Analyze Energy Using Conservation Laws - Grade 11 Physics

1. A ball hits the floor at speed $v$ and rebounds at $0.80v$. What fraction of its kinetic energy is lost?
   Answer: 0.36 Kinetic energy scales with speed squared, so after rebounding at 0.80v the final KE is $(0.80)^2$ = 0.64 of the original. That means 1 - 0.64 = 0.36 of the kinetic energy is lost. The tempting 0.20 is just the speed loss, not the energy loss, because KE isn't proportional to speed directly.
2. A block on a smooth table launched by a spring compressed $x$. At half compression, what fraction of energy is kinetic?
   Answer: $\frac{3}{4}$ Total energy is the spring's stored energy at full compression. At half compression, potential energy is one-quarter of total because spring energy depends on compression squared. So kinetic energy is the remaining three-quarters. The trap is choosing 1/2 by assuming energy scales linearly with compression.
3. A $20\text{ kg}$ sled moves $12\text{ m}$ down a snowy slope while dropping vertically by $3.0\text{ m}$. Its speed increases from $4.0\text{ m/s}$ to $6.0\text{ m/s}$. Assuming the average friction force is constant and using $g=9.8\text{ m/s}^2$, what is its magnitude?
   Answer: $32.3\text{ N}$, found from the mechanical energy transformed into thermal energy Whenever you see a problem involving a moving object on a slope with friction, your go-to framework is the work-energy theorem: the net work done on an object equals its change in kinetic energy. Equivalently, you can think in terms of energy conservation — gravitational potential energy lost goes into either kinetic energy gained or thermal energy from friction. Start by calculating each energy quantity. The change in kinetic energy is $\Delta KE = \frac{1}{2}(20)(6.0^2 - 4.0^2) = \frac{1}{2}(20)(20) = 200\text{ J}$. The gravitational potential energy lost is $\Delta PE = mgh = (20)(9.8)(3.0) = 588\text{ J}$. By energy conservation, the energy converted to thermal energy by friction is $W_f = \Delta PE - \Delta KE = 588 - 200 = 388\text{ J}$. Since friction acts over $12\text{ m}$, the friction force is $f = \frac{388}{12} \approx 32.3\text{ N}$, confirming D. Choice A divides only the kinetic energy gain (200 J) by distance, ignoring that gravity also does work — this severely underestimates friction. Choice B adds the KE increase to the gravitational loss rather than subtracting, which would imply more energy was lost to friction than gravity even supplied — a physical impossibility here. Choice C assigns all of the gravitational PE loss to friction, ignoring that some of that energy legitimately converted into increased kinetic energy. Study tip: Always account for all energy sources and sinks. Set up $W_{gravity} = \Delta KE + W_{friction}$ at the start of any slope-with-friction problem so you never accidentally omit a term.
4. A $0.50\text{ kg}$ pendulum bob on a $1.6\text{ m}$ string is released from rest at a position $0.80\text{ m}$ above its lowest point. Air resistance is negligible. Using $g=9.8\text{ m/s}^2$, what is the tension in the string as the bob passes through its lowest point?
   Answer: $9.8\text{ N}$, because the string must support the weight and provide centripetal force When a pendulum bob swings through its lowest point, two key physics ideas intersect: energy conservation and circular motion. At the bottom of the swing, the string must both support the bob's weight and supply the centripetal force needed to keep it moving in a circular arc — so tension will always exceed the bob's weight at that point. Start with energy conservation to find the speed at the bottom. The bob falls $h = 0.80\text{ m}$, so: $v^2 = 2gh = 2(9.8)(0.80) = 15.68\text{ m}^2/\text{s}^2$ At the lowest point, applying Newton's second law along the string (upward positive): $T - mg = \frac{mv^2}{r}$ $T = mg + \frac{mv^2}{r} = (0.50)(9.8) + \frac{(0.50)(15.68)}{1.6} = 4.9 + 4.9 = 9.8\text{ N}$ This confirms C is correct. A is wrong because it ignores centripetal acceleration entirely — simply setting $T = mg$ would only be true if the bob were hanging stationary, not swinging. B suggests only a "fraction" of weight contributes to the net force, which misunderstands the geometry; at the lowest point, the full weight opposes the tension, and the full centripetal term adds to it. D incorrectly uses the drop height $0.80\text{ m}$ as the radius of circular motion — the radius is the string length $1.6\text{ m}$, not the vertical displacement. As a study tip: whenever a problem involves circular motion at a specific point on a curved path, always draw a free-body diagram, identify which direction is "toward the center," and write $\Sigma F = \frac{mv^2}{r}$ in that direction before plugging in numbers.
