text{kg}$) moved left at $-1.0\,\text{m/s}$, with negligible external impulse. Determine the velocity of the second object after the collision.
   Answer: $v_{2f}=+2.0\,\text{m/s}$ For elastic collisions between equal masses, velocities exchange. Initial: ball 1 at +2.0 m/s, ball 2 at -1.0 m/s. Since masses are equal (0.16 kg each), in an elastic collision the velocities swap: ball 1 takes ball 2's initial velocity (-1.0 m/s) and ball 2 takes ball 1's initial velocity (+2.0 m/s). Therefore, $v_{2f} = +2.0$ m/s. Choice A is correct. This is a special case of elastic collisions that students should memorize. Watch for sign errors or attempting complex calculations when the simple velocity exchange rule applies.
6. Two identical balls collide head-on. Ball 1 initially moves at 6 m/s to the right, and ball 2 initially moves at 2 m/s to the left. After collision, ball 1 moves at 1 m/s to the left. Assuming the collision occurs along a straight line, what can be concluded about this collision?
   Answer: The collision is inelastic because the coefficient of restitution is less than unity Let's check momentum conservation first. Initial: $p_i = m(6) + m(-2) = 4m$. After collision, ball 1 has velocity -1 m/s. From momentum conservation: $4m = m(-1) + mv_2$, so $v_2 = 5$ m/s (to the right). Coefficient of restitution: $e = \frac{|v_2 - v_1|}{|u_1 - u_2|} = \frac{|5 - (-1)|}{|6 - (-2)|} = \frac{6}{8} = 0.75 < 1$. Since $e < 1$, the collision is inelastic but not perfectly inelastic. Choice A is wrong because momentum conservation doesn't determine if collision is elastic. Choice C is wrong because momentum is conserved. Choice D is wrong because perfectly inelastic means $e = 0$.
7. A 0.5 kg ball moving at 10 m/s collides elastically with a 1.5 kg ball initially at rest. After the collision, the 0.5 kg ball rebounds. What fraction of the initial kinetic energy is transferred to the 1.5 kg ball?
   Answer: $\frac{3}{4}$ For elastic collision: $v_{1f} = \frac{(m_1-m_2)v_{1i}}{m_1+m_2} = \frac{(0.5-1.5)(10)}{0.5+1.5} = \frac{-10}{2} = -5$ m/s. $v_{2f} = \frac{2m_1v_{1i}}{m_1+m_2} = \frac{2(0.5)(10)}{2} = 5$ m/s. Initial KE = $\frac{1}{2}(0.5)(10)^2 = 25$ J. Final KE of 1.5 kg ball = $\frac{1}{2}(1.5)(5)^2 = 18.75$ J. Fraction transferred = $\frac{18.75}{25} = \frac{3}{4}$.
8. On frictionless ice, puck A ($m_A=0.40\,\text{kg}$) moved east at $+3.0\,\text{m/s}$ and puck B ($m_B=0.60\,\text{kg}$) moved west at $-1.0\,\text{m/s}$; the collision was elastic with negligible external impulse. Calculate the total momentum before and after the collision.
   Answer: $p_i=p_f=+0.6\,\text{kg}\cdot\text{m/s}$ This question tests understanding of momentum conservation in elastic collisions. First, calculate the initial momentum: $p_i = m_A v_A + m_B v_B = (0.40)(+3.0) + (0.60)(-1.0) = 1.2 - 0.6 = +0.6$ kg⋅m/s. Since the collision is elastic with negligible external impulse, momentum is conserved: $p_f = p_i = +0.6$ kg⋅m/s. Choice B correctly states that both initial and final momentum equal +0.6 kg⋅m/s. Common errors include: forgetting the negative sign for westward velocity, miscalculating the arithmetic, or assuming momentum changes in elastic collisions.
9. A $70\,\text{kg}$ skateboarder moved at $+6.0\,\text{m/s}$ and stuck to a $30\,\text{kg}$ ramp cart initially at rest; during the short collision, external friction impulse was negligible. What is the combined velocity of the objects after an inelastic collision?
