Conservation of Linear Momentum - AP Physics C Mechanics

1. Two ice skaters initially at rest on frictionless ice push off each other: skater A ($60\,\text{kg}$) and skater B ($40\,\text{kg}$). Afterward, A moves at $\vec v_A=-2.0\,\text{m/s}\,\hat\imath$. With negligible external forces so total momentum remains $\vec 0$, calculate skater B's velocity.
   Answer: $\vec v_B=+3.0\,\text{m/s}\,\hat\imath$ This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, two skaters initially at rest push off each other on frictionless ice, creating an isolated system with zero initial momentum that must remain zero. Choice A correctly applies conservation: initial momentum = 0; final momentum = (60 kg)(-2.0 m/s) + (40 kg)(vB) = -120 + 40vB = 0, giving vB = +3.0 m/s. Choice B incorrectly calculates the velocity ratio without considering the mass difference. To help students: Emphasize that when a system starts at rest, the total momentum must remain zero. Practice problems with different mass ratios to reinforce that lighter objects gain proportionally higher velocities.
2. Two skaters on frictionless ice push off from rest: skater 1 ($75\,\text{kg}$) moves at $\vec v_1=+1.6\,\text{m/s}\,\hat\imath$ after the push. With negligible external forces so $\sum \vec p=\vec 0$, calculate skater 2's velocity if $m_2=50\,\text{kg}$.
   Answer: $\vec v_2=-2.4\,\text{m/s}\,\hat\imath$ This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, two skaters push off from rest on frictionless ice, creating an isolated system where initial momentum is zero and must remain zero. Choice A correctly applies conservation: initial momentum = 0; final momentum = (75 kg)(+1.6 m/s) + (50 kg)(v2) = 120 + 50v2 = 0, giving v2 = -120/50 = -2.4 m/s. Choice C incorrectly calculates the velocity magnitude without proper consideration of the mass ratio. To help students: Emphasize that action-reaction pairs during push-offs create equal and opposite momentum changes. Practice problems with different mass ratios to reinforce the inverse relationship between mass and velocity.
3. A railroad car of mass $M$ moves at constant speed $v_0$ on a horizontal track. Rain falls vertically downward at a rate of $\lambda$ kg/s into the car. Assuming the rainwater accumulates in the car and moves with it, what is the speed of the car as a function of time $t$, considering that the horizontal component of momentum must be conserved?
   Answer: $v(t) = \frac{Mv_0}{M + \lambda t}$ for all $t \geq 0$ The rain falls vertically, so it has zero horizontal momentum. The horizontal momentum of the system (car + accumulated rain) must remain constant at $Mv_0$. At time $t$, the total mass is $M + \lambda t$ (original car mass plus accumulated rainwater). By conservation of horizontal momentum: $(M + \lambda t)v(t) = Mv_0$, which gives $v(t) = \frac{Mv_0}{M + \lambda t}$. Choice A is incorrect as it's only a linear approximation valid for small times. Choice C incorrectly treats this as uniform acceleration. Choice D incorrectly applies energy considerations instead of momentum conservation.
4. A 4.0 kg block sliding at 6.0 m/s on a frictionless surface collides elastically with a 2.0 kg block moving at 3.0 m/s in the opposite direction. After the collision, the 2.0 kg block moves at 7.0 m/s in its original direction. A student calculates that the 4.0 kg block's final velocity should be 1.0 m/s in its original direction. Which statement best describes this result?
   Answer: The calculation violates conservation of kinetic energy; this collision cannot be elastic with the given final velocity of the 2.0 kg block Check both conservation laws. Taking the 4.0 kg block's initial direction as positive: Initial momentum = $(4.0)(6.0) + (2.0)(-3.0) = 24 - 6 = 18\text{ kg}\cdot\text{m/s}$. Final momentum = $(4.0)(1.0) + (2.0)(7.0) = 4 + 14 = 18\text{ kg}\cdot\text{m/s}$. Momentum is conserved. Initial KE = $\frac{1}{2}(4.0)(6.0)^2 + \frac{1}{2}(2.0)(-3.0)^2 = 72 + 9 = 81\text{ J}$. Final KE = $\frac{1}{2}(4.0)(1.0)^2 + \frac{1}{2}(2.0)(7.0)^2 = 2 + 49 = 51\text{ J}$. Kinetic energy is not conserved (81 J ≠ 51 J), so this cannot be an elastic collision. The student's calculation satisfies momentum conservation but violates energy conservation for an elastic collision.
5. A $0.60\,\text{kg}$ firework traveling at $\vec v_i=+5.0\,\text{m/s}\,\hat\imath$ explodes into $0.20\,\text{kg}$ and $0.40\,\text{kg}$ pieces. The $0.40\,\text{kg}$ piece has $\vec v=+2.0\,\text{m/s}\,\hat\imath$. With $\sum \vec p$ conserved, calculate the $0.20\,\text{kg}$ piece's velocity.