5. A $0.50\text{ kg}$ object is launched vertically upward from a platform $1.2\text{ m}$ above the ground by a spring with constant $800\text{ N/m}$ compressed by $0.25\text{ m}$. While the spring expands, a constant friction force of $4.0\text{ N}$ acts over the full $0.25\text{ m}$ distance. Neglect air resistance and use $g=9.8\text{ m/s}^2$. What maximum height above the ground does the object reach?
   Answer: $6.10\text{ m}$, by correctly subtracting frictional energy from spring energy and then adding the platform height When a spring launches an object against friction, energy conservation is your tool — but you must account for every energy transformation carefully. Start by identifying all energy inputs and losses before converting to height. The spring stores elastic potential energy: $E_{spring} = \frac{1}{2}kx^2 = \frac{1}{2}(800)(0.25)^2 = 25\text{ J}$. Friction removes energy over the compression distance: $W_{friction} = (4.0)(0.25) = 1.0\text{ J}$. The net kinetic energy leaving the spring is $25 - 1 = 24\text{ J}$. This energy converts entirely to gravitational potential energy as the object rises: $\Delta h = \frac{24}{mg} = \frac{24}{(0.50)(9.8)} \approx 4.90\text{ m}$ above the platform. Adding the platform height gives $4.90 + 1.2 = 6.10\text{ m}$ — confirming C is correct. Choice A gets the rise above the platform right ($4.90\text{ m}$) but forgets to add the $1.2\text{ m}$ platform height, leaving out the object's initial elevation. Choice B incorrectly mixes the platform height into the energy calculation — adding $1.2\text{ m}$ to the spring energy before dividing confuses meters with joules, producing a meaningless hybrid. Choice D skips subtracting friction entirely, using the full $25\text{ J}$: $\frac{25}{4.9} + 1.2 \approx 6.30\text{ m}$ — a classic trap when friction seems "small enough to ignore." A reliable strategy: always convert net energy to height first, then add any baseline height. Keep energy and geometry as separate steps, and never skip friction just because the force looks small.
6. A wave pulse carrying $40\text{ J}$ reaches a boundary between two media. The boundary absorbs no energy. The reflected pulse travels back through the original medium with one-half the amplitude of the incident pulse and the same shape and duration. How much energy is reflected, and how much is transmitted?
   Answer: $10\text{ J}$ is reflected, and $30\text{ J}$ is transmitted When a wave pulse reaches a boundary, its energy depends on amplitude. The key relationship to remember is that wave energy is proportional to the square of the amplitude: $E \propto A^2$. This means if the amplitude is halved, the energy becomes $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ of the original. Here, the incident pulse carries $40\text{ J}$ and the reflected pulse has half the amplitude. Applying the squared relationship, the reflected energy is $\frac{1}{4} \times 40\text{ J} = 10\text{ J}$. Since the boundary absorbs no energy, conservation of energy requires that the transmitted pulse carries the remaining $40 - 10 = 30\text{ J}$. That confirms answer A is correct. Looking at the wrong choices: B assumes energy splits in the same ratio as amplitude — that is, half the amplitude means half the energy, giving $20\text{ J}$ reflected. This is the most common trap; students forget that energy scales with amplitude squared, not linearly. C makes a similar linear error for the reflected pulse (getting $10\text{ J}$ correctly by coincidence of a different flawed method) but then ignores conservation of energy entirely, claiming the transmitted pulse somehow retains the full $40\text{ J}$ — impossible when energy must be conserved. D combines the linear-amplitude error ($20\text{ J}$ reflected) with correct conservation math, but the starting reflection value is wrong. Study tip: Whenever amplitude changes, always square the ratio to find the energy change — $E \propto A^2$ is one of the most frequently tested wave relationships. Then use conservation of energy to find the transmitted portion.
7. Two electric cars start from rest on level roads with negligible resistance. Each motor delivers the same constant mechanical power. Car B has twice the mass of car A. If car A reaches speed $v$ in time $t$, how long will car B take to reach speed $1.5v$?
