Change in Momentum and Impulse - AP Physics C Mechanics

An object's momentum along a straight line is plotted as a function of time. The graph is a straight line passing through the points $(t=1 \text{ s}, p=5 \text{ kg⋅m/s})$ and $(t=4 \text{ s}, p=14 \text{ kg⋅m/s})$ . What is the constant net force acting on the object?
Answer: $3$ N The net force is the rate of change of momentum, $F_{net} = \frac{dp}{dt}$ . For a linear momentum-time graph, this is the slope of the line. $F_{net} = \frac{\Delta p}{\Delta t} = \frac{14 \text{ kg⋅m/s} - 5 \text{ kg⋅m/s}}{4 \text{ s} - 1 \text{ s}} = \frac{9 \text{ kg⋅m/s}}{3 \text{ s}} = 3$ N.
A force $F(t) = 20\cos(\pi t)$ N acts on a 2.0 kg object for 1.0 second, starting from t = 0. If the object is initially at rest, what is its final momentum?
Answer: 0 kg⋅m/s The impulse equals $\int_0^1 F(t) dt = \int_0^1 20\cos(\pi t) dt = 20[\frac{\sin(\pi t)}{\pi}]_0^1 = \frac{20}{\pi}[\sin(\pi) - \sin(0)] = \frac{20}{\pi}[0 - 0] = 0$ . Since the object starts at rest, its final momentum equals the impulse, which is 0 kg⋅m/s. Choice A assumes constant maximum force. Choice B uses average force incorrectly. Choice D uses half the maximum force.
A $3.0\ \text{kg}$ object starts from rest and experiences a constant net force of $18\ \text{N}$ to the right for $0.50\ \text{s}$ . Using the given data, what is the final velocity of the object after the collision?
Answer: $v_f = 3.0\ \text{m/s}$ 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.
A particle of mass $m$ has a momentum vector given by the function $\vec{p}(t) = (at^2 - b)\hat{i} + (ct)\hat{j}$ , where $a, b,$ and $c$ are positive constants. What is the net force vector $\vec{F}$ acting on the particle as a function of time?
Answer: $\vec{F}(t) = (2at)\hat{i} + c\hat{j}$ The net force is the time derivative of the momentum vector, $\vec{F} = \frac{d\vec{p}}{dt}$ . Taking the derivative of each component with respect to time gives $\frac{d}{dt}(at^2 - b) = 2at$ and $\frac{d}{dt}(ct) = c$ . Therefore, the force vector is $\vec{F}(t) = (2at)\hat{i} + c\hat{j}$ .
What is the magnitude of the average force exerted by the stream of water on the wall?
Answer: $\lambda v$ The force on the wall is equal in magnitude to the rate of change of momentum of the water. In a time interval $\Delta t$ , a mass of water $\Delta m = \lambda \Delta t$ hits the wall. Its momentum changes by $\Delta p = 0 - (\Delta m)v = -\lambda v \Delta t$ . The force exerted on the water is $F_{on\ water} = \frac{\Delta p}{\Delta t} = -\lambda v$ . By Newton's third law, the force on the wall is $F_{on\ wall} = -F_{on\ water} = \lambda v$ .
A rocket of initial mass 1000 kg ejects exhaust at a constant rate of 5.0 kg/s with exhaust velocity 2000 m/s relative to the rocket. What is the magnitude of the thrust force on the rocket?
Answer: 10,000 N When you encounter rocket propulsion problems, you're dealing with Newton's third law and the concept of thrust force. The key insight is that thrust equals the rate at which momentum is being expelled from the rocket. The thrust force on a rocket is calculated using the formula: $F_{thrust} = \frac{dm}{dt} \times v_{exhaust}$ , where $\frac{dm}{dt}$ is the mass flow rate of exhaust and $v_{exhaust}$ is the exhaust velocity relative to the rocket. Given: mass flow rate = 5.0 kg/s and exhaust velocity = 2000 m/s $F_{thrust} = 5.0 \text{ kg/s} \times 2000 \text{ m/s} = 10,000 \text{ N}$ This confirms answer D is correct. Let's examine why the other choices are wrong: Choice A (15,000 N) might result from incorrectly including the rocket's initial mass in the calculation somehow. Choice B (5,000 N) could come from mistakenly using only the mass flow rate and multiplying by 1000, confusing units. Choice C (2,000 N) would result from using only the exhaust velocity value without proper multiplication by the mass flow rate. Remember this key pattern for AP Physics C: rocket thrust problems always involve the product of mass flow rate and exhaust velocity. Don't get distracted by the rocket's total mass—that's often given as a red herring. Focus on how fast mass is leaving (kg/s) and how fast it's moving when it leaves (m/s).
