Middle School Science Quiz: Graph Energy And Mass
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Graph Energy And MassQuestion 1 of 20

A cart moves at constant speed v=5 m/sv=5\ \text{m/s}. Based on the KE vs mass data, what is the slope of a graph of kinetic energy (J) vs mass (kg)?

2.5 J/kg
10 J/kg
12.5 J/kg
25 J/kg
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Middle School Science Quiz

Middle School Science Quiz: Graph Energy And Mass

Practice Graph Energy And Mass in Middle School Science 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 Graph Energy And Mass, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Science.

How to use this quiz

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.

All questions

Question 1

A cart moves at constant speed v=5 m/sv=5\ \text{m/s}. Based on the KE vs mass data, what is the slope of a graph of kinetic energy (J) vs mass (kg)?

  1. 2.5 J/kg
  2. 10 J/kg
  3. 12.5 J/kg (correct answer)
  4. 25 J/kg
Explanation: This question tests understanding of how to calculate the slope of a kinetic energy versus mass graph at constant speed. 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 slope is ½v² = 12.5 J/kg, calculated from any two points: using (2, 25) and (6, 75), slope = (75-25)/(6-2) = 50/4 = 12.5 J/kg, and since this equals ½v², we can determine v²=25, so v=5 m/s (the constant speed for this data). Choice C is correct because it correctly identifies the slope as 12.5 J/kg or calculates it from rise/run using the data points. Choice A is wrong because it undercalculates the slope (perhaps using incorrect points); Choice B uses a value not matching ½v² for v=5; Choice D doubles the correct slope, confusing v² with something else. Graphing relationships helps visualize patterns: the slope of KE vs mass line quantitatively represents ½v², allowing you to verify or find the constant speed from the graph. Interpreting graph: (1) identify shape (straight line = linear), (2) check origin (through origin = proportional), (3) find slope (rise/run = ΔKE/Δm, gives ½v² value).

Question 2

Three carts all move at the same constant speed of v=3 m/sv=3\ \text{m/s}. If you extend the pattern, what kinetic energy would you expect for a cart with mass 7 kg7\ \text{kg} at the same speed?

  1. 10.5 J
  2. 18.0 J
  3. 31.5 J (correct answer)
  4. 49.0 J
Explanation: 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.

Question 3

A cart moves at a constant speed of v=5 m/sv=5\ \text{m/s}. On a graph of KE vs mass, what is the slope (rate of change) ΔKE/Δm\Delta \text{KE}/\Delta m based on the data?

  1. 2 J/kg
  2. 6.25 J/kg
  3. 12.5 J/kg (correct answer)
  4. 25 J/kg
Explanation: 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 calculating slope from the data: Using any two points, slope = ΔKE/Δm = rise/run; taking points (2 kg, 25 J) and (4 kg, 50 J): slope = (50-25)/(4-2) = 25/2 = 12.5 J/kg, or using (1 kg, 12.5 J) and (6 kg, 75 J): slope = (75-12.5)/(6-1) = 62.5/5 = 12.5 J/kg—the slope is constant at 12.5 J/kg for all point pairs, confirming linear relationship. Choice C is correct because the slope is 12.5 J/kg, calculated from the constant rate of change in the data. Choice A (2 J/kg) is too small—this would mean only 2 J added per kg; Choice B (6.25 J/kg) is half the correct value; Choice D (25 J/kg) is double the correct value—perhaps confusing with the KE at 2 kg rather than the rate of change.

Question 4

Objects move at a constant speed of v=5 m/sv = 5\ \text{m/s}. The KE vs mass graph is a straight line through the origin.

If the kinetic energy is 62.5 J62.5\ \text{J}, what mass does the graph show (in kg)?

