Explain Energy Transfer Through Interactions - Grade 11 Physics

1. How much energy is transferred from the carts' kinetic energy to internal energy during the collision?
   Answer: $18\text{ J}$, because the joined carts retain half of the initial kinetic energy. When you see a collision where objects stick together, that's your signal: this is a perfectly inelastic collision. Momentum is conserved, but kinetic energy is not — some energy converts to internal energy (heat, sound, deformation). Your job is to calculate how much KE is lost. Start by applying conservation of momentum to find the final velocity. The moving cart ($m = 2.0\text{ kg}$, $v = 6.0\text{ m/s}$) hits an identical cart at rest: $m_1v_1 = (m_1 + m_2)v_f$ $(2.0)(6.0) = (4.0)v_f \implies v_f = 3.0\text{ m/s}$ Now compare kinetic energies: $KE_i = \frac{1}{2}(2.0)(6.0)^2 = 36\text{ J}$ $KE_f = \frac{1}{2}(4.0)(3.0)^2 = 18\text{ J}$ Energy lost to internal energy: $36 - 18 = 18\text{ J}$. That confirms B is correct — exactly half the initial KE is retained, and half is transferred to internal energy. A is wrong because it confuses perfectly inelastic collisions with elastic collisions — kinetic energy is not conserved when objects stick together. C is a trap: yes, each cart moves at half the original speed (3.0 m/s), but the system's final KE is $18\text{ J}$, not $24\text{ J}$ — doubling the mass while halving the speed doesn't keep KE constant. D is wrong because the carts are still moving after impact, so kinetic energy clearly remains. Study tip: Always calculate both $KE_i$ and $KE_f$ separately — never assume the energy split is obvious from the speed ratio alone, since KE depends on mass × speed².
2. Which energy account is consistent with conservation of energy?
   Answer: The battery transfers $60\text{ J}$; mechanical energy increases by $48\text{ J}$, and internal energy increases by $12\text{ J}$. When you see a question like this, think of it as an energy accounting problem: the total energy supplied by the battery must equal the sum of all energy increases in the system. Conservation of energy means nothing is created or destroyed — every joule must be accounted for. Start by calculating what the battery supplies. The electrical energy transferred is $W = qV = (5.0\text{ C})(12\text{ V}) = 60\text{ J}$. That eliminates choice C immediately, since it claims $72\text{ J}$ — no calculation supports that figure. Next, calculate the mechanical energy gained. The load gains gravitational potential energy: $\Delta PE = mgh = (2.0\text{ kg})(10\text{ m/s}^2)(2.0\text{ m}) = 40\text{ J}$. The kinetic energy increase is given as $8.0\text{ J}$. So the total mechanical energy increase is $40 + 8 = 48\text{ J}$. The remaining energy goes to internal energy (heat from friction, motor resistance, etc.): $60 - 48 = 12\text{ J}$. This matches choice A perfectly. Choice B incorrectly states mechanical energy increases by only $40\text{ J}$, forgetting to include the $8\text{ J}$ kinetic energy increase — a classic trap of ignoring one component of mechanical energy. Choice D claims mechanical energy increases by $52\text{ J}$, which would require an unexplained extra $4\text{ J}$ beyond what's calculated. Choice C, as noted, inflates the battery's energy output without justification. Study tip: Always list every form of energy separately — gravitational PE, kinetic energy, and internal energy — and verify they sum to the total energy input. Missing any one term is the most common error on conservation-of-energy problems.
