AP Physics C Mechanics Quiz: Change In Momentum And Impulse
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Change In Momentum And ImpulseQuestion 1 of 20

A constant force F⃗=(6.0 N)i^−(8.0 N)j^\vec{F} = (6.0 \text{ N})\hat{i} - (8.0 \text{ N})\hat{j} acts on a particle for 2.02.0 s.

What is the magnitude of the change in the particle's momentum?

1010 kg⋅m/s
1414 kg⋅m/s
2020 kg⋅m/s
2828 kg⋅m/s
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Change In Momentum And Impulse

Practice Change In Momentum And Impulse in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Change In Momentum And Impulse, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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Question 1

A constant force F⃗=(6.0 N)i^−(8.0 N)j^\vec{F} = (6.0 \text{ N})\hat{i} - (8.0 \text{ N})\hat{j} acts on a particle for 2.02.0 s.

What is the magnitude of the change in the particle's momentum?

  1. 1010 kg⋅m/s
  2. 1414 kg⋅m/s
  3. 2020 kg⋅m/s (correct answer)
  4. 2828 kg⋅m/s
Explanation: The impulse is the change in momentum, Δp⃗=J⃗=F⃗Δt\Delta \vec{p} = \vec{J} = \vec{F} \Delta t. Δp⃗=((6.0i^−8.0j^) N)(2.0 s)=(12i^−16j^)\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 ∣Δp⃗∣=122+(−16)2=144+256=400=20|\Delta \vec{p}| = \sqrt{12^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 kg⋅m/s.

Question 2

An object of mass mm is at rest. A force is applied to the object for a time interval TT. The impulse of the force is JJ.

If the same force is applied to an object of mass 2m2m, also initially at rest, for the same time interval TT, what is the impulse of the force on the second object?

  1. J2\frac{J}{2}
  2. JJ (correct answer)
  3. 2J\sqrt{2}J
  4. 2J2J
Explanation: Impulse is defined as the integral of force over time, J=∫FdtJ = \int F dt. Since the applied force FF and the time interval TT 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.

Question 3

An object of mass mm, starting from rest, is acted upon by a constant force FF for a time tt. Its change in momentum is Δp\Delta p and its change in kinetic energy is ΔK\Delta K. If the time interval is doubled to 2t2t while the force remains the same, how do the new change in momentum Δp′\Delta p' and change in kinetic energy ΔK′\Delta K' compare to the originals?

  1. Δp′=2Δp\Delta p' = 2 \Delta p and ΔK′=2ΔK\Delta K' = 2 \Delta K
  2. Δp′=2Δp\Delta p' = 2 \Delta p and ΔK′=4ΔK\Delta K' = 4 \Delta K (correct answer)
  3. Δp′=4Δp\Delta p' = 4 \Delta p and ΔK′=2ΔK\Delta K' = 2 \Delta K
  4. Δp′=Δp\Delta p' = \Delta p and ΔK′=2ΔK\Delta K' = 2 \Delta K
Explanation: The change in momentum (impulse) is Δp=J=Ft\Delta p = J = F t. If tt is doubled, Δp′=F(2t)=2(Ft)=2Δp\Delta p' = F(2t) = 2(Ft) = 2 \Delta p. Kinetic energy is K=p2/(2m)K = p^2/(2m). Since the initial kinetic energy is zero, ΔK=Kf\Delta K = K_f. As the momentum doubles, the kinetic energy becomes K′=(p′)2/(2m)=(2p)2/(2m)=4(p2/(2m))=4KK' = (p')^2/(2m) = (2p)^2/(2m) = 4(p^2/(2m)) = 4K. Thus, ΔK′=4ΔK\Delta K' = 4 \Delta K.

Question 4

A 1500 kg1500\,\text{kg} rocket in deep space starts at rest and fires its engine producing a constant 9000 N9000\,\text{N} thrust for 6.0 s6.0\,\text{s}. External forces are negligible. Using the given data, determine the change in momentum of the system.