  1. 2 kg
  2. 3 kg
  3. 5 kg (correct answer)
  4. 10 kg
Explanation: 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 completed graph: The graph shows straight line through origin indicating direct proportionality: KE ∝ m (kinetic energy proportional to mass at constant speed); reading the graph: for KE=62.5 J, go from y=62.5 to the line, then down to x-axis at m=5 kg (since slope=12.5 J/kg, m=62.5/12.5=5 kg), demonstrating how to read values from graph. Choice C is correct because it correctly reads the value from the graph: at KE=62.5 J, the mass is 5 kg using the line position and proportional relationship. Choice A is wrong because 2 kg corresponds to KE=25 J (too low); Choice B is wrong because 3 kg is 37.5 J; Choice D is wrong because 10 kg would be 125 J (too high). Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin (regardless of what speed—just different slopes), confirming KE ∝ m proportionality visually, and allowing quantitative reading (find KE for any mass, or mass for any KE by using the line). Interpreting graph: (1) identify shape (straight line = linear, curved = nonlinear), (2) check origin (through origin = proportional, offset = linear but not proportional), (3) find slope (rise/run = ΔKE/Δm, gives ½v² value), (4) read values (pick mass, go up to line, over to KE axis), (5) describe relationship (KE ∝ m: proportional because straight through origin).

Question 5

Four carts all move at a constant speed of v=5 m/sv=5\ \text{m/s}. If a cart has kinetic energy 62.5 J62.5\ \text{J}, what is its mass?

  1. 2 kg
  2. 4 kg
  3. 5 kg (correct answer)
  4. 6 kg
Explanation: 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.

Question 6

A cart moves at a constant speed of v=5m/sv=5 \text{m/s}. Using the data, what is the kinetic energy when the mass is 3kg3 \text{kg}?

  1. 18.75 J
  2. 37.5 J (correct answer)
  3. 62.5 J
  4. 75 J
Explanation: 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=12mv2KE = \frac{1}{2} m v^2, which becomes KE=(12v2)mKE = \left( \frac{1}{2} v^2 \right) m when v is constant—this is a linear equation in the form y=mxy = m x (where y=KE, x=m, slope=\frac{1}{2} v^2),meaningagraphofKEversusmassproducesastraightlinepassingthroughtheoriginwithslopeequalto), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to \frac{1}{2} v^2.Thestraightlineindicatesproportionality(doublemass→doubleKE),theoriginpassageconfirmsthatzeromasshaszeroKE(sensiblephysically),andtheslopevaluetellsyoutheconstantspeed:steeperslopemeanshigherspeed(sinceslope=. 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,largersloperequireslargerv),whilegentlerslopemeanslowerspeed.Forinterpolatingbetweendatapoints:Thedatashowsalinearrelationshipwhereeachkgofmassadds12.5Jofkineticenergy(slope=12.5J/kg),sofor3kg(betweenthegiven2kgand4kgdatapoints),wecancalculate:, 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.

Question 7

A cart moves at a constant speed of v=5 m/sv=5\ \text{m/s}. A student graphs kinetic energy (J) vs mass (kg) and draws a best-fit line through the points. What is the slope of the line (in J/kg\text{J/kg})?

  1. 2.5 J/kg
  2. 6.25 J/kg
  3. 12.5 J/kg (correct answer)
  4. 25 J/kg
Explanation: 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.

Question 8

A cart moves at a constant speed of v=5 m/sv=5\ \text{m/s}. Using the data below, what is the kinetic energy when the mass is 3 kg3\ \text{kg}? (Assume the KE vs mass graph is a straight line through the origin.)

  1. 18.75 J
  2. 37.5 J (correct answer)
  3. 50 J
  4. 62.5 J
Explanation: This question tests understanding of how to interpret a graph of kinetic energy versus mass at constant speed to find intermediate values. 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 completed graph: The graph shows straight line through origin indicating direct proportionality: KE ∝ m (kinetic energy proportional to mass at constant speed). Reading the graph: at mass 3 kg (between plotted points 2 and 4 kg), the line passes through KE ≈ 37.5 J (can interpolate: halfway between 25 J and 50 J), demonstrating how to read values from graph. Choice B is correct because it properly reads the value from the graph using the line position, interpolating to 37.5 J for 3 kg based on the proportional relationship. Choice A is wrong because it underestimates the value (perhaps misreading the scale); Choice C reads a value for a different mass like 4 kg; Choice D overestimates beyond the pattern. Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin, allowing accurate interpolation for unplotted values like 3 kg by finding the point on the line.

