Kinetic energy is $$K = \frac{1}{2}mv^2$$ and momentum is $$p = mv$$. We can express velocity as $$v = p/m$$. Substituting this into the kinetic energy equation gives $$K = \frac{1}{2}m(\frac{p}{m})^2 = \frac{1}{2}m\frac{p^2}{m^2} = \frac{p^2}{2m}$$.

We are given that $$K_A = K_B$$. Therefore, $$\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2$$. Substituting $$m_A = 4m_B$$ gives $$(4m_B) v_A^2 = m_B v_B^2$$. The $$m_B$$ terms cancel, leaving $$4v_A^2 = v_B^2$$. Taking the square root of both sides gives $$2v_A = v_B$$, so the ratio $$v_A/v_B = 1/2$$.

The total kinetic energy is the scalar sum of the individual kinetic energies. $$K_A = \frac{1}{2}m_A v_A^2 = \frac{1}{2}(1000 \text{ kg})(20 \text{ m/s})^2 = 200,000 \text{ J}$$. $$K_B = \frac{1}{2}m_B v_B^2 = \frac{1}{2}(1500 \text{ kg})(10 \text{ m/s})^2 = 75,000 \text{ J}$$. The total kinetic energy is $$K_{total} = K_A + K_B = 200,000 \text{ J} + 75,000 \text{ J} = 275,000 \text{ J}$$.

Translational kinetic energy ($$K = \frac{1}{2}mv^2$$) is a scalar quantity that depends on the magnitude of the velocity (speed), not its direction. Since the speed v remains the same, the kinetic energy is unchanged. Therefore, the change in kinetic energy is zero.

To find the kinetic energy of Car B from Car A's frame of reference, we need the relative velocity. Let east be the positive direction. $$v_A = +20 \text{ m/s}$$ and $$v_B = -10 \text{ m/s}$$. The velocity of B relative to A is $$v_{B/A} = v_B - v_A = -10 \text{ m/s} - 20 \text{ m/s} = -30 \text{ m/s}$$. The relative speed is 30 m/s. The kinetic energy of B in A's frame is $$K_{B/A} = \frac{1}{2}m_B v_{B/A}^2 = \frac{1}{2}(1500 \text{ kg})(30 \text{ m/s})^2 = 675,000 \text{ J}$$.

Using the kinetic energy formula $$K = \frac{1}{2}mv^2$$, we can solve for v: $$v = \sqrt{\frac{2K}{m}}$$. Substituting the values gives $$v = \sqrt{\frac{2(100 \text{ J})}{2.0 \text{ kg}}} = \sqrt{100} = 10 \text{ m/s}$$.

This question tests AP Physics C: Mechanics skills, specifically understanding and calculating translational kinetic energy from power and time relationships. Power is the rate of energy transfer, so the total energy delivered equals power multiplied by time: E = P×t. In this scenario, the engine delivers 80 kW (80,000 W) for 5.0 seconds, and neglecting losses means all this energy converts to kinetic energy: KE = (80,000 W)(5.0 s) = 400,000 J = 4.0×10⁵ J. Choice A is correct because it properly calculates the total energy delivered as kinetic energy gain. Choice B might incorrectly divide by time again, while choices C and D show calculation errors. To help students: Emphasize the relationship between power, energy, and time (P = E/t). Practice problems involving energy conversions and power calculations to reinforce unit consistency and the meaning of 'neglecting losses'.

This question tests AP Physics C: Mechanics skills, specifically understanding and calculating translational kinetic energy using the work-energy theorem. The work-energy theorem states that the net work done on an object equals its change in kinetic energy: W_net = ΔKE = KE_f - KE_i. In this scenario, the skateboarder starts from rest (KE_i = 0) and gravity does 980 J of net work, so the final kinetic energy equals the work done: KE_f = 980 J. Choice A is correct because it directly applies the work-energy theorem, recognizing that all the work done converts to kinetic energy. Choice C might incorrectly assume the work is split between kinetic and potential energy, while choice B might confuse mass with energy. To help students: Emphasize that net work equals change in kinetic energy regardless of the object's mass. Practice identifying when to use work-energy theorem versus conservation of energy approaches.

This question tests AP Physics C: Mechanics skills, specifically understanding and calculating translational kinetic energy. Kinetic energy (KE) is the energy an object possesses due to its motion, calculated using KE = 1/2 mv², where m is mass and v is velocity. In this scenario, a 0.16 kg billiard ball moves at 4.0 m/s just after a collision, requiring calculation of its kinetic energy immediately after impact. Choice B is correct because it correctly applies the kinetic energy formula: KE = 1/2 × 0.16 kg × (4.0 m/s)² = 1/2 × 0.16 × 16 = 1.28 J. Choice A is incorrect because it doubles the correct answer, indicating the student likely forgot the 1/2 factor in the formula. To help students: Use collision problems to reinforce energy calculations before and after impacts. Practice with small masses and velocities to build confidence with decimal calculations and proper unit tracking.

This question tests AP Physics C: Mechanics skills, specifically calculating changes in translational kinetic energy. The change in kinetic energy is calculated as ΔKE = KEf - KEi = 1/2 m(vf² - vi²), where m is mass and v represents velocities. In this scenario, a 60 kg skateboarder increases speed from 5.0 m/s to 11 m/s, requiring calculation of the kinetic energy change. Choice A is correct because ΔKE = 1/2 × 60 kg × [(11 m/s)² - (5.0 m/s)²] = 1/2 × 60 × (121 - 25) = 1/2 × 60 × 96 = 2,880 J = 2.88×10³ J. Choice C is incorrect as it likely calculated the final kinetic energy only, forgetting to subtract the initial kinetic energy. To help students: Emphasize the importance of calculating the difference between final and initial states. Practice problems with both increases and decreases in speed to reinforce proper subtraction order.