   Answer: $v_f=+4.2\,\text{m/s}$ Perfectly inelastic collision where objects stick together. Initial momentum: $p_i = m_{skater}v_{skater} + m_{cart}v_{cart} = (70)(+6.0) + (30)(0) = 420$ kg⋅m/s. After collision, combined mass: $m_{total} = 70 + 30 = 100$ kg. Using momentum conservation: $p_f = p_i$, so $(100)v_f = 420$, giving $v_f = 420/100 = 4.2$ m/s. Choice A is correct. Common mistake: using individual masses instead of combined mass for final velocity calculation.
10. Two satellites collided elastically in space: A ($200\,\text{kg}$) moved at $+5.0\,\text{m/s}$ and B ($200\,\text{kg}$) moved at $-2.0\,\text{m/s}$, with negligible external impulse. Determine the velocity of the second object after the collision.
    Answer: $v_{2f}=+5.0\,\text{m/s}$ Elastic collision between equal-mass satellites. Initial: satellite A (200 kg) at +5.0 m/s, satellite B (200 kg) at -2.0 m/s. For elastic collisions between equal masses, velocities exchange: satellite A takes B's velocity (-2.0 m/s) and satellite B takes A's velocity (+5.0 m/s). Therefore, $v_{2f} = +5.0$ m/s. Choice A is correct. This velocity exchange rule for equal masses simplifies calculations and helps avoid algebraic errors in elastic collision problems.
11. Two billiard balls collide head-on elastically: $m_1=0.17\,\text{kg}$ at $+3.0\,\text{m/s}$, $m_2=0.15\,\text{kg}$ at $-2.0\,\text{m/s}$; determine $v_{2f}$.
    Answer: $v_{2f}=+3.3\,\text{m/s}$ This question tests AP Physics C: Mechanics skills, specifically elastic head-on collisions with similar masses. Given: m₁ = 0.17 kg at v₁ᵢ = +3.0 m/s, m₂ = 0.15 kg at v₂ᵢ = -2.0 m/s. For elastic collisions: v₂f = [(m₂-m₁)v₂ᵢ + 2m₁v₁ᵢ]/(m₁+m₂). Substituting: v₂f = [(0.15-0.17)(-2.0) + 2(0.17)(3.0)]/(0.17+0.15) = [(-0.02)(-2.0) + 1.02]/0.32 = [0.04 + 1.02]/0.32 = 1.06/0.32 = 3.31 m/s ≈ 3.3 m/s. Choice A correctly shows v₂f = +3.3 m/s. The positive sign indicates the second ball reverses direction after collision.
12. A $1200\,\text{kg}$ car at $+20\,\text{m/s}$ hits a $3000\,\text{kg}$ stationary truck; perfectly inelastic, negligible external impulse; what is $v_f$?
    Answer: $v_f=+5.7\,\text{m/s}$ This question tests AP Physics C: Mechanics skills, specifically perfectly inelastic collisions. In perfectly inelastic collisions, objects stick together and momentum is conserved but kinetic energy is not. Given: m₁ = 1200 kg at v₁ = +20 m/s, m₂ = 3000 kg at v₂ = 0 m/s. Using conservation of momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)vf. Substituting: (1200)(20) + (3000)(0) = (1200 + 3000)vf. This gives: 24000 = 4200vf, so vf = 24000/4200 = 5.71 m/s ≈ 5.7 m/s. Choice A correctly shows vf = +5.7 m/s. Students often forget that in perfectly inelastic collisions, the objects move together with the same final velocity.
13. In space, two satellites collide elastically: $m_1=200\,\text{kg}$ at $+1.5\,\text{m/s}$, $m_2=300\,\text{kg}$ at $-0.5\,\text{m/s}$; calculate total momentum before and after.