Question 9

Different carts all move at the same constant speed of v=3 m/sv=3\ \text{m/s}. Based on the data, which cart has the greatest kinetic energy?

  1. Cart A (1 kg)
  2. Cart B (3 kg)
  3. Cart C (5 kg) (correct answer)
  4. All carts have the same kinetic energy because their speeds are equal.
Explanation: 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.

Question 10

Two straight-line graphs show kinetic energy (KE) vs mass (m) for objects moving at constant speeds. Graph A is for v=5 m/sv=5\ \text{m/s} and Graph B is for v=2 m/sv=2\ \text{m/s}. Both pass through the origin.

What does the slope of each KE vs mass line represent?

  1. The speed vv (in m/s).
  2. The mass mm (in kg).
  3. The value 12v2\tfrac{1}{2}v^2 (in J/kg) for that graph's speed. (correct answer)
  4. The value 12m2\tfrac{1}{2}m^2 (in kg2^2).
Explanation: This question tests understanding of what the slope represents in a kinetic energy versus mass graph at constant speed. 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 comparing graphs at different speeds: Two graphs both show straight lines through origin, but different slopes: at v=2 m/s, slope = ½(4) = 2 J/kg, while at v=5 m/s, slope = ½(25) = 12.5 J/kg—the slope value is ½v² for that graph's speed. Choice C is correct because it correctly identifies the slope as ½v² (in J/kg) for that graph's speed. Choice A is wrong because slope isn't v directly (it's ½v²); Choice B confuses it with mass, which varies; Choice D uses ½m², which isn't relevant. Graphing relationships helps visualize patterns: the slope of KE vs mass line quantitatively represents ½v², allowing you to determine the constant speed from the graph (v = sqrt(2 × slope)). Interpreting graph: find slope (rise/run = ΔKE/Δm, gives ½v² value), which differs between graphs based on their speeds.

Question 11

A cart moves at constant speed v=5 m/sv=5 \text{ m/s}. Using the data, what mass would have a kinetic energy of 100 J100 \text{ J} at this same speed?

  1. 4 kg
  2. 6 kg
  3. 8 kg (correct answer)
  4. 10 kg
Explanation: 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=12mv2KE = \frac{1}{2} m v^2, which becomes KE=(12v2)mKE = \left( \frac{1}{2} v^2 \right) m when v is constant—this is a linear equation in the form y=mxy = m x (where y=KE, x=m, slope=\frac{1}{2} v^2),meaningagraphofKEversusmassproducesastraightlinepassingthroughtheoriginwithslopeequalto), meaning a graph of KE versus mass produces a straight line passing through the origin with slope equal to \frac{1}{2} v^2.Thestraightlineindicatesproportionality(doublemass→doubleKE),theoriginpassageconfirmsthatzeromasshaszeroKE(sensiblephysically),andtheslopevaluetellsyoutheconstantspeed:steeperslopemeanshigherspeed(sinceslope=. 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,largersloperequireslargerv),whilegentlerslopemeanslowerspeed.ForfindingmassgivenKE:Fromthedatapattern,therateis12.5Jperkg(sinceeachkgadds12.5JofKEat, 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}),sotohave100J:mass=KE/rate=100J÷12.5J/kg=8kg;alternatively,using), so to have 100 J: mass = KE/rate = 100 J ÷ 12.5 J/kg = 8 kg; alternatively, using KE = \frac{1}{2} m v^2withKE=100Jandv=5m/s:100=with KE = 100 J and v = 5 m/s: 100 =\frac{1}{2} m (25),som=200/25=8kg.ChoiceCiscorrectbecause8kgmasswouldhave100Jkineticenergyat, 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.

Question 12

Two straight-line graphs show KE (J) vs mass (kg), both passing through the origin.

  • Line 1 is for constant speed v=2 m/sv=2\ \text{m/s}.
  • Line 2 is for constant speed v=5 m/sv=5\ \text{m/s}.

Which statement is true?

