Energy Speed Patterns - Middle School Physical Science

1. Identify how $KE$ changes when speed is halved (mass constant) using $KE \propto v^2$.

   $KE$ becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving speed gives $(\frac{1}{2})^2 = \frac{1}{4}$ the energy.

2. Using slope $2.5$ on a $KE$ vs $v^2$ graph, what mass $m$ does the line represent?

   $m=5$. Since slope $= \frac{1}{2}m = 2.5$, then $m = 5$ kg.

3. If mass stays constant, what type of relationship should a $KE$ vs. $v$ graph show?

   Quadratic: $KE \propto v^2$. KE is proportional to the square of velocity, creating a parabola.

4. Which change produces a larger increase in $KE$: increasing speed from $2$ to $4\ \text{m/s}$ or from $4$ to $6\ \text{m/s}$?

   From $4$ to $6\ \text{m/s}$. $KE$ changes by $6^2 - 4^2 = 20$ vs $4^2 - 2^2 = 12$ (in $\frac{1}{2}m$ units).

5. If $KE=18\,J$ at $v=3\,m/s$, what is $KE$ at $v=6\,m/s$ for the same object?

   $72\,J$. Since speed doubles, $KE$ increases by $2^2 = 4$, so $18 \times 4 = 72\,J$.

6. Identify the units typically used for kinetic energy and speed on graphs.

   $KE$ in J; $v$ in $\text{m/s}$. Standard SI units: joules for energy, meters per second for velocity.

7. A $KE$ vs. $v$ graph is a straight line through the origin. Identify the most likely error.

   It should be $KE$ vs. $v^2$, not $v$. A straight line suggests linear relationship, but $KE$ varies with $v^2$, not $v$.

8. Identify the speed factor when a graph shows kinetic energy increases from $20\,\text{J}$ to $80\,\text{J}$ (same mass).

   Speed doubles ($\times 2$). $KE$ increased by factor of $4$, so speed doubled ($\sqrt{4} = 2$).

9. If mass halves while speed stays constant, by what factor does kinetic energy change?

   It becomes $\frac{1}{2}$ as large. Since $KE \propto m$ when $v$ is constant, halving $m$ halves $KE$.

10. Identify which graph is expected for $KE$ vs. $v$ (constant mass): straight line or upward-curving parabola?

    Upward-curving parabola. The quadratic relationship $KE \propto v^2$ creates a parabola.

11. A $KE$ vs $v$ graph shows one curve above another at all speeds. What does the higher curve indicate?

    Greater mass. At any given speed, higher KE means the object has more mass.

12. Choose the correct unit for kinetic energy on a graph axis: $\text{J}$ or $\text{N}$.

    $\text{J}$. Joules (J) is the SI unit for energy; Newtons (N) measure force.

13. Identify the physical meaning of the slope on a $KE$ versus $v^2$ graph for one object.

    Slope $=\frac{1}{2}m$. From $KE = \frac{1}{2}mv^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.

14. If speed is cut in half and mass stays constant, what happens to kinetic energy?

    It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving $v$ gives $(\frac{1}{2})^2 = \frac{1}{4}$ times the energy.

15. What is the kinetic energy formula in terms of mass $m$ and speed $v$?

    $KE=\frac{1}{2}mv^2$. Shows kinetic energy depends on mass and speed squared.

16. If mass doubles and speed stays constant, by what factor does kinetic energy change?

    $2\times$. KE is directly proportional to mass when speed is constant.

17. Identify the speed factor when a graph shows kinetic energy increases from $10\,\text{J}$ to $90\,\text{J}$ (same mass).

    Speed triples ($\times 3$). $KE$ increased by factor of $9$, so speed tripled ($\sqrt{9} = 3$).

18. Identify the correct pattern: as $v$ increases, does $KE$ increase linearly or faster than linear?

    Faster than linear (quadratic increase). KE increases with the square of velocity, not linearly.

19. If two objects have the same speed $v$, which has greater kinetic energy: larger mass or smaller mass?

    The object with larger mass has greater $KE$. At same $v$, $KE \propto m$, so larger mass means more kinetic energy.