  1. Δp=5.4×104 kg\cdotpm/s\Delta p=5.4\times10^4\,\text{kg·m/s} (correct answer)
  2. Δp=1.5×103 kg\cdotpm/s\Delta p=1.5\times10^3\,\text{kg·m/s}
  3. Δp=9.0×103 kg\cdotpm/s\Delta p=9.0\times10^3\,\text{kg·m/s}
  4. Δp=5.4×103 kg\cdotpm/s\Delta p=5.4\times10^3\,\text{kg·m/s}
Explanation: 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.

Question 5

A 70 kg70\ \text{kg} sprinter starts from rest and experiences an average forward net force of 420 N420\ \text{N} for 0.80 s0.80\ \text{s}. Using the given data, determine the change in momentum of the sprinter.

  1. Δp=3.4×102 kg\cdotpm/s\Delta p = 3.4\times10^2\ \text{kg·m/s} (correct answer)
  2. Δp=5.3×102 kg\cdotpm/s\Delta p = 5.3\times10^2\ \text{kg·m/s}
  3. Δp=3.4×102 N\Delta p = 3.4\times10^2\ \text{N}
  4. Δp=4.8×102 kg\cdotpm/s\Delta p = 4.8\times10^2\ \text{kg·m/s}
Explanation: 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.

Question 6

A 2.0 kg2.0\ \text{kg} cart moves right at 6.0 m/s6.0\ \text{m/s} and experiences a constant leftward force of 10 N10\ \text{N} for 0.50 s0.50\ \text{s}. Using the given data, determine the change in momentum of the cart.

  1. Δp=+5.0 kg\cdotpm/s\Delta p = +5.0\ \text{kg·m/s}
  2. Δp=−5.0 kg\cdotpm/s\Delta p = -5.0\ \text{kg·m/s} (correct answer)
  3. Δp=−20 kg\cdotpm/s\Delta p = -20\ \text{kg·m/s}
  4. Δp=−10 kg\cdotpm/s\Delta p = -10\ \text{kg·m/s}
Explanation: 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.

Question 7

A 3.0 kg3.0\ \text{kg} object starts from rest and experiences a constant net force of 18 N18\ \text{N} to the right for 0.50 s0.50\ \text{s}. Using the given data, what is the final velocity of the object after the collision?

  1. vf=1.5 m/sv_f = 1.5\ \text{m/s}
  2. vf=3.0 m/sv_f = 3.0\ \text{m/s} (correct answer)
  3. vf=9.0 m/sv_f = 9.0\ \text{m/s}
  4. vf=0.33 m/sv_f = 0.33\ \text{m/s}
Explanation: 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.

Question 8

A 0.50 kg0.50\ \text{kg} ball falls straight down at −8.0 m/s-8.0\ \text{m/s} and rebounds at +6.0 m/s+6.0\ \text{m/s} after floor contact lasting 0.040 s0.040\ \text{s}. Based on the problem above, calculate the impulse experienced by the ball.

  1. Impulse = +7.0 N\cdotps+7.0\ \text{N·s} (correct answer)
  2. Impulse = +1.0 N\cdotps+1.0\ \text{N·s}
  3. Impulse = −7.0 N\cdotps-7.0\ \text{N·s}
  4. Impulse = +14 N\cdotps+14\ \text{N·s}
Explanation: 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.

Question 9

A 0.20 kg0.20\ \text{kg} hockey puck slides right at 10 m/s10\ \text{m/s} and slows to 4.0 m/s4.0\ \text{m/s} due to a constant leftward force over 0.60 s0.60\ \text{s}. Based on the problem above, calculate the impulse experienced by the puck.