3. How does doubling the amplitude affect the energy transferred from the motor to the rope during the fixed time?
   Answer: The transferred energy quadruples because wave energy is proportional to amplitude squared. Whenever you see a question about wave energy, your first instinct should be to recall how energy in a wave depends on amplitude. This is the central formula being tested here. The power transmitted by a sinusoidal wave on a rope is given by: $P = \frac{1}{2} \mu \omega^2 A^2 v$ where $\mu$ is linear mass density, $\omega$ is angular frequency, $A$ is amplitude, and $v$ is wave speed. Notice that energy is proportional to $A^2$, not $A$. When you double the amplitude, you replace $A$ with $2A$, giving $(2A)^2 = 4A^2$. Since frequency and wave speed are unchanged, the power — and therefore the total energy transferred over the same fixed time — quadruples. D is correct. Choice A is wrong because wave speed and frequency alone do not determine energy. They set the wave's spatial and temporal structure, but amplitude controls how vigorously each rope element oscillates, which directly scales the energy. Choice B falls into a common linear-thinking trap. It assumes energy scales directly with amplitude (doubling $A$ doubles energy), ignoring the squared relationship. Each element does oscillate through twice the displacement, but the energy of simple harmonic motion goes as displacement squared, not displacement. Choice C invents a conservation argument that doesn't apply. Energy isn't a fixed quantity being "spread out" — the motor inputs more energy into the rope when the amplitude is larger. Dissipative effects are also explicitly negligible. Study tip: Whenever a wave question involves amplitude changes, immediately write down $E \propto A^2$. The quadratic relationship is one of the most frequently tested wave properties — don't let your intuition default to a linear relationship.
4. What is the block's speed immediately after it crosses the rough section?
   Answer: $5.0\text{ m/s}$, because $5.0\text{ J}$ of the spring's stored energy remains as kinetic energy after friction acts. Whenever you see a block launched by a spring that then crosses a rough patch, reach for conservation of energy with a loss term: $KE_{final} = PE_{spring} - W_{friction}$. Start by calculating the spring's stored potential energy: $PE = \frac{1}{2}kx^2 = \frac{1}{2}(400)(0.20)^2 = 8.0\text{ J}$. Since the surface before the rough section is smooth, all $8.0\text{ J}$ converts to kinetic energy before friction acts. Friction then removes $3.0\text{ J}$, leaving $8.0 - 3.0 = 5.0\text{ J}$ as kinetic energy. Solving $\frac{1}{2}mv^2 = 5.0\text{ J}$ gives $v = \sqrt{\frac{2(5.0)}{0.40}} = \sqrt{25} = 5.0\text{ m/s}$, confirming D. Choice A claims friction removes "half" of the stored energy, leaving $4.0\text{ J}$. This is arithmetic carelessness — friction removes exactly $3.0\text{ J}$, not $4.0\text{ J}$. Choice B suggests speed decreases proportionally to energy removed, which confuses energy and speed. Because $KE \propto v^2$, you cannot scale speed linearly with energy loss — you must recalculate $v$ from the remaining kinetic energy. Choice C invents an energy gain during the rough section, which contradicts the physics: friction only removes mechanical energy, it never adds it. Your strategy tip: always find the spring's potential energy first, subtract any energy lost to friction, then use $v = \sqrt{\frac{2KE}{m}}$ to find speed. Never subtract energy losses directly from speed — always work through energy, then convert to speed at the end.
5. Which statement correctly accounts for the energy transferred through the crate-ramp friction interaction?
   Answer: Friction increases internal energy by $50\text{ J}$, while total mechanical energy increases by $210\text{ J}$. When a question involves work, friction, and multiple forms of energy, your go-to tool is the work-energy theorem extended to all energy forms: $W_{\text{applied}} = \Delta KE + \Delta PE + \Delta E_{\text{internal}}$. The energy input must account for every change happening in the system. Start by calculating each energy change. The change in kinetic energy is $\Delta KE = \frac{1}{2}(10)(4.0^2 - 2.0^2) = \frac{1}{2}(10)(12) = 60 \text{ J}$. The change in gravitational potential energy is $\Delta PE = mgh = (10)(10)(1.5) = 150 \text{ J}$. So total mechanical energy increases by $\Delta KE + \Delta PE = 60 + 150 = 210 \text{ J}$. Since the person inputs $260 \text{ J}$ and only $210 \text{ J}$ goes into mechanical energy, friction must account for the remaining $260 - 210 = 50 \text{ J}$, converted into internal (thermal) energy. This confirms C: friction increases internal energy by $50 \text{ J}$ while total mechanical energy increases by $210 \text{ J}$. Choice A correctly finds $\Delta PE = 150 \text{ J}$ but incorrectly assigns $110 \text{ J}$ to friction, ignoring the kinetic energy change entirely. Choice B arrives at $\Delta KE = 60 \text{ J}$ correctly but inflates friction to $200 \text{ J}$, which doesn't satisfy the energy balance. Choice D claims friction decreases internal energy — this is physically impossible, since friction always converts mechanical energy into heat, never the reverse. Your strategy: always compute all energy changes first ($\Delta KE$, $\Delta PE$), sum them, then subtract from work input to isolate the friction term. Don't skip any energy category.