20. Which graph shape best matches $KE$ versus $v$ for constant mass: straight line or upward curve?

    Upward curve (parabola-like). The quadratic relationship $KE \propto v^2$ creates a parabolic curve.

21. If speed increases from $2\ \text{m/s}$ to $6\ \text{m/s}$ (mass constant), by what factor does $KE$ change?

    $9\times$. Speed triples ($\frac{6}{2} = 3$), so KE increases by $3^2 = 9$ times.

22. Which graph becomes linear if you plot kinetic energy on the y-axis and $v^2$ on the x-axis (mass constant)?

    $KE$ vs $v^2$. Since $KE = \frac{1}{2}m \cdot v^2$, plotting vs $v^2$ gives a straight line.

23. Choose the correct comparison for constant mass: at $v=2$ and $v=6$, what is $\frac{KE_6}{KE_2}$?

    $9$. $\frac{KE_6}{KE_2} = \frac{v_6^2}{v_2^2} = \frac{6^2}{2^2} = \frac{36}{4} = 9$.

24. If speed triples and mass stays constant, by what factor does kinetic energy change?

    $9\times$. Since $KE \propto v^2$, tripling $v$ gives $3^2 = 9$ times the energy.

25. What power of speed $v$ appears in $KE=\frac{1}{2}mv^2$, and what graph shape does it imply?

    $v^2$; a curved (parabolic) $KE$ vs $v$ graph. Speed squared creates a parabola when graphed against $KE$.

26. If speed triples from $v$ to $3v$, by what factor does kinetic energy change?

    Kinetic energy becomes $9$ times as large. Since $KE \propto v^2$, tripling $v$ gives $(3)^2 = 9$ times the energy.

27. What happens to kinetic energy when speed triples and mass stays constant?

    It becomes $9\times$ larger. Since $KE \propto v^2$, tripling speed gives $3^2 = 9$ times the energy.

28. What power relationship between kinetic energy and speed is shown by $KE=\frac{1}{2}mv^2$?

    $KE\propto v^2$. KE is proportional to the square of velocity, not velocity itself.

29. If the slope of a $KE$ vs $v^2$ graph is $6$, what is the mass $m$?

    $m=12$. Since slope $= \frac{1}{2}m$, then $6 = \frac{1}{2}m$, so $m = 12$.

30. Identify which change increases $KE$ more: doubling $v$ or doubling $m$ (starting from same $m$ and $v$).

    Doubling $v$ increases $KE$ more. Doubling $v$ quadruples $KE$ while doubling $m$ only doubles $KE$.

31. A graph shows $KE = 8\,\text{J}$ at $v = 2\,\text{m/s}$. Find $KE$ at $v = 1\,\text{m/s}$ for the same mass.

    $2\,\text{J}$. Halving speed gives $\frac{1}{4}$ the $KE$: $8 \times \frac{1}{4} = 2\,\text{J}$.

32. Two points on a $KE$ vs. $v$ graph are $(2,10)$ and $(4,40)$. Identify the pattern that fits these points.

    $KE \propto v^2$. KE quadruples when speed doubles: $40 = 4 \times 10$.

33. Which graph becomes a straight line for constant mass: $KE$ vs. $v$ or $KE$ vs. $v^2$?

    $KE$ vs. $v^2$. Since $KE = \frac{1}{2}m \cdot v^2$, plotting against $v^2$ gives a linear relationship.

34. A $KE$ vs. $v^2$ graph shows a line through the origin. Identify what this implies about $KE$ when $v=0$.

    $KE = 0$ at $v = 0$. Line through origin means zero $KE$ when speed is zero.

35. Find the ratio $\frac{KE_2}{KE_1}$ when speed increases from $5\,m/s$ to $10\,m/s$ (same mass).

    $4$. $\frac{KE_2}{KE_1} = \frac{v_2^2}{v_1^2} = \frac{10^2}{5^2} = \frac{100}{25} = 4$.

36. Identify the pattern between kinetic energy and mass when speed stays constant.

    $KE\propto m$. KE is directly proportional to mass when velocity is constant.

37. Which option best describes a $KE$ versus $v^2$ graph for constant mass: linear, quadratic, or inverse?

    Linear. Plotting $KE$ against $v^2$ linearizes the quadratic relationship.