A variable force acts on a 3.0 kg object initially at rest. The force varies linearly with time according to F(t) = 12t N for the first 4.0 seconds, then drops to zero. What is the final momentum of the object?
Answer: 96 kg⋅m/s The impulse equals the area under the force-time graph. For F(t) = 12t from t = 0 to t = 4 s, this forms a triangle with base 4 s and height F(4) = 48 N. Area = ½(4)(48) = 96 N⋅s. Since the object starts at rest, the impulse equals the final momentum: 96 kg⋅m/s. Choice B uses only half the triangular area. Choice C incorrectly multiplies by the mass. Choice D uses the average force times time incorrectly.
A rocket experiences a constant thrust of $2500\ \text{N}$ for $4.0\ \text{s}$ while moving in a straight line. Using the given data, determine the change in momentum of the rocket.
Answer: $\Delta p = 1.0\times10^4\ \text{kg}\cdot\text{m/s}$ 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.
The impulse-momentum theorem is a direct consequence of which fundamental law of physics?
Answer: Newton's second law of motion Newton's second law states that the net force on an object is equal to the rate of change of its momentum ( $\vec{F}_{net} = \frac{d\vec{p}}{dt}$ ). Integrating this equation over a time interval $\Delta t$ yields $\int \vec{F}_{net} dt = \int d\vec{p}$ , which simplifies to $\vec{J} = \Delta \vec{p}$ . This is the impulse-momentum theorem.
A 2.0 kg object moving at 8.0 m/s collides with a wall and rebounds elastically. If the collision lasts 0.15 s, what is the magnitude of the average force exerted by the wall on the object?
Answer: 107 N For an elastic collision with a wall, the object's velocity changes from +8.0 m/s to -8.0 m/s. The change in momentum is Δp = m(vf - vi) = 2.0 kg(-8.0 - 8.0) m/s = -32 kg⋅m/s. The impulse magnitude is |Δp| = 32 N⋅s. Using F = |Δp|/Δt = 32 N⋅s / 0.15 s = 107 N. Choice B uses only the final momentum magnitude. Choice C uses the wrong time calculation. Choice D adds initial and final momentum magnitudes incorrectly.
What is the magnitude of the impulse delivered to the piece of mass $2m$ during the explosion?
Answer: $mv$ By conservation of momentum, the total momentum remains zero. The momentum of the mass $m$ piece is $mv$ . Therefore, the momentum of the mass $2m$ piece must be $-mv$ . Since the piece of mass $2m$ started from rest (momentum=0), its change in momentum (impulse) is equal to its final momentum, $-mv$ . The magnitude of this impulse is $mv$ .
What is the magnitude of the change in the particle's momentum?
Answer: $20$ kg⋅m/s 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.
A force acting on an object is graphed as a function of time. The graph is a horizontal line at $F = 10$ N from $t=0$ to $t=2$ s, and then a straight line from $(2 \text{ s}, 10 \text{ N})$ to $(4 \text{ s}, 0 \text{ N})$ . What is the total change in momentum of the object from $t=0$ to $t=4$ s?
Answer: $30$ kg⋅m/s The total change in momentum is the total impulse, which is the area under the force-time graph. The area is composed of a rectangle and a triangle. Area of rectangle (0 to 2 s) = $\$10 \text{ N} \times 2 \text{ s} = 20$ N⋅s. Area of triangle (2 to 4 s) = $\frac{1}{2} \times (4-2) \text{ s} \times 10 \text{ N} = 10$ N⋅s. Total impulse = $\$20 + 10 = 30$ N⋅s, which equals a change in momentum of $30$ kg⋅m/s.
A $1500\,\text{kg}$ rocket in deep space starts at rest and fires its engine producing a constant $9000\,\text{N}$ thrust for $6.0\,\text{s}$ . External forces are negligible. Using the given data, determine the change in momentum of the system.
Answer: $\Delta p=5.4\times10^4\,\text{kg}\cdot\text{m/s}$ 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.
