Graph Energy and Mass - Middle School Physical Science

Two straight-line graphs show kinetic energy (KE) vs mass (m) for objects moving at constant speeds. Graph A is for $v=5\ \text{m/s}$ and Graph B is for $v=2\ \text{m/s}$ . Both pass through the origin. What does the slope of each KE vs mass line represent?
Answer: The value $\tfrac{1}{2}v^2$ (in J/kg) for that graph's speed. 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.
At constant speed $v=5\ \text{m/s}$ , a graph of kinetic energy (KE) vs mass (m) is a straight line through the origin. Using the data pattern below, what mass would have a kinetic energy of $50\ \text{J}$ ?
Answer: 4 kg This question tests understanding of how to find mass from a given kinetic energy using the graph or data pattern 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. Reading the graph: for KE=50 J, trace horizontally to the line, then down to mass axis to find m=4 kg (matches data pattern: at slope=12.5 J/kg, m=50/12.5=4 kg). Choice C is correct because it correctly identifies the mass as 4 kg for 50 J using the line position or proportional calculation. Choice A is wrong because it underestimates mass (for lower KE like 25 J); Choice B mismatches; Choice D overestimates for higher KE. Graphing relationships helps visualize patterns: the straight line allows reading mass for any KE by going from y-axis to line to x-axis. Interpreting graph: (1) identify shape (straight line = linear), (2) check origin (through origin = proportional), (3) read values inversely (pick KE, go across to line, down to mass axis).
A cart moves at a constant speed of $v=5\ \text{m/s}$ . 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 $\text{J/kg}$ )?
Answer: 12.5 J/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. 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.
A cart moves at constant speed $v=5\ \text{m/s}$ . Based on the KE vs mass data, what is the slope of a graph of kinetic energy (J) vs mass (kg)?
Answer: 12.5 J/kg 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).
A student is told to make a graph of kinetic energy vs mass for objects moving at the same speed. Which set of axis labels is correct for a KE vs mass graph?
Answer: x-axis: mass (kg); y-axis: KE (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 = 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 graph from data: Graphing requires: (1) set up axes with mass (kg) horizontal (x-axis, independent variable) and KE (J) vertical (y-axis, dependent variable), both starting at 0, (2) plot each point as dots on graph, (3) draw straight line through all points—should pass through origin (0, 0) because zero mass has zero kinetic energy, and (4) label axes with units and title. Choice B is correct because it has the independent variable (mass in kg) on x-axis and dependent (KE in J) on y-axis, following standard graphing conventions. Choice A is wrong because it swaps axes (KE on x, mass on y, which reverses independent/dependent); Choice C is wrong because it uses speed on x instead of mass; Choice D is wrong because units are swapped and incorrect (mass in J, KE in kg doesn't make sense). Creating graph from data: (1) identify variables (independent: mass on x-axis, dependent: KE on y-axis), (2) set scale including origin and all data, (3) plot carefully (each data point as dot at correct coordinates), (4) draw best-fit line (should be straight through all points and origin for this relationship), (5) check: does line pass through origin? (must for KE-mass), is it straight? (must for linear relationship), does it include all data points? (should if data is accurate). 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).
A cart moves at a constant speed of $v=5\ \text{m/s}$ . If you graph KE (J) on the vertical axis and mass (kg) on the horizontal axis using the data, which graph would you expect?
Answer: A straight line that goes through the origin and rises as mass increases. 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 creating the graph from data: Plotting the points (1, 12.5), (2, 25), (4, 50), and (6, 75) on axes with mass horizontal and KE vertical produces a straight line that passes through the origin (0, 0) and rises steadily—this is because KE = 12.5m, a linear equation with y-intercept of 0 and slope of 12.5 J/kg. Choice C is correct because it accurately describes "a straight line that goes through the origin and rises as mass increases," matching the linear proportional relationship KE ∝ m. Choice A is wrong because it describes a line that "curves upward more and more," but the relationship is linear not curved; Choice B incorrectly states the line "slopes downward" when KE clearly increases with mass; Choice D wrongly claims a "horizontal line" suggesting KE doesn't change with mass, when the data shows KE increasing from 12.5 J to 75 J as mass increases.
A cart moves at constant speed $v=5\ \text{m/s}$ . If you graph kinetic energy (vertical axis) vs mass (horizontal axis) using the data below, which description matches the correct graph?
Answer: A straight line through the origin (0,0). This question tests understanding of the shape and features of a graph plotting 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. For creating graph from data: Given data at v=5 m/s (masses 1, 2, 4, 6 kg with KE 12.5, 25, 50, 75 J), graphing requires: (1) set up axes with mass (kg) horizontal and KE (J) vertical, both starting at 0 and extending to at least 6 kg and 75 J to include all data, (2) plot each point (1, 12.5), (2, 25), (4, 50), (6, 75) as dots on graph, (3) draw straight line through all points—should pass through origin (0, 0) because zero mass has zero kinetic energy, and (4) label axes with units and title. Choice B is correct because it correctly describes the graph as a straight line through the origin. Choice A is wrong because it describes a curved line, but the data show a straight line (mass effect is linear, not parabolic like speed effect); Choice C claims the line doesn't pass through the origin when KE=0 at m=0 requires origin passage; Choice D describes a horizontal line, but KE increases with mass. The resulting graph is a straight line from origin with slope = rise/run = (50-25)/(4-2) = 25/2 = 12.5 J/kg, which equals ½(5²) = 12.5 J/kg, confirming the data represents constant 5 m/s speed. Graphing relationships helps visualize patterns: KE vs mass at constant speed always produces a straight line through origin, confirming KE ∝ m proportionality visually.