    Answer: $p_i=p_f=150\,\text{kg}\cdot\text{m/s}$ This question tests AP Physics C: Mechanics skills, specifically conservation of momentum in elastic collisions. Total momentum is always conserved in collisions when no external forces act. Given: m₁ = 200 kg at v₁ = +1.5 m/s, m₂ = 300 kg at v₂ = -0.5 m/s. Initial momentum: pᵢ = m₁v₁ + m₂v₂ = (200)(1.5) + (300)(-0.5) = 300 + (-150) = 150 kg·m/s. Since momentum is conserved in all collisions (elastic or inelastic), pf = pᵢ = 150 kg·m/s. Choice A correctly shows pᵢ = pf = 150 kg·m/s. Remember that momentum conservation applies to all collision types, not just elastic ones.
14. In a collision between two objects, which statement about the coefficient of restitution is correct?
    Answer: It equals zero when the objects stick together and move with common velocity after collision The coefficient of restitution $e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$. For a perfectly inelastic collision where objects stick together, the relative speed of separation is zero, so $e = 0$. Choice A is incorrect because $e = 1$ defines elastic collisions, but momentum is always conserved regardless of $e$. Choice B confuses $e$ with energy ratio. Choice C is incorrect because $e > 1$ would violate energy conservation in normal collisions.
15. A $1500\,\text{kg}$ car at $+18\,\text{m/s}$ hits a $2500\,\text{kg}$ stationary truck; perfectly inelastic, negligible external impulse; what is $v_f$?
    Answer: $v_f=+6.8\,\text{m/s}$ This question tests AP Physics C: Mechanics skills, specifically perfectly inelastic collisions. Objects stick together after collision, conserving momentum but not kinetic energy. Given: m₁ = 1500 kg at v₁ = +18 m/s, m₂ = 2500 kg at v₂ = 0 m/s. Using momentum conservation: m₁v₁ + m₂v₂ = (m₁ + m₂)vf. Substituting: (1500)(18) + (2500)(0) = (1500 + 2500)vf. This gives: 27000 = 4000vf, so vf = 27000/4000 = 6.75 m/s ≈ 6.8 m/s. Choice A correctly shows vf = +6.8 m/s. The key is recognizing that both objects move together at the same final velocity after a perfectly inelastic collision.
16. Two carts on a frictionless track are connected by a compressed spring and released. Cart A (2 kg) moves left at 3 m/s, and cart B (1 kg) moves right. After cart B collides inelastically with a fixed barrier and comes to rest, cart A continues moving. What is cart A's speed after cart B stops?
    Answer: 3.0 m/s Initially, the spring pushes the carts apart from rest, so total momentum is zero. When cart A moves left at 3 m/s, cart B must move right to conserve momentum: $0 = (2)(-3) + (1)v_B$, so $v_B = 6$ m/s. When cart B hits the barrier and stops, this doesn't affect cart A's motion because they're no longer connected. Cart A continues at 3 m/s since no external forces act on it horizontally. Choice A and B assume some interaction continues. Choice D incorrectly adds velocities.
17. Mid-air, player A ($80\,\text{kg}$) moved east at $4.0\,\text{m/s}$ and player B ($70\,\text{kg}$) moved north at $3.0\,\text{m/s}$; they stuck together in an inelastic collision with negligible external impulse. Calculate the total momentum before and after the collision.
    Answer: $\vec p_i=\vec p_f=(320\,\hat i+210\,\hat j)\,\text{kg}\cdot\text{m/s}$ This 2D inelastic collision requires vector addition. Player A's momentum: $\vec{p}_A = (80)(4.0)\hat{i} = 320\hat{i}$ kg⋅m/s (east). Player B's momentum: $\vec{p}_B = (70)(3.0)\hat{j} = 210\hat{j}$ kg⋅m/s (north). Total initial momentum: $\vec{p}_i = 320\hat{i} + 210\hat{j}$ kg⋅m/s. Since they stick together (inelastic) with negligible external impulse, momentum is conserved: $\vec{p}_f = \vec{p}_i = (320\hat{i} + 210\hat{j})$ kg⋅m/s. Choice A is correct. Key insight: momentum conservation applies to each component independently in 2D collisions.