   Answer: $4.5t$, because both the larger mass and squared speed increase the energy Whenever you see a question about constant power and changing speeds, your instinct should be to use the work-energy theorem, not kinematics. Constant power means the engine delivers energy at a fixed rate, so the time to reach a speed depends entirely on how much kinetic energy must be built up. Start with the key relationship: since power $P$ is constant, $P = \frac{\Delta KE}{t}$, which gives $t = \frac{\Delta KE}{P} = \frac{\frac{1}{2}mv^2}{P}$. For car A starting from rest: $t = \frac{\frac{1}{2}m_A v^2}{P}$. For car B, with mass $m_B = 2m_A$ reaching speed $1.5v$: $t_B = \frac{\frac{1}{2}(2m_A)(1.5v)^2}{P} = \frac{\frac{1}{2}(2m_A)(2.25v^2)}{P} = \frac{2 \times 2.25}{1} \cdot \frac{\frac{1}{2}m_A v^2}{P} = 4.5t$. The answer is D. Choice A is wrong because time isn't simply proportional to speed — kinetic energy involves $v^2$, so doubling the speed quadruples the energy needed, not doubles it. Choice B correctly identifies that speed is squared, giving the factor of $2.25$ from $(1.5)^2$, but it ignores the mass doubling entirely, giving only $2.25t$. Choice C accounts for both mass and speed linearly ($2 \times 1.5 = 3$), but speed enters as $v^2$, not $v$ — this is the most tempting trap. Study tip: Anytime you see "constant power" on an exam, immediately write $t = \frac{mv^2}{2P}$ and identify which variables change. Mass and $v^2$ multiply together — miss either one and you'll fall into one of the designed traps.
8. A roller-coaster car of mass $1200\text{ kg}$ is moving at $5.0\text{ m/s}$ at the top of a hill that is $20\text{ m}$ above the bottom. During the descent, friction converts $30\text{ kJ}$ of mechanical energy into thermal energy. Using $g=9.8\text{ m/s}^2$, what is the car's speed at the bottom?
   Answer: $19.2\text{ m/s}$, after accounting for the energy transformed by friction Whenever you see a roller-coaster or falling-object problem involving friction, reach for the Work-Energy Theorem with energy losses: the mechanical energy at the bottom equals the mechanical energy at the top minus the energy lost to friction. Start by calculating the total mechanical energy at the top of the hill: $KE_i = \frac{1}{2}(1200)(5.0)^2 = 15{,}000 \text{ J}$ $PE_i = (1200)(9.8)(20) = 235{,}200 \text{ J}$ $E_{top} = 15{,}000 + 235{,}200 = 250{,}200 \text{ J}$ Subtract the 30,000 J lost to friction: $E_{bottom} = 250{,}200 - 30{,}000 = 220{,}200 \text{ J}$ At the bottom, all remaining energy is kinetic, so solve for speed: $v = \sqrt{\frac{2(220{,}200)}{1200}} = \sqrt{367} \approx 19.2 \text{ m/s}$ This confirms B is correct. A is wrong because it claims the initial kinetic energy is "lost" — in reality, the car retains and gains energy through descent; only friction removes energy from the system. C ignores friction entirely and assumes a perfectly conservative system, giving an inflated speed of about 20.4 m/s. D makes the opposite error, adding the frictional energy rather than subtracting it — friction always removes mechanical energy, never adds it. A reliable strategy: when friction is present, always subtract the thermal energy from total mechanical energy before solving. If your final speed is higher than the frictionless case, you've made an error in direction.
9. A block slides down a rough slope. Gravitational PE drops 50 J; KE rises 30 J. How much energy becomes thermal?
   Answer: 20 J Gravitational PE drops 50 J, but KE rises only 30 J. The missing 20 J is converted to thermal energy by friction. The tempting -20 J is the change in mechanical energy, not the amount of energy that becomes thermal, which is positive 20 J.
10. A $0.40\text{ kg}$ cart moving at $6.0\text{ m/s}$ collides with a stationary $0.80\text{ kg}$ cart. The carts stick together and then compress an ideal spring with spring constant $300\text{ N/m}$ on a frictionless track. What is the maximum spring compression?