6. Which comparison best explains energy transfer through the gravitational interaction in the two situations?
   Answer: Gravity transfers energy to the falling stone but not to the circular-orbit satellite because only the stone's displacement has a component along gravity. Whenever you see a question about work done by a force, the key concept to apply is the work-energy theorem: a force transfers energy (does work) only when there is a displacement component parallel to that force. Mathematically, $W = F \cdot d \cdot \cos\theta$, where θ is the angle between the force and displacement vectors. For the falling stone, gravity pulls downward and the stone moves downward — θ = 0°, so cos θ = 1. Gravity does positive work, transferring energy into the stone's kinetic energy, which is exactly why it speeds up. For the orbiting satellite, gravity points toward Earth's center (centripetal direction), but at every instant the satellite's displacement is tangential — perpendicular to gravity. That makes θ = 90°, cos 90° = 0, and therefore zero work is done. The satellite's speed stays constant because no energy is transferred. This makes C the correct answer. A is wrong because a nonzero force does not automatically increase kinetic energy — it only does so when it has a component along the displacement. A force perpendicular to motion (like in circular orbits) changes direction, not speed. B is wrong because it falsely treats gravitational potential energy as a separate, isolated conserved quantity. In reality, gravitational PE and kinetic energy can freely convert into each other — they're not independently conserved. D is wrong because centripetal acceleration describes a change in direction, not a transfer of energy. Acceleration alone doesn't imply work done. Your study tip: always ask "what angle exists between force and displacement?" before deciding whether a force transfers energy — that single question unlocks most work-energy problems.
7. Compared with the first stop, what happens to the stopping distance and the energy transferred from the car's motion to thermal energy during the second stop?
   Answer: The distance becomes $160\text{ m}$, and the transferred energy becomes $800\text{ kJ}$. When a car stops, two key physics relationships govern what happens: the work-energy theorem and the dependence of kinetic energy on speed squared. The braking force does negative work equal to the car's initial kinetic energy, so $F \cdot d = \frac{1}{2}mv^2$. This tells you that stopping distance $d = \frac{mv^2}{2F}$, meaning distance scales with $v^2$, not $v$. When speed doubles from $20\text{ m/s}$ to $40\text{ m/s}$, the kinetic energy quadruples: $KE_1 = \frac{1}{2}(1000)(20)^2 = 200\text{ kJ}$, while $KE_2 = \frac{1}{2}(1000)(40)^2 = 800\text{ kJ}$. Since the same braking force must absorb all of that energy, the stopping distance also quadruples: $d_2 = \frac{800{,}000}{5000} = 160\text{ m}$. The energy transferred to thermal energy equals the kinetic energy lost, so it becomes $800\text{ kJ}$. This confirms C. Choice A incorrectly assumes distance only doubles (linear relationship with speed), and underestimates the thermal energy at $400\text{ kJ}$. Choice B makes an even more serious error — claiming distance stays the same — while at least correctly quadrupling the energy; but if energy quadruples and force is constant, distance must quadruple too, since $W = Fd$. Choice D gets the distance right at $160\text{ m}$ but incorrectly claims the transferred energy stays at $200\text{ kJ}$, ignoring that all kinetic energy converts to thermal energy. Study tip: Whenever speed doubles, remember kinetic energy quadruples (because of $v^2$), and so does stopping distance for a constant braking force. The $v^2$ relationship is the most commonly missed trap on kinematics-energy questions.