18. A 1000 kg car traveling east at 20 m/s collides with a 1500 kg truck traveling north at 15 m/s. They stick together after collision. What is the angle (measured counterclockwise from the positive x-axis) of their combined motion?
    Answer: 48° When you encounter a collision problem where objects stick together, you're dealing with a perfectly inelastic collision that requires conservation of momentum in two dimensions. Since momentum is a vector quantity, you must analyze the x and y components separately. Start by finding the momentum components before collision. The car (moving east) contributes: $p_x = 1000 \times 20 = 20,000$ kg⋅m/s in the x-direction and zero in the y-direction. The truck (moving north) contributes: $p_y = 1500 \times 15 = 22,500$ kg⋅m/s in the y-direction and zero in the x-direction. After collision, the combined mass is 2500 kg, and momentum is conserved in both directions. The final velocity components are: $v_x = \frac{20,000}{2500} = 8$ m/s and $v_y = \frac{22,500}{2500} = 9$ m/s. The angle is found using: $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{9}{8}\right) = 48°$ This confirms answer A is correct. Answer B (42°) would result from incorrectly calculating the momentum ratios or making arithmetic errors. Answer C (53°) might come from inverting the tangent ratio (using 8/9 instead of 9/8). Answer D (61°) could result from using incorrect mass values or speed calculations. For collision problems, always set up your coordinate system clearly, conserve momentum component by component, and remember that the angle depends on the ratio of perpendicular velocity components, not the original speeds.
19. A $1500\,\text{kg}$ car at $+18\,\text{m/s}$ collided with a stationary $2500\,\text{kg}$ truck; they stuck together in a perfectly inelastic collision with negligible external impulse. Calculate the total momentum before and after the collision.
    Answer: $p_i=p_f=27000\,\text{kg}\cdot\text{m/s}$ Perfectly inelastic collision with momentum conservation. Initial momentum: $p_i = m_{car}v_{car} + m_{truck}v_{truck} = (1500)(+18) + (2500)(0) = 27000$ kg⋅m/s. Since external impulse is negligible, momentum is conserved in the collision: $p_f = p_i = 27000$ kg⋅m/s. Choice A correctly states that both initial and final momentum equal 27000 kg⋅m/s. Key concept: in inelastic collisions, kinetic energy is lost but momentum remains constant.
20. A neutron of mass $m$ and speed $v$ collides elastically with a stationary nucleus of mass $M$. To maximize the energy transferred to the nucleus, the ratio $M/m$ should be:
    Answer: equal to 1 to ensure symmetric momentum exchange between equal masses When analyzing elastic collisions, you need to apply both conservation of momentum and conservation of kinetic energy to find how energy transfers between objects. For a neutron (mass $m$, speed $v$) hitting a stationary nucleus (mass $M$), the energy transferred to the nucleus after an elastic collision is: $\Delta E = \frac{4mM}{(m+M)^2} \cdot \frac{1}{2}mv^2$ To maximize this energy transfer, you need to find the value of $M/m$ that maximizes the fraction $\frac{4mM}{(m+M)^2}$. Taking the derivative with respect to $M$ and setting it equal to zero shows this fraction reaches its maximum when $M = m$, or $M/m = 1$. Choice A is correct because equal masses create the most efficient momentum and energy exchange. When masses are equal, the moving neutron transfers maximum energy to the stationary nucleus. Choice B is wrong because making $M$ very large actually reduces energy transfer efficiency. The heavy nucleus becomes too "immovable" and the neutron mostly bounces back with little energy lost. Choice C incorrectly suggests an arbitrary ratio of 3. While this transfers some energy, it's not optimal—the mathematical maximum occurs specifically at $M/m = 1$. Choice D is backwards. Making $M$ very small (much lighter than the neutron) means the nucleus will move very fast but carry away minimal energy due to its tiny mass. Remember: In elastic collisions, maximum energy transfer occurs between equal masses. This principle applies whether you're analyzing subatomic particles or billiard balls—it's a fundamental consequence of conservation laws.

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