This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is force applied across a time period, expressed as J = F Δt. It corresponds exactly to Δp, the vector change in momentum. For the cart, 9 N left for 0.20 s results in Δp of 1.8 kg·m/s left. Choice D omits time, stating just the force. Calculate impulse first and set it equal to Δp for consistent results.

This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse occurs when a force acts over time, quantified as $J = F \Delta t$. It directly corresponds to the momentum change, $\Delta p = J$, including the vector direction. The $5 , \text{N}$ right for $0.40 , \text{s}$ results in $\Delta p$ of $2.0 , \text{N!\cdot!s}$ right for the puck. Choice C might come from dividing force by time instead of multiplying, a calculation error. Consistently use $J = F \Delta t = \Delta p$ to verify answers in impulse problems.

This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is calculated as the constant force times the duration, $J = F \Delta t$. This equals the change in momentum, $\Delta p$, which is a vector pointing in the force's direction. The 3.0 N right for 0.60 s gives $\Delta p$ of 1.8 kg·m/s right, independent of initial motion. Choice D incorrectly uses force without time, missing the impulse. Apply $J = F \Delta t = \Delta p$ routinely for any force-time scenario.

This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is the integral of force over time, but for constant force, it's simply $J = F \Delta t$. It equates to the momentum shift, $\Delta p = J$, with direction opposite to the ball's initial motion here. The 12 N left for 0.10 s causes $\Delta p$ of $1.2 , \text{N·s}$ left. Choice B lists only the force, forgetting the time multiplication. Use the formula $J = F \Delta t = \Delta p$ as a key step in all impulse-related questions.

This question assesses change in momentum from impulse in AP Physics 1. Impulse is force over time, providing the net effect that alters momentum. The impulse-momentum theorem links them directly: Δp⃗ = F⃗ Δt. The leftward -12 N force for 0.25 s yields Δp⃗ of -3.0 kg·m/s to the left, focusing only on the force applied. Choice C is a distractor, maybe from using initial momentum without the sign. Always isolate impulse calculation from initial conditions for accurate Δp⃗ determination.

This problem evaluates finding force from given impulse in AP Physics 1. Impulse equals force multiplied by time for constant forces, embodying the cumulative impact. It corresponds to change in momentum, so J⃗ = F⃗ Δt = Δp⃗. With J = -1.8 N·s over 0.30 s, F = -6.0 N to the left. Choice D is a distractor, perhaps from confusing impulse with momentum units. Rearrange the impulse formula to solve for unknowns like force in such scenarios.

This question examines the relationship between impulse and change in momentum in AP Physics 1. Impulse is force applied over time, serving as a vector quantity that changes an object's momentum. The theorem equates impulse directly to Δp⃗, meaning the change matches the impulse's magnitude and direction. Thus, a rightward impulse of 3.0 N·s causes Δp⃗ of 3.0 kg·m/s to the right, irrespective of initial motion. Choice A is a distractor as it wrongly assigns a leftward direction, perhaps confusing with initial velocity. Consistently use Δp⃗ = J⃗ to handle direction correctly in momentum problems.

This question asks for the magnitude of momentum change given force and time. Using the impulse-momentum theorem, the magnitude of impulse is |J| = |F|·Δt = (8 N)(0.75 s) = 6 N·s. Since impulse equals change in momentum, |Δp| = 6 kg·m/s. The problem states the force acts in the direction of motion, so both impulse and momentum change are positive. Choice A incorrectly divides force by time, while choice D gives just the time value. When calculating momentum change from constant force, multiply force magnitude by time duration to get impulse magnitude.

This problem directly states that an impulse is delivered and asks for change in momentum. By the impulse-momentum theorem, change in momentum equals impulse: Δp = J. Since the impulse is 2.4 N·s to the left (opposite the cart's rightward motion), Δp = -2.4 kg·m/s (leftward). The negative sign indicates leftward direction in our coordinate system. Choice A incorrectly uses positive sign for rightward, while choice D gives units of force instead of momentum. Remember that impulse and change in momentum are always equal—they're the same physical quantity expressed in equivalent units.

This problem tests understanding of change in momentum and impulse. Impulse is the product of force and time: J = F·Δt = (-3.0 N)(0.40 s) = -1.2 N·s, where negative indicates leftward direction. The impulse-momentum theorem states that impulse equals change in momentum: J = Δp. Therefore, Δp = -1.2 N·s (to the left). Choice C incorrectly multiplies force by initial velocity instead of time. When calculating impulse, always multiply force by the time duration, not by velocity or mass.