38. Identify the mass ratio if two objects have equal $KE$ but speeds $v_1=2\,m/s$ and $v_2=4\,m/s$.

    $\frac{m_1}{m_2}=4$. Equal $KE$ means $m_1v_1^2 = m_2v_2^2$, so $\frac{m_1}{m_2} = \frac{v_2^2}{v_1^2} = \frac{16}{4} = 4$.

39. A point on a $KE$ vs $v^2$ graph is $(16,40)$; what is the slope of the line through the origin?

    Slope $=\frac{40}{16}=2.5$. Slope equals rise over run: $\frac{40-0}{16-0} = 2.5$.

40. A $KE$ vs. $v^2$ line has slope $3$. What is the mass $m$ in kilograms?

    $m = 6\ \text{kg}$. Since slope $= \frac{1}{2}m = 3$, solving gives $m = 6$ kg.

41. What happens to kinetic energy when mass doubles and speed stays constant?

    It becomes $2\times$ larger. Since $KE \propto m$, doubling mass doubles the kinetic energy.

42. What happens to kinetic energy when speed is cut in half and mass stays constant?

    It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving speed gives $(\frac{1}{2})^2 = \frac{1}{4}$ the energy.

43. A graph shows $KE$ is constant while speed increases. What does this imply about the mass (idealized)?

    Mass is decreasing as speed increases. For constant $KE = \frac{1}{2}mv^2$, if $v$ increases, $m$ must decrease.

44. A graph shows $KE = 18\,\text{J}$ at $v = 3\,\text{m/s}$. Find $KE$ at $v = 6\,\text{m/s}$ for the same mass.

    $72\,\text{J}$. Doubling speed quadruples $KE$: $18 \times 4 = 72\,\text{J}$.

45. What is the formula for kinetic energy of a moving object in terms of mass and speed?

    $KE=\frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.

46. If a $KE$ vs. $v^2$ graph has slope $6$, identify the mass $m$ using slope $=\frac{1}{2}m$.

    $m = 12\,\text{kg}$. If slope $= \frac{1}{2}m = 6$, then $m = 12\,\text{kg}$.

47. Which option best describes a $KE$ versus $v$ graph for constant mass: linear, quadratic, or inverse?

    Quadratic. The $v^2$ relationship creates a parabolic curve.

48. What pattern should a $KE$ vs. $v$ graph show for constant mass: linear or curved upward?

    Curved upward (quadratic increase). The $v^2$ relationship creates a parabolic curve.

49. On a $KE$ vs. $v$ graph for one object, what is $KE$ when $v=0$?

    $KE = 0$. An object at rest has zero kinetic energy.

50. On a $KE$ vs. $v^2$ graph, what does the slope represent for one object of mass $m$?

    Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}mv^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.

51. Which graph shape best represents kinetic energy vs speed for constant mass: linear, quadratic, or inverse?

    Quadratic (upward-curving parabola). The $v^2$ relationship creates a parabolic curve opening upward.

52. Find the kinetic energy for $m=2\ \text{kg}$ and $v=3\ \text{m/s}$.

    $KE=9\ \text{J}$. $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9$ J.

53. A graph shows $KE=20\,J$ at $v=2\,m/s$. For the same mass, what is $KE$ at $v=4\,m/s$?

    $80\,J$. Doubling speed quadruples KE: $20J \times 4 = 80J$.

54. Find the ratio $\frac{KE_2}{KE_1}$ when speed increases from $4\,m/s$ to $12\,m/s$ (same mass).

    $9$. $\frac{KE_2}{KE_1} = \frac{v_2^2}{v_1^2} = \frac{12^2}{4^2} = \frac{144}{16} = 9$.

55. If speed is cut in half and mass stays constant, by what factor does kinetic energy change?

    $\frac{1}{4}$. Halving speed gives $(\frac{1}{2})^2 = \frac{1}{4}$ of the original KE.

56. If mass stays constant, what type of relationship should a $KE$ vs. $v^2$ graph show?

    Linear: $KE \propto v^2$ gives a straight line. Plotting $KE$ against $v^2$ linearizes the quadratic relationship.

57. If two objects have the same speed, how does kinetic energy compare when one has twice the mass?

    The heavier object has $2\times$ the $KE$. Since $KE \propto m$ at constant speed, doubling mass doubles KE.