  1. Line 1 is steeper because lower speed gives more kinetic energy per kilogram.
  2. Line 2 is steeper because higher speed gives more kinetic energy per kilogram. (correct answer)
  3. Both lines have the same slope because KE depends only on mass.
  4. Neither line should pass through the origin because objects can have KE at zero mass.
Explanation: 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.

Question 13

A graph of kinetic energy (KE) vs mass is a straight line through the origin. The plotted points include (1 kg, 12.5 J), (2 kg, 25 J), and (4 kg, 50 J).

What is the kinetic energy at a mass of 5 kg5\ \text{kg} according to the line?

  1. 50 J
  2. 62.5 J (correct answer)
  3. 75 J
  4. 100 J
Explanation: 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 reading values from the graph: The points show a constant rate of 12.5 J per kg (since 12.5 J/1 kg = 25 J/2 kg = 50 J/4 kg = 12.5 J/kg), so for 5 kg on this line: KE = 12.5 J/kg × 5 kg = 62.5 J, which can also be found by noting 5 kg is halfway between 4 kg (50 J) and 6 kg (which would be 75 J), so KE is halfway between 50 J and 75 J = 62.5 J. Choice B is correct because 62.5 J is the kinetic energy at 5 kg mass, found by extending the linear pattern shown by the plotted points. Choice A (50 J) is the KE for 4 kg shown in the data, not 5 kg; Choice C (75 J) would be the KE for 6 kg, not 5 kg; Choice D (100 J) would require 8 kg at this rate of 12.5 J/kg.

Question 14

A cart moves at a constant speed of v=5 m/sv=5\ \text{m/s}. You graph kinetic energy (J) on the vertical axis and mass (kg) on the horizontal axis using the data shown. Which graph description is correct?

  1. A straight line through the origin (0,0). (correct answer)
  2. A curved line that gets steeper as mass increases.
  3. A straight line that does NOT pass through the origin.
  4. A horizontal line because speed is constant.
Explanation: 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 creating a graph from the given data at v=5 m/s: plotting points (1, 12.5), (2, 25), (4, 50), (6, 75) on axes with mass horizontal and KE vertical, we see all points fall on a straight line that passes through the origin (0, 0)—this origin passage is required because zero mass must have zero kinetic energy, and the straight line confirms the linear relationship KE = (constant)×m. Choice A is correct because it accurately describes the graph as a straight line through the origin, which matches both the mathematical relationship (KE = ½mv² gives a linear equation in m when v is constant) and the plotted data points. Choice B incorrectly describes a curved line, but the relationship is linear not quadratic in mass; Choice C incorrectly states the line doesn't pass through origin, but it must because KE=0 when m=0; Choice D incorrectly suggests a horizontal line, confusing constant speed with constant KE—speed is constant but KE still varies with mass. The straight line through origin visually confirms the proportional relationship: any point on the line shows that KE/m ratio is constant (equal to ½v²). This graphical representation helps students see that doubling mass doubles KE, tripling mass triples KE, and so on, making the proportional relationship clear.

Question 15

Three carts move at the same constant speed of v=3 m/sv=3\ \text{m/s}. Which statement best describes the pattern between mass and kinetic energy shown by the data?

  1. Kinetic energy increases by a constant amount of 9 J each time mass increases by 1 kg.
  2. Kinetic energy is directly proportional to mass (a straight-line relationship through the origin). (correct answer)
  3. Kinetic energy is proportional to the square of mass.
  4. Kinetic energy decreases as mass increases.
Explanation: 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. Analyzing the data pattern at v=3 m/s: for 2 kg cart, KE = 9 J; for 4 kg cart, KE = 18 J; for 6 kg cart, KE = 27 J—notice that when mass doubles from 2 to 4 kg, KE doubles from 9 to 18 J, and when mass triples from 2 to 6 kg, KE triples from 9 to 27 J, confirming direct proportionality where KE/m = constant = 4.5 J/kg for all data points. Choice B is correct because it accurately describes the pattern as kinetic energy being directly proportional to mass, which means a straight-line relationship through the origin—this matches both the data (constant KE/m ratio) and the physics (KE = ½mv² linear in m). Choice A seems correct noting the 9 J increase per kg, but this describes a linear relationship, not specifically a proportional one—the key distinction is that proportional relationships pass through origin while merely linear ones might not; Choice C incorrectly suggests KE ∝ m², which would show a parabolic curve; Choice D incorrectly claims KE decreases with mass, opposite of the data trend. The proportional relationship means the ratio KE/m is constant, equaling ½v² = ½(3²) = 4.5 J/kg. This demonstrates that at any constant speed, mass and kinetic energy maintain a proportional relationship, graphically shown as a straight line through the origin.