Objects move at a constant speed of $v = 5\ \text{m/s}$ . The KE vs mass graph is a straight line through the origin. If the kinetic energy is $62.5\ \text{J}$ , what mass does the graph show (in kg)?
Answer: 5 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. 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).
Three carts move at the same constant speed of $v=3\ \text{m/s}$ . Which statement best describes the pattern between mass and kinetic energy shown by the data?
Answer: Kinetic energy is directly proportional to mass (a straight-line relationship through the origin). 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.
A cart moves at a constant speed of $v=5\ \text{m/s}$ . What does the slope of a KE vs mass graph represent for this situation?
Answer: The value of $\tfrac{1}{2}v^2$ (how many joules per kilogram 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. 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 understanding the physical meaning of slope: Since KE = ½mv² = (½v²)m, the slope of KE vs m graph equals ½v², which represents the kinetic energy per unit mass at the given speed—with v = 5 m/s, slope = ½(5²) = ½(25) = 12.5 J/kg, meaning each kilogram of mass contributes 12.5 J of kinetic energy at this speed. Choice B is correct because the slope represents "the value of ½v² (how many joules per kilogram at this speed)," which is the coefficient of m in the equation KE = (½v²)m. Choice A is wrong—the slope is not the mass but rather KE per unit mass; Choice C incorrectly states the slope is the speed v, but slope = ½v² not v; Choice D wrongly suggests the slope is m², but the slope is a constant ½v² independent of mass.
Three carts all move at the same constant speed of $v=3\ \text{m/s}$ . If you extend the pattern, what kinetic energy would you expect for a cart with mass $7\ \text{kg}$ at the same speed?
Answer: 31.5 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 = 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.
A cart moves at a constant speed of $v=5\ \text{m/s}$ . 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?
Answer: A straight line through the origin (0,0). 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.
Different carts all move at the same constant speed of $v=3\ \text{m/s}$ . If you graph kinetic energy (J) vs mass (kg), what pattern should the points make?
Answer: A straight line through the origin because KE is proportional to mass at constant speed. 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).
A student plots kinetic energy (KE) vs mass for an object moving at a constant speed. The plotted points include $(1\ \text{kg},\ 12.5\ \text{J})$ , $(2\ \text{kg},\ 25\ \text{J})$ , and $(4\ \text{kg},\ 50\ \text{J})$ . Which statement best describes the relationship shown by the graph?
Answer: KE is directly proportional to mass: doubling mass doubles 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 = 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).
Two straight-line graphs show KE (J) vs mass (kg), both passing through the origin. - Line 1 is for constant speed $v=2\ \text{m/s}$ . - Line 2 is for constant speed $v=5\ \text{m/s}$ . Which statement is true?
Answer: Line 2 is steeper because higher speed gives more kinetic energy per kilogram. 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.
Two straight-line graphs show kinetic energy (KE) vs mass for objects moving at constant speeds. One line is for $v=2\ \text{m/s}$ and the other is for $v=5\ \text{m/s}$ . Both lines pass through the origin. Which line should be steeper, and why?
Answer: The $v=5\ \text{m/s}$ line, because at constant speed the slope is $\tfrac{1}{2}v^2$ , which is larger for 5 m/s. 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).
A cart moves at a constant speed of $v=5\ \text{m/s}$ . Using the data below, what is the kinetic energy when the mass is $3\ \text{kg}$ ? (Assume kinetic energy is proportional to mass.)
Answer: 37.5 J 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.
Four carts all move at a constant speed of $v=5\ \text{m/s}$ . If a cart has kinetic energy $62.5\ \text{J}$ , what is its mass?
Answer: 5 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. 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.
Two straight-line graphs show kinetic energy (KE) vs mass (m). Graph 1 is for constant speed $v=5\ \text{m/s}$ and Graph 2 is for constant speed $v=2\ \text{m/s}$ . Both lines go through the origin. Which graph should have the steeper slope?
Answer: Graph 1 ( $v=5\ \text{m/s}$ ), because higher speed makes KE increase faster with mass. This question tests understanding of how constant speed affects the slope of kinetic energy versus mass graphs when comparing different speeds. 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 higher speed makes KE increase faster with mass, resulting in a steeper slope for Graph 1 (v=5 m/s). 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); Choice C ignores that slopes differ by speed; Choice D states neither is straight, but both are linear. 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). Graphing relationships helps visualize patterns: comparing slopes shows how speed influences the KE-mass relationship quantitatively.
At a constant speed of $v = 5\ \text{m/s}$ , a line graph of KE (J) vs mass (kg) includes the points $(1, 12.5)$ , $(2, 25)$ , and $(4, 50)$ and is a straight line through the origin. What is the kinetic energy when the mass is $3\ \text{kg}$ ?
Answer: 37.5 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 = 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).

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