This question tests understanding of how mass affects kinetic energy when speed is constant across different objects. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². The straight line indicates proportionality (double mass → double KE), so the cart with the largest mass will have the greatest KE at the same speed. Choice C is correct because it identifies the cart with the greatest mass (5 kg) as having the highest KE, following the direct proportionality. Choice D is wrong because it states all have the same KE, but KE increases with mass at constant speed; Choices A and B select lighter carts, which would have less KE. Graphing relationships helps visualize patterns: plotting KE vs mass for these carts would show points on a straight line through origin, with the highest point (greatest KE) at the largest mass. Interpreting graph: the position along the line shows that larger x (mass) corresponds to larger y (KE), confirming the heaviest has the most KE.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following $KE = \frac{1}{2} m v^2$, which becomes $KE = \left( \frac{1}{2} v^2 \right) m$ when v is constant—this is a linear equation in the form $y = m x$ (where y=KE, x=m, slope=\frac{1}{2} v^2$), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to $\frac{1}{2} v^2$. The straight line indicates proportionality (double mass → double KE), the origin passage confirms that zero mass has zero KE (sensible physically), and the slope value tells you the constant speed: steeper slope means higher speed (since slope = $\frac{1}{2} v^2$, larger slope requires larger v), while gentler slope means lower speed. For finding mass given KE: From the data pattern, the rate is 12.5 J per kg (since each kg adds 12.5 J of KE at $v = 5 \text{ m/s}$), so to have 100 J: mass = KE/rate = 100 J ÷ 12.5 J/kg = 8 kg; alternatively, using $KE = \frac{1}{2} m v^2$ with KE = 100 J and v = 5 m/s: 100 = $\frac{1}{2} m (25)$, so m = 200/25 = 8 kg. Choice C is correct because 8 kg mass would have 100 J kinetic energy at $v = 5 \text{ m/s}$, found by dividing desired KE by the rate of 12.5 J/kg. Choice A (4 kg) would only have 50 J as shown in data; Choice B (6 kg) would have 75 J as shown in data; Choice D (10 kg) would have 125 J, exceeding the target of 100 J.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following $KE = \frac{1}{2} m v^2$, which becomes $KE = \left( \frac{1}{2} v^2 \right) m$ when v is constant—this is a linear equation in the form $y = m x$ (where y=KE, x=m, slope=\frac{1}{2} v^2$), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to $\frac{1}{2} v^2$. The straight line indicates proportionality (double mass → double KE), the origin passage confirms that zero mass has zero KE (sensible physically), and the slope value tells you the constant speed: steeper slope means higher speed (since slope = $\frac{1}{2} v^2$, larger slope requires larger v), while gentler slope means lower speed. For interpolating between data points: The data shows a linear relationship where each kg of mass adds 12.5 J of kinetic energy (slope = 12.5 J/kg), so for 3 kg (between the given 2 kg and 4 kg data points), we can calculate: $KE = \text{slope} \times \text{mass} = 12.5 , \text{J/kg} \times 3 , \text{kg} = 37.5 , \text{J}$, or alternatively, 3 kg is halfway between 2 kg (25 J) and 4 kg (50 J), so KE is halfway between 25 J and 50 J, which is 37.5 J. Choice B is correct because 37.5 J is the kinetic energy at 3 kg mass, found by linear interpolation or using the constant rate of 12.5 J per kg. Choice A (18.75 J) is wrong—this would be the KE for 1.5 kg, not 3 kg; Choice C (62.5 J) would be the KE for 5 kg, not 3 kg; Choice D (75 J) is the KE for 6 kg as shown in the data, not for 3 kg.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². The straight line indicates proportionality (double mass → double KE), the origin passage confirms that zero mass has zero KE (sensible physically), and the slope value tells you the constant speed: steeper slope means higher speed (since slope = ½v², larger slope requires larger v), while gentler slope means lower speed. For comparing graphs at different speeds: Line 1 at v = 2 m/s has slope = ½(2²) = 2 J/kg (gentle slope), while Line 2 at v = 5 m/s has slope = ½(5²) = 12.5 J/kg (steeper slope)—the higher speed produces a steeper line because each kilogram of mass contributes more kinetic energy when moving faster (KE per kg = ½v² increases with v). Choice B is correct because it states "Line 2 is steeper because higher speed gives more kinetic energy per kilogram," accurately describing how slope = ½v² makes higher speeds produce steeper lines. Choice A is backwards—lower speed gives less KE per kg and gentler slope; Choice C wrongly claims same slope when slopes differ by speed (2 J/kg vs 12.5 J/kg); Choice D incorrectly states lines shouldn't pass through origin, but KE = 0 when m = 0 requires origin passage for both speeds.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². The straight line indicates proportionality (double mass → double KE), the origin passage confirms that zero mass has zero KE (sensible physically), and the slope value tells you the constant speed: steeper slope means higher speed (since slope = ½v², larger slope requires larger v), while gentler slope means lower speed. For verifying proportionality from data: At v = 3 m/s, the data shows mass 2 kg has KE 9 J, mass 4 kg has KE 18 J (doubled), and mass 6 kg has KE 27 J (tripled from 2 kg)—this confirms direct proportionality where tripling mass from 2 kg to 6 kg triples KE from 9 J to 27 J. Choice A is correct because it states "KE is proportional to mass (if mass triples, KE triples)," which matches the linear relationship KE ∝ m shown in the data. Choice B wrongly claims KE ∝ m² which would make tripling mass increase KE by 9×, but data shows only 3× increase; Choice C incorrectly suggests inverse proportionality where KE decreases as mass increases; Choice D wrongly states KE is the same for all masses, ignoring that data shows different KE values (9 J, 18 J, 27 J) for different masses.