Question 16

Two straight-line graphs show kinetic energy (KE) vs mass for objects moving at constant speeds. One line is for v=2 m/sv=2\ \text{m/s} and the other is for v=5 m/sv=5\ \text{m/s}. Both lines pass through the origin.

Which line should be steeper, and why?

  1. The v=2 m/sv=2\ \text{m/s} line, because lower speed gives more KE per kilogram.
  2. The v=5 m/sv=5\ \text{m/s} line, because at constant speed the slope is 12v2\tfrac{1}{2}v^2, which is larger for 5 m/s. (correct answer)
  3. The lines have the same slope, because KE depends only on mass.
  4. The v=2 m/sv=2\ \text{m/s} line, because KE is proportional to vv (not v2v^2).
Explanation: 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 comparing graphs at different speeds: Two graphs both show straight lines through origin (both demonstrate KE ∝ m), but different slopes: at v=2 m/s, slope = ½(4) = 2 J/kg (gentle slope: KE increases slowly with mass), while at v=5 m/s, slope = ½(25) = 12.5 J/kg (steep slope: KE increases rapidly with mass)—higher speed creates steeper line because same mass increase adds more KE when speed is higher (KE = ½mv² means KE per kg is ½v², larger for larger v). Choice B is correct because it correctly identifies slope as ½v² or calculates it from rise/run. Choice A is wrong because it compares graphs incorrectly: claims gentler slope at higher speed when steeper slope indicates higher speed (slope = ½v², larger v means larger slope). Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin (regardless of what speed—just different slopes), confirming KE ∝ m proportionality visually, and allowing quantitative reading (find KE for any mass, or mass for any KE by using the line). Both pass through origin (both speeds give KE=0 for m=0), and both are straight lines (linearity of mass effect universal), but vertical separation increases with mass (at m=4 kg: high speed has much more KE than low speed).

Question 17

At a constant speed of v=5 m/sv = 5\ \text{m/s}, a line graph of KE (J) vs mass (kg) includes the points (1,12.5)(1, 12.5), (2,25)(2, 25), and (4,50)(4, 50) and is a straight line through the origin.

What is the kinetic energy when the mass is 3 kg3\ \text{kg}?

  1. 18.75 J
  2. 37.5 J (correct answer)
  3. 62.5 J
  4. 75 J
Explanation: 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 completed graph: The graph shows straight line through origin indicating direct proportionality: KE ∝ m (kinetic energy proportional to mass at constant speed); reading the graph: at mass 3 kg (between plotted points 2 and 4 kg), the line passes through KE = 37.5 J (can interpolate: halfway between 25 J and 50 J since slope is constant), demonstrating how to read values from graph; the slope is ½v² = 12.5 J/kg, calculated from any two points: using (2, 25) and (4, 50), slope = (50-25)/(4-2) = 25/2 = 12.5 J/kg, and since this equals ½v², we can determine v²=25, so v=5 m/s (the constant speed for this data). Choice B is correct because it properly reads the value from the graph using the line position at 3 kg, which is 37.5 J based on the proportional relationship. Choice A is wrong because it reads a wrong value from the graph (18.75 J is too low, perhaps misidentifying point location or extrapolating incorrectly); Choice C is wrong because 62.5 J would correspond to a higher mass like 5 kg; Choice D is wrong because 75 J is for 6 kg, not 3 kg. Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin (regardless of what speed—just different slopes), confirming KE ∝ m proportionality visually, and allowing quantitative reading (find KE for any mass, or mass for any KE by using the line). Interpreting graph: (1) identify shape (straight line = linear, curved = nonlinear), (2) check origin (through origin = proportional, offset = linear but not proportional), (3) find slope (rise/run = ΔKE/Δm, gives ½v² value), (4) read values (pick mass, go up to line, over to KE axis), (5) describe relationship (KE ∝ m: proportional because straight through origin).