  1. Impulse = +1.2 N\cdotps+1.2\ \text{N·s}
  2. Impulse = −1.2 N\cdotps-1.2\ \text{N·s} (correct answer)
  3. Impulse = −0.20 N\cdotps-0.20\ \text{N·s}
  4. Impulse = −6.0 N\cdotps-6.0\ \text{N·s}
Explanation: This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating impulse when an object slows down. Impulse equals the change in momentum, J = m(vf - vi), where proper sign convention is crucial for objects changing speed in one direction. In this scenario, the 0.20 kg puck moving right slows from 10 m/s to 4.0 m/s, so J = 0.20(4.0 - 10) = 0.20(-6.0) = -1.2 N·s. Choice B is correct because it shows impulse = -1.2 N·s, with the negative sign indicating the leftward direction of the force (opposing the rightward motion). Choice A incorrectly shows positive impulse, while choices C and D show incorrect magnitudes. When teaching deceleration problems, emphasize that slowing down in the positive direction requires negative impulse. Students should recognize that friction or resistance forces oppose motion direction, creating negative impulse when motion is positive.

Question 10

A 1200 kg1200\,\text{kg} car moving east at 20 m/s20\,\text{m/s} collides with a 900 kg900\,\text{kg} car moving east at 5 m/s5\,\text{m/s}. After the collision, the 1200 kg1200\,\text{kg} car moves east at 8 m/s8\,\text{m/s}. Based on the problem above, calculate the impulse experienced by the 1200 kg1200\,\text{kg} car.

  1. Impulse = +1.4×104 N\cdotps+1.4\times10^4\,\text{N·s}
  2. Impulse = −1.4×104 N\cdotps-1.4\times10^4\,\text{N·s} (correct answer)
  3. Impulse = −9.6×103 N\cdotps-9.6\times10^3\,\text{N·s}
  4. Impulse = +9.6×103 N\cdotps+9.6\times10^3\,\text{N·s}
Explanation: This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating impulse experienced by one object in a collision. Impulse equals the change in momentum: J = m(vf - vi), where positive direction is east. In this scenario, the 1200 kg car changes velocity from +20 m/s to +8 m/s, so Δv = 8 - 20 = -12 m/s, resulting in impulse J = 1200 × (-12) = -14,400 N·s. Choice B (-1.4×10⁴ N·s) is correct because it accurately applies the impulse-momentum theorem with proper sign convention, showing the car experienced a negative impulse (opposing its initial motion). Choice A (+1.4×10⁴ N·s) is incorrect due to a sign error, failing to recognize that a decrease in eastward velocity represents negative impulse. When teaching collisions, emphasize consistent coordinate systems and that impulse direction matters. Have students practice identifying initial and final velocities carefully and checking that their impulse sign makes physical sense.

Question 11

A 1000 kg1000\,\text{kg} car traveling east at 12 m/s12\,\text{m/s} hits a barrier and rebounds west at 2.0 m/s2.0\,\text{m/s}. The collision time is 0.15 s0.15\,\text{s}. Based on the problem above, calculate the impulse experienced by the object.

  1. Impulse = +1.0×104 N\cdotps+1.0\times10^4\,\text{N·s}
  2. Impulse = −1.4×104 N\cdotps-1.4\times10^4\,\text{N·s} (correct answer)
  3. Impulse = −1.0×104 N\cdotps-1.0\times10^4\,\text{N·s}
  4. Impulse = +1.4×104 N\cdotps+1.4\times10^4\,\text{N·s}
Explanation: This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on impulse in a collision with velocity reversal. Impulse equals the change in momentum: J = m(vf - vi), where eastward is positive. In this scenario, the car has vi = +12 m/s (east) and vf = -2.0 m/s (west), so Δv = -2.0 - 12 = -14 m/s, giving impulse J = 1000 × (-14) = -14,000 N·s = -1.4×10⁴ N·s. Choice B (-1.4×10⁴ N·s) is correct because it accurately calculates the momentum change with proper sign convention for the velocity reversal. Choice D (+1.4×10⁴ N·s) is incorrect due to a sign error, failing to recognize that the impulse opposes the initial motion. When teaching collision problems, emphasize establishing clear coordinate systems and that negative impulse indicates force opposing initial motion. Practice with various collision scenarios reinforces proper sign usage.

Question 12

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?