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². The straight line indicates proportionality (double mass → double KE), the origin passage confirms that zero mass has zero KE (sensible physically), and the slope value tells you the constant speed: steeper slope means higher speed (since slope = ½v², larger slope requires larger v), while gentler slope means lower speed. For extending the pattern: The data shows a constant rate of 4.5 J per kg (since 9 J/2 kg = 18 J/4 kg = 27 J/6 kg = 4.5 J/kg), which equals ½v² = ½(3²) = 4.5 J/kg confirming v = 3 m/s; therefore, for 7 kg: KE = 4.5 J/kg × 7 kg = 31.5 J, or using the formula directly: KE = ½mv² = ½(7)(3²) = ½(7)(9) = 31.5 J. Choice C is correct because 31.5 J is the kinetic energy for 7 kg mass at v = 3 m/s, found by extending the linear pattern. Choice A (10.5 J) is too small—this would be closer to 2.3 kg; Choice B (18.0 J) is the KE for 4 kg shown in the data; Choice D (49.0 J) is too large—this would require about 10.9 kg at this speed.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². When mass increases from 2 kg to 6 kg at constant v=5 m/s: initial KE = ½(2)(25) = 25 J, final KE = ½(6)(25) = 75 J, so KE increases by factor of 75/25 = 3; alternatively, since KE ∝ m at constant speed, when mass triples (6/2 = 3), KE also triples—this is the defining characteristic of direct proportionality. Choice B is correct because kinetic energy becomes 3 times as large when mass triples from 2 kg to 6 kg, since KE is directly proportional to mass at constant speed (if mass × 3, then KE × 3). Choice A incorrectly suggests KE becomes 1/3 as large, which would indicate inverse proportionality; Choice C incorrectly claims KE becomes 9 times larger, confusing this with the effect of tripling speed (where KE ∝ v² would give 9× increase); Choice D incorrectly states KE doesn't change, missing that KE varies with mass even at constant speed. The factor change in KE equals the factor change in mass because of the proportional relationship. This demonstrates the power of proportional relationships: knowing how one variable changes immediately tells us how the other changes by the same factor.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². For calculating the slope from the graphed data: using any two points from the table, such as (2 kg, 25 J) and (4 kg, 50 J), slope = rise/run = ΔKE/Δm = (50-25)/(4-2) = 25/2 = 12.5 J/kg; we can verify with other points like (1 kg, 12.5 J) and (6 kg, 75 J): slope = (75-12.5)/(6-1) = 62.5/5 = 12.5 J/kg—the consistent slope confirms the linear relationship. Choice C is correct because the slope of the KE vs mass line is 12.5 J/kg, calculated from rise over run using any two data points, and this value equals ½v² = ½(5²) = 12.5 J/kg as expected from the physics. Choice A (2.5 J/kg) would correspond to v ≈ 2.2 m/s; Choice B (6.25 J/kg) would correspond to v ≈ 3.5 m/s; Choice D (25 J/kg) would correspond to v ≈ 7.1 m/s—none match the given speed of 5 m/s. The slope has physical meaning: it represents the kinetic energy per kilogram of mass at the given constant speed, and equals ½v² from the kinetic energy formula. Understanding that slope = ½v² allows us to determine the speed from a KE vs mass graph or predict the slope if we know the speed.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². For interpreting values from the data: since KE is directly proportional to mass at constant speed, we can find KE at 3 kg by recognizing that 3 kg is halfway between 2 kg and 4 kg in the table, so KE at 3 kg must be halfway between 25 J and 50 J, which is (25+50)/2 = 37.5 J; alternatively, using the proportional relationship KE = (½v²)m with ½v² = 12.5 J/kg (found from any data point like 25 J ÷ 2 kg), we get KE = 12.5 × 3 = 37.5 J. Choice B is correct because 37.5 J is the kinetic energy when mass is 3 kg, found either by interpolation between the 2 kg and 4 kg values or by using the proportional relationship KE = 12.5m. Choice A (18.75 J) would be the KE at 1.5 kg, not 3 kg; Choice C (50 J) is the KE at 4 kg from the table; Choice D (75 J) is the KE at 6 kg from the table—these are all valid points but for different masses. The linear relationship means we can find KE for any mass value, not just those in the table, by using the constant ratio KE/m = ½v². This interpolation skill is important for reading values from graphs and understanding that the relationship holds for all mass values, not just the measured ones.

This question tests understanding of how to graph or interpret the relationship between kinetic energy and mass at constant speed, which produces a straight line through the origin. At constant speed, kinetic energy is directly proportional to mass following KE = ½mv², which becomes KE = (½v²)m when v is constant—this is a linear equation in the form y = mx (where y=KE, x=m, slope=½v²), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to ½v². To find mass given KE = 62.5 J at v = 5 m/s: using KE = ½mv² with KE = 62.5 J and v = 5 m/s, we get 62.5 = ½m(25), so 62.5 = 12.5m, therefore m = 62.5/12.5 = 5 kg; alternatively, since KE/m = ½v² = 12.5 J/kg at this speed, we have m = KE/(12.5 J/kg) = 62.5/12.5 = 5 kg. Choice C is correct because a cart with kinetic energy 62.5 J at speed 5 m/s must have mass 5 kg, found by rearranging KE = ½mv² to solve for m = 2KE/v² = 2(62.5)/25 = 5 kg. Choice A (2 kg) would have KE = 25 J at this speed; Choice B (4 kg) would have KE = 50 J; Choice D (6 kg) would have KE = 75 J—we can verify these using KE = 12.5m where 12.5 is the ½v² value. The ability to solve for mass given KE demonstrates understanding of the proportional relationship and how to use it in reverse. This problem reinforces that the KE-mass relationship at constant speed allows us to find any variable given the other, using the constant ratio ½v².