Question 18

A student makes a KE (J) vs mass (kg) graph for an object moving at constant speed. The line goes through the origin and has a slope of 12.5 J/kg12.5\ \text{J/kg}.

What constant speed was the object moving at?

  1. v=2 m/sv=2\ \text{m/s}
  2. v=4 m/sv=4\ \text{m/s}
  3. v=5 m/sv=5\ \text{m/s} (correct answer)
  4. v=10 m/sv=10\ \text{m/s}
Explanation: 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 slope is ½v² = 12.5 J/kg, and since this equals ½v², we can determine v²=25, so v=5 m/s (the constant speed for this data). Choice C is correct because it correctly identifies slope as ½v² or calculates it from rise/run. Choice D is wrong because it states slope is mass or KE when slope is ½v² (units: J/kg confirms this). Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin (regardless of what speed—just different slopes), confirming KE ∝ m proportionality visually, and allowing quantitative reading (find KE for any mass, or mass for any KE by using the line). Interpreting graph: (1) identify shape (straight line = linear, curved = nonlinear), (2) check origin (through origin = proportional, offset = linear but not proportional), (3) find slope (rise/run = ΔKE/Δm, gives ½v² value), (4) read values (pick mass, go up to line, over to KE axis), (5) describe relationship (KE ∝ m: proportional because straight through origin).

Question 19

Different carts all move at the same constant speed of v=3 m/sv=3\ \text{m/s}. If you graph kinetic energy (J) vs mass (kg), what pattern should the points make?

  1. A straight line through the origin because KE is proportional to mass at constant speed. (correct answer)
  2. A curved line because KE is proportional to m2m^2.
  3. A horizontal line because speed is constant.
  4. A line sloping downward because larger mass means less KE.
Explanation: This question tests understanding of the expected pattern in a graph of kinetic energy versus mass at constant speed. 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 graph shows straight line through origin indicating direct proportionality: KE ∝ m (kinetic energy proportional to mass at constant speed). Choice A is correct because it accurately describes the graph as a straight line through the origin because KE is proportional to mass at constant speed. Choice B is wrong because it confuses mass and speed relationships: applies parabolic curve for mass when that's for speed (KE vs v curved, KE vs m straight); Choice C describes a horizontal line, but KE increases with mass; Choice D claims a downward slope, but the relationship is positive. Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin, confirming KE ∝ m proportionality visually. Comparing to KE vs speed graph: very different shapes (mass gives straight line, speed gives parabola—curved), demonstrates mass and speed affect KE differently (m linear factor, v squared factor).

Question 20

A student plots kinetic energy (KE) vs mass for an object moving at a constant speed. The plotted points include (1 kg, 12.5 J)(1\ \text{kg},\ 12.5\ \text{J}), (2 kg, 25 J)(2\ \text{kg},\ 25\ \text{J}), and (4 kg, 50 J)(4\ \text{kg},\ 50\ \text{J}).

Which statement best describes the relationship shown by the graph?

  1. KE is inversely proportional to mass: doubling mass halves KE.
  2. KE is directly proportional to mass: doubling mass doubles KE. (correct answer)
  3. KE is proportional to the square of mass: doubling mass quadruples KE.
  4. KE does not depend on mass at constant speed.
Explanation: 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. Choice B is correct because it accurately interprets linear relationship as KE ∝ m. Choice C is wrong because it interprets relationship as inverse or squared when clearly direct proportional (linear). Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin (regardless of what speed—just different slopes), confirming KE ∝ m proportionality visually, and allowing quantitative reading (find KE for any mass, or mass for any KE by using the line). Comparing to KE vs speed graph: very different shapes (mass gives straight line, speed gives parabola—curved), demonstrates: mass and speed affect KE differently (m linear factor, v squared factor), both graphs through origin (both give KE=0 when respective variable is 0), provides visual understanding: straight vs curved shows different mathematical relationships (KE=½mv² has m linearly, v squared).