   Answer: $\vec v=+11\,\text{m/s}\,\hat\imath$ This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, a firework explodes into two pieces with total momentum conserved throughout the explosion process. Choice A correctly applies conservation: initial momentum = (0.60 kg)(+5.0 m/s) = 3.0 kg·m/s; after explosion, (0.20 kg)(v) + (0.40 kg)(+2.0 m/s) = 3.0, giving 0.20v + 0.80 = 3.0, so v = 2.2/0.20 = +11 m/s. Choice C incorrectly assumes the lighter piece moves at an intermediate velocity. To help students: Emphasize that in explosions, lighter pieces often achieve higher velocities to conserve momentum. Practice problems with varying mass ratios to build intuition about velocity distributions.
6. A $3.0\,\text{kg}$ cart moving at $\vec v_i=+2.0\,\text{m/s}\,\hat\imath$ explodes into two pieces of $1.0\,\text{kg}$ and $2.0\,\text{kg}$. The $1.0\,\text{kg}$ piece leaves at $\vec v_1=+8.0\,\text{m/s}\,\hat\imath$. With external impulse negligible, calculate the $2.0\,\text{kg}$ piece's velocity.
   Answer: $\vec v_2=-1.0\,\text{m/s}\,\hat\imath$ This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, a cart explodes into two pieces with negligible external impulse, making this an internal force problem where total momentum before equals total momentum after. Choice A correctly applies conservation: initial momentum = (3.0 kg)(+2.0 m/s) = 6.0 kg·m/s; after explosion, (1.0 kg)(+8.0 m/s) + (2.0 kg)(v2) = 6.0, giving 8.0 + 2.0v2 = 6.0, so v2 = -2.0/2.0 = -1.0 m/s. Choice B incorrectly assumes both pieces must move in the same direction. To help students: Emphasize that in explosions, pieces can move in opposite directions while conserving total momentum. Practice setting up equations systematically and checking that calculated velocities make physical sense.
7. A firework of mass $3.0\,\text{kg}$ moves horizontally at $+4.0\,\text{m/s}$ (no air resistance). It explodes into two pieces: piece 1 has mass $1.0\,\text{kg}$ and piece 2 has mass $2.0\,\text{kg}$. Immediately after, piece 2 moves at $+1.0\,\text{m/s}$. With external impulse negligible, momentum is conserved: $\vec p_i=(3.0)(+4.0)\hat\imath$ and $\vec p_f=(1.0\,\vec v_1+2.0\cdot +1.0\hat\imath)$. Calculate the velocity of piece 1 after the event given the initial conditions.
   Answer: $\vec v_1=+10\,\hat\imath\,\text{m/s}$ This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, a firework explodes into two pieces with no air resistance, making it an isolated system where momentum conservation applies. Choice A correctly calculates piece 1's velocity: initial momentum = (3.0 kg)(+4.0 m/s) = 12.0 kg·m/s, final momentum = (1.0 kg)(v1) + (2.0 kg)(+1.0 m/s), so 12.0 = v1 + 2.0, giving v1 = +10.0 m/s. Choice B incorrectly adds rather than conserves momentum. To help students: Emphasize that explosions are internal forces that don't violate momentum conservation. Practice problems with fragmenting objects helps students recognize that pieces can have different velocities while total momentum remains constant.
8. Two ice skaters initially at rest on level ice (friction negligible) push off each other. Skater 1 has mass $50\,\text{kg}$ and skater 2 has mass $75\,\text{kg}$. After the push, skater 1 moves at $+3.0\,\text{m/s}$ along $+\hat\imath$. The system is isolated horizontally, so $\vec p_i=\vec 0$ and $\vec p_f=50(+3.0)\hat\imath+75\,\vec v_2$. Calculate the velocity of skater 2 after the event given the initial conditions.
   Answer: $\vec v_2=-2.0\,\hat\imath\,\text{m/s}$ This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, two ice skaters initially at rest push off each other on frictionless ice, creating an isolated system with zero initial momentum. Choice A correctly calculates skater 2's velocity: initial momentum = 0, final momentum = (50 kg)(+3.0 m/s) + (75 kg)(v2) = 0, so 150 + 75v2 = 0, giving v2 = -150/75 = -2.0 m/s. Choice B has the wrong sign, failing to recognize that the skaters must move in opposite directions. To help students: Emphasize that when starting from rest, objects must move in opposite directions to conserve zero momentum. Practice problems with different mass ratios helps students see the inverse relationship between mass and velocity.

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