58. What happens to kinetic energy when speed is cut in half and mass stays constant?

    It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving $v$ gives $(\frac{1}{2})^2 = \frac{1}{4}$ times the KE.

59. A graph shows $KE$ increases by a factor of $4$ when $v$ increases by a factor of $2$. Identify the relationship.

    $KE \propto v^2$. Factor of 4 increase for factor of 2 in speed indicates $v^2$ relationship.

60. A $KE$ vs $v^2$ graph shows points $(4,10)$ and $(8,20)$. What mass $m$ does the data imply?

    $m=5\ \text{kg}$. Slope $= \frac{20-10}{8-4} = 2.5$, so $m = 2 \times 2.5 = 5$ kg.

61. Which option indicates an incorrect $KE$ vs $v$ pattern for constant mass: linear or quadratic?

    Linear is incorrect; it should be quadratic. KE varies quadratically with $v$, not linearly.

62. Identify which speed gives greater kinetic energy for the same mass: $6\,m/s$ or $8\,m/s$.

    $8\,m/s$. Higher speed means more kinetic energy for the same mass.

63. What happens to $KE$ when speed is cut in half from $v$ to $\frac{v}{2}$ for the same mass?

    $KE$ becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving $v$ gives $(\frac{v}{2})^2 = \frac{v^2}{4}$.

64. Which object has greater kinetic energy at the same speed: mass $2\,\text{kg}$ or mass $6\,\text{kg}$?

    $6\,\text{kg}$ object. Greater mass means greater $KE$ at same speed ($KE \propto m$).

65. If a $KE$ vs $v^2$ line has slope $6$, what is the mass $m$?

    $m=12$. Since slope $= \frac{1}{2}m = 6$, multiply by 2 to get mass.

66. On a $KE$ vs $v^2$ graph, what does the slope equal for an object of mass $m$?

    $\frac{1}{2}m$. From $KE = \frac{1}{2}m \cdot v^2$, the coefficient of $v^2$ is the slope.

67. What does the slope represent on a graph of $KE$ (y) versus $v^2$ (x) for one object?

    Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}m(v^2)$, the coefficient of $v^2$ is $\frac{1}{2}m$.

68. Which option best describes the relationship between $KE$ and $v$ for constant mass: $KE \propto v$, $KE \propto v^2$, or $KE \propto \frac{1}{v}$?

    $KE \propto v^2$. Kinetic energy varies with the square of velocity.

69. Identify the graph shape of $KE$ versus $v^2$ for a constant mass object.

    A straight line through the origin. Since $KE = \frac{1}{2}m(v^2)$, it's linear in $v^2$.

70. Identify the best plot to make $KE$ vs speed appear linear for constant mass: $KE$ vs $v$ or $KE$ vs $v^2$?

    Plot $KE$ vs $v^2$. Since $KE = \frac{1}{2}mv^2$, plotting against $v^2$ gives a straight line.

71. Identify how $KE$ changes when speed triples (mass constant) using $KE \propto v^2$.

    $KE$ becomes $9$ times larger. Since $KE \propto v^2$, tripling speed gives $3^2 = 9$ times the energy.

72. Which plot best reveals a straight-line pattern for constant mass: $KE$ vs $v$, or $KE$ vs $v^2$?

    $KE$ vs $v^2$. The linear relationship $KE = \frac{1}{2}m \cdot v^2$ appears when plotting vs $v^2$.

73. Choose the correct comparison at the same speed: if mass doubles from $m$ to $2m$, how does $KE$ change?

    $KE$ doubles. KE is directly proportional to mass when speed is constant.

74. A $KE$ vs $v$ curve becomes steeper at higher speeds. What pattern does this show about $KE$ changes?

    $KE$ increases faster as $v$ increases (quadratic growth). The $v^2$ relationship means each speed increase adds more energy than the last.

75. Two objects have the same kinetic energy. Which has greater speed: the one with larger $m$ or smaller $m$?

    The one with smaller $m$. From $v = \sqrt{\frac{2KE}{m}}$, smaller $m$ gives larger $v$ when $KE$ is constant.