  1. 15,000 N
  2. 5,000 N
  3. 2,000 N
  4. 10,000 N (correct answer)
Explanation: 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: Fthrust=dmdt×vexhaustF_{thrust} = \frac{dm}{dt} \times v_{exhaust}, where dmdt\frac{dm}{dt} is the mass flow rate of exhaust and vexhaustv_{exhaust} is the exhaust velocity relative to the rocket. Given: mass flow rate = 5.0 kg/s and exhaust velocity = 2000 m/s Fthrust=5.0 kg/s×2000 m/s=10,000 NF_{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).

Question 13

A particle is subject to a force F(t)=F0e−t/τF(t) = F_0 e^{-t/\tau} where F0F_0 and τ\tau are constants. The particle starts from rest at t=0t=0. What is the total impulse delivered to the particle as tt approaches infinity?

  1. 00
  2. F0/τF_0 / \tau
  3. F0τF_0 \tau (correct answer)
  4. Infinite
Explanation: The total impulse is the integral of the force from t=0t=0 to t=∞t=\infty. J=∫0∞F0e−t/τdt=F0[−τe−t/τ]0∞=−F0τ[e−∞−e0]=−F0τ[0−1]=F0τJ = \int_0^\infty F_0 e^{-t/\tau} dt = F_0 [-\tau e^{-t/\tau}]_0^\infty = -F_0 \tau [e^{-\infty} - e^0] = -F_0 \tau [0 - 1] = F_0 \tau.

Question 14

A particle of mass mm has a momentum vector given by the function p⃗(t)=(at2−b)i^+(ct)j^\vec{p}(t) = (at^2 - b)\hat{i} + (ct)\hat{j}, where a,b,a, b, and cc are positive constants. What is the net force vector F⃗\vec{F} acting on the particle as a function of time?

  1. F⃗(t)=(2at)i^+cj^\vec{F}(t) = (2at)\hat{i} + c\hat{j} (correct answer)
  2. F⃗(t)=(13at3−bt)i^+(12ct2)j^\vec{F}(t) = (\frac{1}{3}at^3 - bt)\hat{i} + (\frac{1}{2}ct^2)\hat{j}
  3. F⃗(t)=(2at−b)i^+cj^\vec{F}(t) = (2at - b)\hat{i} + c\hat{j}
  4. F⃗(t)=(at2)i^+cj^\vec{F}(t) = (at^2)\hat{i} + c\hat{j}
Explanation: The net force is the time derivative of the momentum vector, F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}. Taking the derivative of each component with respect to time gives ddt(at2−b)=2at\frac{d}{dt}(at^2 - b) = 2at and ddt(ct)=c\frac{d}{dt}(ct) = c. Therefore, the force vector is F⃗(t)=(2at)i^+cj^\vec{F}(t) = (2at)\hat{i} + c\hat{j}.

Question 15

The net force on a particle varies with time according to the equation F(t)=12t−3t2F(t) = 12t - 3t^2, where FF is in newtons and tt is in seconds. What is the total impulse delivered to the particle from t=0t=0 s to t=2t=2 s?

  1. 88 N⋅s
  2. 1616 N⋅s (correct answer)
  3. 1818 N⋅s
  4. 2424 N⋅s
Explanation: Impulse is the integral of force with respect to time: J=∫t1t2F(t)dtJ = \int_{t_1}^{t_2} F(t) dt. Integrating from 0 to 2 seconds: J=∫02(12t−3t2)dt=[6t2−t3]02=(6(2)2−(2)3)−(0)=(24−8)=16J = \int_0^2 (12t - 3t^2) dt = [6t^2 - t^3]_0^2 = (6(2)^2 - (2)^3) - (0) = (24 - 8) = 16 N⋅s.

Question 16

A 0.150.15 kg baseball is pitched horizontally at 4040 m/s. The batter hits it, and the ball leaves the bat horizontally in the opposite direction at 6060 m/s. If the bat and ball are in contact for 2.02.0 ms, what is the magnitude of the average force the bat exerts on the ball?