8. Which statement best explains the energy transfers that occurred?
   Answer: Its gravitational potential energy is the same there, but some mechanical energy was transferred to the air and pivot as internal energy. Whenever you see a question involving a real-world pendulum with air and friction mentioned, your first instinct should be to think about **conservation of energy — and where energy leaks out of the mechanical system**. The pendulum starts at a certain height, so it has a specific gravitational potential energy (GPE) determined by $E_{grav} = mgh$. When it returns to that same height, its GPE is identical — height hasn't changed, neither has mass or gravity. However, the pendulum is moving slower, meaning its kinetic energy is lower than before. So where did that energy go? The passage tells you: the air and the nonideal pivot. These resistive interactions convert mechanical energy into internal (thermal) energy — essentially heat dissipated into the surroundings. The total energy is still conserved globally, but the mechanical energy of the pendulum system has decreased. This makes A the correct answer. B is wrong because GPE is not smaller — the pendulum returned to the same height, so GPE is unchanged. Missing energy wasn't converted into kinetic energy; it was lost to resistive forces. C is wrong because it contradicts the scenario. If kinetic energy were unchanged despite the slower speed, that would violate basic physics. The question explicitly states the pendulum moves more slowly, meaning kinetic energy did change. D is partially right in saying kinetic energy is smaller, but its claim that energy was "removed from the system entirely" is misleading — energy wasn't destroyed, it was transferred as internal energy to the air and pivot. Energy is always conserved. A reliable strategy: when a problem mentions real-world resistive forces, always track where mechanical energy converts to, not where it disappears.
9. Which result correctly describes the energy transferred through the thermal interaction?
   Answer: The final temperature is $32.5^\circ\text{C}$, and $3750\text{ J}$ transfers from A to B. When two objects reach thermal equilibrium inside an insulated container, heat lost by the hotter object equals heat gained by the cooler one. The key equation is $Q = C \cdot \Delta T$, where $C$ is heat capacity (already given per degree, so no mass conversion needed here). Setting heat lost by A equal to heat gained by B gives you $C_A(T_A - T_f) = C_B(T_f - T_B)$, so $100(70 - T_f) = 300(T_f - 20)$. Expanding: $7000 - 100T_f = 300T_f - 6000$, which yields $13000 = 400T_f$, giving $T_f = 32.5^\circ\text{C}$. The energy transferred from A to B is $Q = 100 \times (70 - 32.5) = 100 \times 37.5 = 3750\text{ J}$. Heat flows from the hotter object (A) to the cooler one (B), confirming answer A is correct. Answer B reaches $45.0^\circ\text{C}$, which would only be correct if both objects had equal heat capacities (a simple average). Ignoring the difference in $C$ values is a classic mistake. Answer C gets the final temperature right at $32.5^\circ\text{C}$ but claims heat flows from B to A — the opposite of physical reality, since heat always flows from hot to cold. It also inflates the energy to $11{,}250\text{ J}$, which is actually $Q = 300 \times 37.5$, confusing which object's heat capacity to use. Answer D misidentifies both the direction of heat flow and the final temperature. A useful habit: always sanity-check your final temperature — it must fall between the two starting temperatures, closer to the object with the larger heat capacity. Here, $32.5^\circ\text{C}$ sits much closer to B's starting temperature of $20^\circ\text{C}$, which makes sense since B has three times the heat capacity and resists temperature change more strongly.
10. Which result correctly describes the second fragment and the energy transferred by the explosion?
    Answer: It moves west at $2.0\text{ m/s}$, and $24\text{ J}$ of chemical energy becomes kinetic energy. When an explosion occurs from rest, two key principles apply simultaneously: conservation of momentum and conservation of energy. The system starts with zero total momentum, so it must end with zero — meaning the fragments must move in opposite directions with equal and opposite momenta. Start with momentum conservation. The initial momentum is zero, so: $m_1v_1 + m_2v_2 = 0$ $(1.0)(+6.0) + (3.0)v_2 = 0$ $v_2 = -2.0 \text{ m/s}$ The negative sign means the 3.0 kg fragment moves west at 2.0 m/s. Now calculate the total kinetic energy released (this equals the chemical energy converted): $KE = \frac{1}{2}(1.0)(6.0)^2 + \frac{1}{2}(3.0)(2.0)^2 = 18 + 6 = 24 \text{ J}$ This confirms B is correct. Choice A fails the momentum test immediately — if both fragments move east, total momentum would be positive, not zero. The system can't gain momentum from an internal explosion. Choice C gives the right direction (west) but assigns the wrong speed (6.0 m/s instead of 2.0 m/s), which would violate momentum conservation since $(3.0)(6.0) \neq (1.0)(6.0)$. The 72 J energy figure in C also comes from incorrectly using 6.0 m/s for the heavy fragment. Choice D combines both errors from A and C — wrong direction and wrong energy. A useful strategy: in any explosion or recoil problem, always apply momentum conservation first to find the velocity, then calculate kinetic energy. Never assume both fragments share the same speed or direction.

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