A $2.0\ \text{kg}$ cart moves right at $6.0\ \text{m/s}$ and experiences a constant leftward force of $10\ \text{N}$ for $0.50\ \text{s}$ . Using the given data, determine the change in momentum of the cart.
Answer: $\Delta p = -5.0\ \text{kg}\cdot\text{m/s}$ 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.
A $2.0$ kg object's motion is governed by the momentum function $p(t) = 10 + 4t^2$ for $t \ge 0$ , where $p$ is in kg⋅m/s and $t$ is in seconds. What is the net force on the object at $t=3.0$ s?
Answer: $24$ N The net force is the time derivative of momentum: $F(t) = \frac{dp}{dt} = \frac{d}{dt}(10 + 4t^2) = 8t$ . At $t=3.0$ s, the force is $F(3) = 8(3.0) = 24$ N.
A $0.15$ kg baseball is pitched horizontally at $40$ m/s. The batter hits it, and the ball leaves the bat horizontally in the opposite direction at $60$ m/s. If the bat and ball are in contact for $2.0$ ms, what is the magnitude of the average force the bat exerts on the ball?
Answer: $7500$ N The change in momentum is $\Delta p = m(v_f - v_i)$ . Choosing the pitched direction as negative, $v_i = -40$ m/s and $v_f = 60$ m/s. So, $\Delta p = 0.15(60 - (-40)) = 0.15(100) = 15$ kg⋅m/s. The average force is $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{15 \text{ kg⋅m/s}}{2.0 \times 10^{-3} \text{ s}} = 7500$ N.
Two objects, a rubber ball and a clay ball of identical mass, are thrown with the same speed toward a brick wall. The rubber ball bounces back with nearly the same speed, while the clay ball sticks to the wall. Which statement correctly compares the impulse delivered by the wall to each ball?
Answer: The rubber ball experiences a greater impulse because its change in momentum is larger. Impulse equals the change in momentum ( $\Delta p$ ). The clay ball's momentum changes from $mv$ to 0, so $|\Delta p_{clay}| = mv$ . The rubber ball's momentum changes from $mv$ to approximately $-mv$ , so $|\Delta p_{rubber}| \approx 2mv$ . Since the rubber ball has a larger change in momentum, it experiences a greater impulse.
A baseball of mass 0.15 kg is pitched horizontally at 25 m/s and is hit by a bat. After being hit, the ball travels at 30 m/s at an angle of 37° above the horizontal. What is the magnitude of the impulse delivered by the bat to the ball?
Answer: 6.8 N⋅s Initial momentum: $\vec{p_i} = (0.15)(25, 0) = (3.75, 0)$ kg⋅m/s. Final momentum: $\vec{p_f} = (0.15)(30\cos 37°, 30\sin 37°) = (0.15)(24, 18) = (3.6, 2.7)$ kg⋅m/s. Change in momentum: $\Delta\vec{p} = (3.6-3.75, 2.7-0) = (-0.15, 2.7)$ kg⋅m/s. Magnitude: $|\Delta\vec{p}| = \sqrt{(-0.15)^2 + (2.7)^2} = \sqrt{0.0225 + 7.29} = 6.8$ N⋅s. Choice A uses incorrect trigonometry. Choice C only considers vertical component. Choice D adds momentum magnitudes instead of vector subtraction.
A $0.50\ \text{kg}$ ball falls straight down at $-8.0\ \text{m/s}$ and rebounds at $+6.0\ \text{m/s}$ after floor contact lasting $0.040\ \text{s}$ . Based on the problem above, calculate the impulse experienced by the ball.
Answer: Impulse = $+7.0\ \text{N}\cdot\text{s}$ This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating impulse during a ball-floor collision. Impulse equals the change in momentum, where J = m(vf - vi), and careful attention to velocity signs is crucial for collisions involving direction reversal. In this scenario, the 0.50 kg ball changes velocity from -8.0 m/s (downward) to +6.0 m/s (upward), resulting in Δp = 0.50(6.0 - (-8.0)) = 0.50(14) = 7.0 kg·m/s. Choice A is correct because it shows the impulse as +7.0 N·s, with the positive sign indicating the upward direction of the impulse from the floor. Choice C incorrectly shows negative impulse, failing to recognize that the floor pushes upward on the ball. When teaching collision problems, emphasize establishing clear sign conventions (e.g., up as positive) and recognizing that impulse direction matches the change in momentum direction. Students should practice problems with rebounds to master sign handling.

Start this quiz