Statics and Dynamics Quiz: Linear Impulse Momentum

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Linear Impulse MomentumQuestion 1 of 3

A rocket (modeled as a particle) of initial mass $M_0 = 1000 \text{ kg}$ ejects mass at a constant rate $\dot{m} = 20 \text{ kg/s}$ with an exhaust speed of $v_e = 500 \text{ m/s}$ relative to the rocket. The rocket moves in free space (no gravity). At $t = 0$ , the rocket is at rest.

Using the variable-mass impulse-momentum principle, what is the rocket's velocity at $t = 10 \text{ s}$ ? Assume the thrust force equals $\dot{m}v_e$ and the mass at time $t$ is $M(t) = M_0 - \dot{m}t$ .

v \approx 105.4 \text{ m/s}

, obtained by correctly integrating the equation of motion for a variable-mass system, equivalent to the Tsiolkovsky rocket equation:

v = v_e \ln\!\left(\dfrac{M_0}{M_0 - \dot{m}t}\right)

v = 100 \text{ m/s}

, obtained by treating the rocket mass as constant at

M_0

throughout the interval and dividing the total thrust impulse by

M_0

v \approx 111 \text{ m/s}

, obtained by dividing the total thrust impulse by the average mass

\bar{M} = (M_0 + M(10))/2 = 900 \text{ kg}

, approximating the variable-mass system with a single effective constant mass.

v = 200 \text{ m/s}

, obtained by incorrectly computing the impulse as the exhaust speed multiplied by the total time and dividing by the ejected mass rather than the retained rocket mass, reversing the roles of ejected and remaining mass.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Linear Impulse Momentum

Practice Linear Impulse Momentum in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Linear Impulse Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

$v \approx 105.4 \text{ m/s}$ , obtained by correctly integrating the equation of motion for a variable-mass system, equivalent to the Tsiolkovsky rocket equation: $v = v_e \ln\!\left(\dfrac{M_0}{M_0 - \dot{m}t}\right)$ . (correct answer)
$v = 100 \text{ m/s}$ , obtained by treating the rocket mass as constant at $M_0$ throughout the interval and dividing the total thrust impulse by $M_0$ .
$v \approx 111 \text{ m/s}$ , obtained by dividing the total thrust impulse by the average mass $\bar{M} = (M_0 + M(10))/2 = 900 \text{ kg}$ , approximating the variable-mass system with a single effective constant mass.
$v = 200 \text{ m/s}$ , obtained by incorrectly computing the impulse as the exhaust speed multiplied by the total time and dividing by the ejected mass rather than the retained rocket mass, reversing the roles of ejected and remaining mass.

Explanation: Whenever you see a rocket (or any variable-mass system) problem, recognize that the mass itself is changing over time — which means you cannot simply divide a fixed impulse by a fixed mass. The governing principle is the Tsiolkovsky rocket equation, derived by integrating Newton's second law with a time-varying mass:

M(t)\frac{dv}{dt} = \dot{m}v_e

, which rearranges to

dv = v_e \frac{d(\dot{m}t)}{M_0 - \dot{m}t}

. Integrating from

t = 0

t = 10\text{ s}

gives

v = v_e \ln\!\left(\dfrac{M_0}{M_0 - \dot{m}t}\right)

. Plugging in:

M(10) = 1000 - 20(10) = 800\text{ kg}

, so

v = 500\ln\!\left(\dfrac{1000}{800}\right) = 500\ln(1.25) \approx 500(0.2231) \approx 105.4\text{ m/s}

. This confirms A is correct. B treats the rocket's mass as the constant

M_0 = 1000\text{ kg}

throughout, computing

v = \frac{\dot{m}v_e \cdot t}{M_0} = \frac{100{,}000}{1000} = 100\text{ m/s}

. This ignores the fact that as mass is ejected, the remaining rocket accelerates more easily — the denominator shrinks over time, so the true velocity is higher. C approximates by using the average mass

\bar{M} = 900\text{ kg}

, giving

v \approx 111\text{ m/s}

. While closer, a simple average mass cannot substitute for proper integration; the logarithmic relationship is nonlinear, and linear averaging introduces systematic error. D reverses the roles of ejected and retained mass entirely, producing a physically nonsensical result of

200\text{ m/s}

. Study tip: On variable-mass problems, your instinct to use

\Delta p = F\Delta t

with a constant mass will always underestimate the final velocity — always integrate, or apply the rocket equation directly.

Question 2

Two particles, A (mass $m_A = 4 \text{ kg}$ ) and B (mass $m_B = 6 \text{ kg}$ ), move along the same line. Particle A has velocity $+9 \text{ m/s}$ and particle B has velocity $-3 \text{ m/s}$ . They collide and stick together (perfectly plastic collision).

An impulsive internal force acts between the particles during collision. Which statement correctly applies the principle of linear impulse-momentum to each particle individually and to the system?