    Answer: $0.126\text{ m}$, using momentum first and energy after the collision When a question involves a collision followed by another energy interaction, you need to treat it in two distinct phases — this is the key insight being tested here. Phase 1: The collision. The carts stick together, making this a perfectly inelastic collision. Momentum is conserved, but kinetic energy is not. Using conservation of momentum: $p_i = (0.40)(6.0) = 2.4 \text{ kg}\cdot\text{m/s}$ $v_f = \frac{2.4}{0.40 + 0.80} = 2.0 \text{ m/s}$ Phase 2: Spring compression. Now the combined 1.20 kg system compresses the spring. Since the track is frictionless, all remaining kinetic energy converts to spring potential energy: $\frac{1}{2}(1.20)(2.0)^2 = \frac{1}{2}(300)x^2$ $x = \sqrt{\frac{2.40}{300}} \approx 0.089... \text{ m}$ Wait — rechecking: $\frac{1}{2}(1.20)(4.0) = 2.40\text{ J}$, so $x = \sqrt{\frac{2(2.40)}{300}} \approx 0.126\text{ m}$. This confirms B is correct. Choice A is wrong because it ignores the lighter moving cart and only uses the heavier cart's energy — there's no physical justification for this. Choice C assumes a perfectly elastic collision, which contradicts the problem statement that the carts stick together. Choice D skips the collision entirely, feeding all of the initial kinetic energy ($\frac{1}{2}(0.40)(6.0)^2 = 7.2\text{ J}$) into the spring — but energy is lost during the inelastic collision and isn't available afterward. Your strategy: whenever a problem has two sequential events (collision, then spring/ramp/etc.), solve them separately. Never carry pre-collision energy past an inelastic collision.
11. A $0.20\text{ kg}$ metal block at $200^\circ\text{C}$ is placed in $0.50\text{ kg}$ of water at $20^\circ\text{C}$ inside an insulated container. The container has a heat capacity of $300\text{ J/}^\circ\text{C}$ and is initially at $20^\circ\text{C}$. The metal's specific heat capacity is $900\text{ J/(kg}\cdot^\circ\text{C)}$, and water's is $4200\text{ J/(kg}\cdot^\circ\text{C)}$. What final equilibrium temperature is reached?
    Answer: $32.6^\circ\text{C}$, after including energy gained by both the water and the container When a hot object is placed in a cooler system, heat flows until everything reaches the same final temperature. The key principle is conservation of energy: heat lost by the hot object equals heat gained by everything else — including the container, which students often forget. Set up the equation as $Q_{lost} = Q_{gained}$: $m_{metal} c_{metal} (T_i - T_f) = m_{water} c_{water} (T_f - T_i) + C_{container}(T_f - T_i)$ $0.20 \times 900 \times (200 - T_f) = 0.50 \times 4200 \times (T_f - 20) + 300 \times (T_f - 20)$ $180(200 - T_f) = 2100(T_f - 20) + 300(T_f - 20)$ $36000 - 180T_f = 2400(T_f - 20)$ $36000 - 180T_f = 2400T_f - 48000$ $84000 = 2580T_f \Rightarrow T_f \approx 32.6^\circ\text{C}$ This confirms A is correct — it properly accounts for heat absorbed by both the water and the container. B is wrong because it ignores the temperature difference when calculating heat transfer; you can't just use mass and specific heat without multiplying by $\Delta T$. C is tempting but incorrect — omitting the container's heat capacity (300 J/°C) means less energy is "absorbed," artificially raising the calculated final temperature to 34.2°C. D is a naive average of 200°C and 20°C, completely ignoring the different heat capacities involved. Your study tip: always list every object in thermal contact — container, calorimeter, thermometer — and include each one's heat capacity in your energy balance. Forgetting even one can cost you the answer.
12. A block is released from height $h$ above the bottom of a frictionless loop of radius $R$. Minimum $h$ to complete loop?
    Answer: $2.5R$ At the top of the loop the block needs enough speed so centripetal force is at least its weight: $v^2$ is at least gR. Its height at the top is 2R, so energy conservation gives gh = g(2R) + (1/2)$v^2$, which is at least 2gR + (1/2)gR = 2.5gR. Thus h = 2.5R. The tempting wrong answer is 2R, which only gets you to the top but forgets that you must still have speed there to stay on the track.
