Linear Impulse-Momentum - Statics and Dynamics

1. 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 $M(t) = M_0 - \dot{m}t$.
   Answer: $v \approx 105.4 \text{ m/s}$, obtained by correctly integrating the equation of motion for a variable-mass system, equivalent to the Tsiolkovsky rocket equation: $v = v_e \ln\!\left(\dfrac{M_0}{M_0 - \dot{m}t}\right)$. 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 $t = 0$ to $t = 10\text{ s}$ 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.
2. 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?
   Answer: 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 $+1.8 \text{ m/s}$. 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: $p_i = m_A v_A + m_B v_B = (4)(+9) + (6)(-3) = 36 - 18 = 18 \text{ kg}\cdot\text{m/s}$ $v_f = \frac{18}{m_A + m_B} = \frac{18}{10} = +1.8 \text{ m/s}$ 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.
3. What is $F_{max}$, and what is the velocity at $t = 2 \text{ s}$ (the moment of peak force)?
   Answer: $F_{max} = 30 \text{ N}$ 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. 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 $F_{max}$ at $t = 2\text{ s}$, 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: $2F_{max} = m \cdot v_f = 10 \times 6 = 60 \text{ N}\cdot\text{s}$, 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: $\frac{1}{2}(2)(30) = 30 \text{ N}\cdot\text{s}$. 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.

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