The internal impulse on A equals $-m_A \Delta v_A$ and on B equals $-m_B \Delta v_B$ ; the system's total momentum changes by the sum of these internal impulses, which is nonzero because the collision is perfectly plastic.
The internal impulse on A has the same magnitude as the internal impulse on B by Newton's third law; the system's total momentum is conserved because the internal impulses are equal and opposite, and the common final velocity is $+1.8 \text{ m/s}$ . (correct answer)
The internal impulse on A equals the internal impulse on B in both magnitude and direction; the system's total momentum is conserved, and the common final velocity is $+1.8 \text{ m/s}$ .
The internal impulse on A has the same magnitude as the internal impulse on B by Newton's third law; the system's total momentum is conserved because the internal impulses are equal and opposite, and the common final velocity is $+2.6 \text{ m/s}$ .

Explanation: Whenever you see a collision problem asking about impulse on individual particles, anchor yourself to two principles working together: Newton's Third Law and conservation of momentum for the system. During any collision, the internal force that A exerts on B is equal and opposite to the force B exerts on A. Since impulse is force × time and both forces act over the same time interval, the impulse on A and the impulse on B are equal in magnitude but opposite in direction. These internal impulses cancel when you sum them across the system, which is exactly why the system's total momentum is conserved — no net external impulse acts on the pair. To find the common final velocity, apply conservation of momentum:

p_i = m_A v_A + m_B v_B = (4)(+9) + (6)(-3) = 36 - 18 = 18 \text{ kg·m/s}

v_f = \frac{18}{m_A + m_B} = \frac{18}{10} = +1.8 \text{ m/s}

This confirms B is correct: Newton's Third Law guarantees equal-and-opposite internal impulses, total momentum is conserved, and

v_f = +1.8 \text{ m/s}

. A is wrong on two counts: it incorrectly states momentum changes due to internal impulses (internal forces never change system momentum), and those impulses are not both negative — they're opposite in direction. C is wrong because it claims the impulses act in the same direction, which violates Newton's Third Law entirely. D gets the physics of impulses right but makes an arithmetic error, arriving at

+2.6 \text{ m/s}

instead of the correct

+1.8 \text{ m/s}

. Your study tip: in any collision, internal impulses always cancel in pairs — they can redistribute momentum between particles but can never change the system's total momentum.

Question 3

A 10 kg particle is subjected to a force that varies with time as shown by the following description: the force increases linearly from $0 \text{ N}$ at $t = 0$ to $F_{max}$ at $t = 2 \text{ s}$ , then decreases linearly back to $0 \text{ N}$ at $t = 4 \text{ s}$ . The particle starts from rest. At $t = 4 \text{ s}$ , its velocity is measured to be $6 \text{ m/s}$ .

What is $F_{max}$ , and what is the velocity at $t = 2 \text{ s}$ (the moment of peak force)?

$F_{max} = 30 \text{ N}$ and $v(2) = 2 \text{ m/s}$ , found by setting the total impulse equal to the final momentum, then computing the first-half impulse using half of $F_{max}$ as the triangle height instead of $F_{max}$ itself.
$F_{max} = 60 \text{ N}$ and $v(2) = 6 \text{ m/s}$ , found by treating the force-time graph as a rectangle ( $F_{max} \times t_{total} = \Delta p$ ) rather than a triangle, which doubles the computed peak force, then assuming the full final velocity is reached at the midpoint.
$F_{max} = 30 \text{ N}$ and $v(2) = 3 \text{ m/s}$ , found by correctly computing the total impulse as the area of the triangular force-time graph, then recognizing that the area of the first half-triangle equals half the total impulse. (correct answer)
$F_{max} = 30 \text{ N}$ and $v(2) = 4 \text{ m/s}$ , found by computing the correct peak force but then assuming velocity is proportional to elapsed time (i.e., $v(2) = v_f \times \frac{2}{4} \times \frac{4}{3}$ ) rather than to accumulated impulse, misapplying a linear-time scaling to a nonlinear momentum accumulation.

Explanation: When a time-varying force acts on a particle, the impulse-momentum theorem is your core tool: the impulse (area under the force-time graph) equals the change in momentum,

J = \Delta p = m \Delta v

. Here, the force-time graph forms a symmetric triangle — rising linearly to

F_{max}

t = 2\text{ s}

, then falling back to zero at

t = 4\text{ s}

. The total area is

\frac{1}{2} \times base \times height = \frac{1}{2}(4)(F_{max}) = 2F_{max}

. Setting this equal to the final momentum change:

2F_{max} = m \cdot v_f = 10 \times 6 = 60 \text{ N·s}

, so

F_{max} = 30 \text{ N}

. For

v(2)

, the impulse delivered in the first two seconds is the area of the first smaller triangle:

\frac{1}{2}(2)(30) = 30 \text{ N·s}

. Thus

v(2) = \frac{30}{10} = 3 \text{ m/s}

. This confirms C is correct. A is wrong because it uses half of

F_{max}

(i.e., 15 N) as the triangle height when computing the first-half impulse — the height is still the full

F_{max} = 30\text{ N}

; you just integrate over half the base. B treats the graph as a rectangle, computing

F_{max} \times 4

instead of the triangular area, which artificially doubles the peak force to 60 N. It then incorrectly assigns the final velocity to the midpoint. D gets

F_{max}

right but assumes velocity scales linearly with time — it doesn't. Velocity scales with accumulated impulse (area), which is nonlinear for a triangular force profile. Strategy tip: Always sketch the force-time graph and compute areas carefully. Remember: equal time intervals don't mean equal impulse when the force isn't constant.