13. A projectile is launched at $30^\circ$ above horizontal. At peak, what fraction of initial kinetic energy is potential?
    Answer: $\frac{1}{4}$ At the peak, only horizontal motion remains, so speed is v cos 30 = (sqrt 3 / 2) v. The kinetic energy there is (3/4) of the launch kinetic energy because $\cos^2$ 30 = 3/4. Since total mechanical energy is conserved, the rest, 1/4, has become gravitational potential energy. Don't pick 3/4: that is the kinetic energy still present, not the potential energy gained.
14. An electric motor operates from a $240\text{ V}$ source and draws a current of $12\text{ A}$ for $15\text{ s}$. The motor is $75\%$ efficient and lifts a $180\text{ kg}$ load at constant speed. Using $g=9.8\text{ m/s}^2$, through what vertical distance is the load lifted?
    Answer: $18.4\text{ m}$, after converting useful electrical energy into gravitational energy When a question involves a motor lifting a load, you're dealing with an energy chain: electrical energy in → useful mechanical energy out → gravitational potential energy stored. Efficiency tells you how much of the input actually becomes useful output. Start by finding the total electrical energy supplied: $E_{input} = V \times I \times t = 240 \times 12 \times 15 = 43{,}200 \text{ J}$. Since the motor is 75% efficient, the useful mechanical energy delivered to the load is $E_{useful} = 0.75 \times 43{,}200 = 32{,}400 \text{ J}$. This useful energy all goes into lifting the load, so $E_{useful} = mgh$, giving $h = \frac{32{,}400}{180 \times 9.8} = \frac{32{,}400}{1764} \approx 18.4 \text{ m}$. That confirms B is correct. A is wrong because it applies the 75% efficiency factor twice — once to get useful energy and again when solving for height. Efficiency is applied only once, to find useful output from input. C ignores efficiency entirely, assuming all 43,200 J becomes gravitational PE. That gives $h \approx 24.5 \text{ m}$, which is the result before accounting for energy lost as heat in the motor. D is physically nonsensical — efficiency loss represents wasted energy (heat, friction), not a bonus. Treating it as extra output would give a height larger than even the 100%-efficiency case, which violates conservation of energy. A useful strategy: always trace the energy flow step by step — input → apply efficiency → useful output → set equal to the relevant energy formula. Never skip or double-apply the efficiency factor.
15. A rechargeable battery has a potential difference of $12\text{ V}$ and delivers $2.4\text{ A}\cdot\text{h}$ of charge before it is depleted. A motor converts $80\%$ of the battery's electrical energy into gravitational potential energy. Neglecting other losses, approximately how high can the motor lift a $50\text{ kg}$ mass? Use $g=9.8\text{ m/s}^2$.
    Answer: $169\text{ m}$, converting charge to energy and then applying efficiency When you see a question combining battery specs with efficiency and gravitational potential energy, your roadmap is: (1) find total electrical energy stored, (2) apply efficiency to find useful energy, (3) set equal to $mgh$ and solve. Start by converting the battery's charge: $2.4 \text{ A}\cdot\text{h} = 2.4 \times 3600 = 8640 \text{ C}$. The total electrical energy stored is $E = QV = 8640 \times 12 = 103{,}680 \text{ J}$. Since the motor is 80% efficient, the useful energy converted to gravitational PE is $E_{useful} = 0.80 \times 103{,}680 = 82{,}944 \text{ J}$. Setting this equal to $mgh$: $h = \frac{82{,}944}{50 \times 9.8} \approx 169 \text{ m}$. That confirms C is correct. A is wrong because it skips the conversion from ampere-hours to coulombs, effectively treating "2.4 A·h" as if it were 2.4 C. This dramatically underestimates the stored charge and yields a much smaller height around 16.9 m — a classic unit-conversion trap. B goes wrong by applying the 80% efficiency factor twice — once to the energy and again somewhere in the calculation — which cuts the available energy to 64% instead of 80%, giving an inflated-looking answer of 135 m (actually too small due to the double reduction). D ignores efficiency entirely, using the full $103{,}680 \text{ J}$ as gravitational PE, producing $\approx 212 \text{ m}$ — a common mistake when students overlook the word "converts 80%." Study tip: Always convert A·h → C (multiply by 3600) before calculating energy, and apply efficiency once at the energy step before solving for height.

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