  1. 15001500 N
  2. 30003000 N
  3. 75007500 N (correct answer)
  4. 1500015000 N
Explanation: The change in momentum is Δp=m(vf−vi)\Delta p = m(v_f - v_i). Choosing the pitched direction as negative, vi=−40v_i = -40 m/s and vf=60v_f = 60 m/s. So, Δp=0.15(60−(−40))=0.15(100)=15\Delta p = 0.15(60 - (-40)) = 0.15(100) = 15 kg⋅m/s. The average force is Favg=ΔpΔt=15 kg⋅m/s2.0×10−3 s=7500F_{avg} = \frac{\Delta p}{\Delta t} = \frac{15 \text{ kg⋅m/s}}{2.0 \times 10^{-3} \text{ s}} = 7500 N.

Question 17

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 s,p=5 kg⋅m/s)(t=1 \text{ s}, p=5 \text{ kg⋅m/s}) and (t=4 s,p=14 kg⋅m/s)(t=4 \text{ s}, p=14 \text{ kg⋅m/s}). What is the constant net force acting on the object?

  1. 33 N (correct answer)
  2. 4.54.5 N
  3. 99 N
  4. 1919 N
Explanation: The net force is the rate of change of momentum, Fnet=dpdtF_{net} = \frac{dp}{dt}. For a linear momentum-time graph, this is the slope of the line. Fnet=ΔpΔt=14 kg⋅m/s−5 kg⋅m/s4 s−1 s=9 kg⋅m/s3 s=3F_{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.

Question 18

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?

  1. The clay ball experiences a greater impulse because it undergoes a perfectly inelastic collision.
  2. The rubber ball experiences a greater impulse because its change in momentum is larger. (correct answer)
  3. Both balls experience the same impulse because their mass and initial speed are identical.
  4. The clay ball experiences a greater impulse because the contact time with the wall is longer.
Explanation: Impulse equals the change in momentum (Δp\Delta p). The clay ball's momentum changes from mvmv to 0, so ∣Δpclay∣=mv|\Delta p_{clay}| = mv. The rubber ball's momentum changes from mvmv to approximately −mv-mv, so ∣Δprubber∣≈2mv|\Delta p_{rubber}| \approx 2mv. Since the rubber ball has a larger change in momentum, it experiences a greater impulse.

Question 19

A rocket ejects fuel at a constant rate RR with a constant exhaust velocity uexu_{ex} relative to the rocket. According to the impulse-momentum theorem for a system of variable mass, what is the magnitude of the thrust produced?

  1. 12Ruex2\frac{1}{2}Ru_{ex}^2
  2. MrocketgM_{rocket}g
  3. RuexRu_{ex} (correct answer)
  4. RuexMrocket\frac{Ru_{ex}}{M_{rocket}}
Explanation: The thrust force on a rocket is given by the rate of change of momentum of the ejected fuel. This is expressed as Fthrust=∣dp⃗dt∣=∣u⃗exdmdt∣F_{thrust} = |\frac{d\vec{p}}{dt}| = |\vec{u}_{ex} \frac{dm}{dt}|. Since RR is the rate of mass ejection (dm/dtdm/dt), the thrust is RuexRu_{ex}.

Question 20

A 2.02.0 kg object's motion is governed by the momentum function p(t)=10+4t2p(t) = 10 + 4t^2 for t≥0t \ge 0, where pp is in kg⋅m/s and tt is in seconds. What is the net force on the object at t=3.0t=3.0 s?

  1. 8.08.0 N
  2. 1212 N
  3. 2424 N (correct answer)
  4. 4646 N
Explanation: The net force is the time derivative of momentum: F(t)=dpdt=ddt(10+4t2)=8tF(t) = \frac{dp}{dt} = \frac{d}{dt}(10 + 4t^2) = 8t. At t=3.0t=3.0 s, the force is F(3)=8(3.0)=24F(3) = 8(3.0) = 24 N.