The impulse is the change in momentum, $$\Delta \vec{p} = \vec{J} = \vec{F} \Delta t$$. $$\Delta \vec{p} = ((6.0\hat{i} - 8.0\hat{j}) \text{ N})(2.0 \text{ s}) = (12\hat{i} - 16\hat{j})$$ kg⋅m/s. The magnitude is $$|\Delta \vec{p}| = \sqrt{12^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20$$ kg⋅m/s.

Impulse is defined as the integral of force over time, $$J = \int F dt$$. Since the applied force $$F$$ and the time interval $$T$$ are the same in both cases, the impulse delivered to the second object is identical to the first. Mass does not appear in the definition of impulse in terms of force and time.

The change in momentum (impulse) is $$\Delta p = J = F t$$. If $$t$$ is doubled, $$\Delta p' = F(2t) = 2(Ft) = 2 \Delta p$$. Kinetic energy is $$K = p^2/(2m)$$. Since the initial kinetic energy is zero, $$\Delta K = K_f$$. As the momentum doubles, the kinetic energy becomes $$K' = (p')^2/(2m) = (2p)^2/(2m) = 4(p^2/(2m)) = 4K$$. Thus, $$\Delta K' = 4 \Delta K$$.

Impulse is equal to the change in momentum, $$\vec{J} = \Delta \vec{p} = m(\vec{v}_f - \vec{v}_i)$$. Let the initial direction be positive. Then $$v_i = 20$$ m/s and $$v_f = -12$$ m/s. The change in momentum is $$0.5 \text{ kg} \times(-12 \text{ m/s} - 20 \text{ m/s}) = 0.5 \text{ kg} \times(-32 \text{ m/s}) = -16$$ kg⋅m/s. The magnitude of the impulse is $$16$$ kg⋅m/s.

This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on momentum change from constant thrust. The change in momentum equals impulse, which is force times time: Δp = FΔt when force is constant. In this scenario, the rocket experiences 9000 N thrust for 6.0 s, giving Δp = 9000 × 6.0 = 54,000 kg·m/s = 5.4×10⁴ kg·m/s. Choice A (5.4×10⁴ kg·m/s) is correct because it accurately applies the impulse-momentum theorem using J = FΔt for constant force. Choice D (5.4×10³ kg·m/s) is incorrect due to an order of magnitude error, possibly from miscalculating the scientific notation. When teaching rocket problems, emphasize that thrust is a force and that momentum change depends only on impulse, not on the object's mass when calculating from force and time. Practice with scientific notation helps avoid common calculation errors.

This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating change in momentum during a collision that brings a car to rest. Change in momentum equals mass times the change in velocity, Δp = m(vf - vi), where careful attention to direction is essential. In this scenario, the 1500 kg car moving east at 12 m/s comes to rest (vf = 0), so Δp = 1500(0 - 12) = -18,000 kg·m/s = $-1.8×10^4$ kg·m/s. Choice B is correct because it shows Δp = $-1.8×10^4$ kg·m/s, with the negative sign indicating the westward direction of the momentum change (opposite to initial motion). Choice A incorrectly shows positive change, while choices C and D show incorrect magnitudes. When teaching momentum change, emphasize that coming to rest means final velocity is zero, and the change in momentum opposes the initial direction of motion. Students should practice problems involving objects brought to rest to reinforce sign conventions.

This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating change in momentum from constant thrust force. The impulse-momentum theorem states that impulse (J = FΔt) equals the change in momentum (Δp), making them interchangeable when one is known. In this scenario, the rocket experiences 2500 N thrust for 4.0 s, resulting in impulse J = (2500)(4.0) = 10,000 N·s = $1.0×10^4$ N·s, which equals the change in momentum. Choice A is correct because Δp = $1.0×10^4$ kg·m/s, properly expressing the change in momentum with correct units (kg·m/s, not N). Choice D incorrectly uses force units (N) instead of momentum units, a common student error. When teaching this concept, emphasize that while impulse has units of N·s and momentum has units of kg·m/s, they are dimensionally equivalent and numerically equal. Students should practice unit analysis to avoid confusion between force and momentum.

This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating change in momentum from average force. The impulse-momentum theorem states that Δp = J = FavgΔt, where the average net force over a time interval determines the momentum change. In this scenario, the sprinter experiences Favg = 420 N for Δt = 0.80 s, resulting in Δp = (420)(0.80) = 336 kg·m/s, which rounds to $3.4×10^2$ kg·m/s. Choice A is correct because it shows Δp = $3.4×10^2$ kg·m/s with proper units for momentum. Choice C incorrectly uses force units (N) instead of momentum units, while choices B and D show incorrect calculations. When teaching impulse from average force, emphasize that the sprinter's mass (70 kg) is not needed for this calculation—only force and time determine impulse. Students should practice distinguishing between problems requiring F = ma versus J = FΔt approaches.

This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating change in momentum from applied force. The impulse-momentum theorem states that impulse (J = FΔt) equals the change in momentum (Δp), where force and momentum are vector quantities. In this scenario, the cart experiences a leftward force of 10 N for 0.50 s, creating an impulse of J = (-10)(0.50) = -5.0 N·s, which equals the change in momentum. Choice B is correct because Δp = -5.0 kg·m/s, with the negative sign indicating the leftward direction opposing the initial rightward motion. Choice A incorrectly shows positive change, while choices C and D show incorrect magnitudes. When teaching this concept, emphasize that the change in momentum depends only on the impulse (force × time), not on the object's initial velocity. Students should practice identifying force directions and applying consistent sign conventions throughout their calculations.

This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on finding final velocity after applying a constant force. The impulse-momentum theorem combined with kinematics allows us to find final velocity: J = FΔt = m(vf - vi), so vf = vi + (FΔt)/m. In this scenario, starting from rest (vi = 0), with F = 18 N, Δt = 0.50 s, and m = 3.0 kg, we get vf = 0 + (18 × 0.50)/3.0 = 9.0/3.0 = 3.0 m/s. Choice B is correct because it shows vf = 3.0 m/s, resulting from proper application of the impulse-momentum theorem. Choice C incorrectly calculates 18/2 = 9.0 m/s, while choice A shows half the correct value. When teaching this concept, emphasize the connection between Newton's second law and the impulse-momentum theorem. Students should practice problems starting from rest to build confidence before tackling more complex scenarios with non-zero initial velocities.