76. A graph shows $KE=45\,J$ at $v=3\,m/s$. For the same mass, what is $KE$ at $v=6\,m/s$?

    $180\,J$. Doubling speed quadruples KE: $45J \times 4 = 180J$.

77. If two points on a $KE$ vs $v$ graph are $(2,8)$ and $(4,32)$, what pattern is indicated?

    Quadratic pattern ($KE$ scales with $v^2$). KE quadruples when $v$ doubles: $32/8 = 4 = 2^2$.

78. Choose the correct statement for a $KE$ versus $v$ graph at $v=0$ for any mass.

    At $v=0$, $KE=0$. No motion means no kinetic energy for any object.

79. Identify the relationship between $KE$ and $v$ when mass is constant.

    $KE \propto v^2$. Since $m$ is constant in $KE = \frac{1}{2}mv^2$, $KE$ is proportional to $v^2$.

80. If kinetic energy increases by a factor of $16$ with constant mass, by what factor did speed change?

    $4\times$. If $KE$ increases $16\times$, then $v$ increases $\sqrt{16} = 4\times$.

81. State the formula for kinetic energy in terms of mass $m$ and speed $v$.

    $KE = \frac{1}{2}mv^2$. Shows kinetic energy equals half the product of mass and velocity squared.

82. If $KE=50\,J$ at $v=10\,m/s$, what is $KE$ at $v=5\,m/s$ for the same object?

    $12.5\,J$. Since speed halves, $KE$ becomes $(\frac{1}{2})^2 = \frac{1}{4}$, so $50 \times \frac{1}{4} = 12.5\,J$.

83. On a $KE$ vs $v$ graph, what does it mean if the curve is higher for one object than another at the same $v$?

    The higher curve corresponds to larger mass. At same speed, $KE \propto m$, so higher curve means larger mass.

84. Choose the correct comparison: at the same speed, which has greater kinetic energy, larger mass or smaller mass?

    Larger mass. At equal speeds, $KE$ is directly proportional to mass.

85. Interpret this: on a $KE$ vs $v$ graph, the curve gets steeper at higher speeds. What pattern does that show?

    Kinetic energy increases faster than speed. The increasing slope shows KE grows more rapidly at higher speeds.

86. If mass doubles while speed stays constant, by what factor does kinetic energy change?

    It becomes $2$ times larger. Since $KE \propto m$ when $v$ is constant, doubling $m$ doubles $KE$.

87. What happens to kinetic energy when speed triples and mass stays constant?

    It becomes $9$ times as large. Since $KE \propto v^2$, tripling $v$ gives $3^2 = 9$ times the KE.

88. If speed doubles and mass stays constant, by what factor does kinetic energy change?

    $4\times$. Since $KE \propto v^2$, doubling $v$ gives $2^2 = 4$ times the energy.

89. For constant mass, what does the slope represent on a graph of $KE$ (y) versus $v^2$ (x)?

    Slope $=\frac{1}{2}m$. From $KE = \frac{1}{2}mv^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.

90. On a $KE$ vs. $v$ graph, which point must be on the curve for any object?

    $(0,0)$. When $v = 0$, $KE = \frac{1}{2}m(0)^2 = 0$, so the curve passes through the origin.

91. Find the slope on a $KE$ versus $v^2$ graph for an object of mass $8\,\text{kg}$.

    Slope $=4$. Slope $= \frac{1}{2}m = \frac{1}{2}(8) = 4$ J/(m²/s²).

92. What happens to kinetic energy when speed triples and mass stays the same?

    It becomes $9$ times larger. Since $KE \propto v^2$, tripling $v$ gives $(3v)^2 = 9v^2$.

93. Identify the pattern: as speed increases, do equal increases in speed add equal increases in kinetic energy?

    No; kinetic energy increases by larger amounts. Due to the $v^2$ term, each speed increase adds more KE than the last.

94. Identify the pattern between kinetic energy and speed when mass stays constant.

    $KE\propto v^2$. KE is proportional to the square of velocity when mass is constant.

95. Which graph becomes a straight line for constant mass: $KE$ vs $v$ or $KE$ vs $v^2$?

    $KE$ vs $v^2$. Since $KE = \frac{1}{2}mv^2$, plotting against $v^2$ linearizes the relationship.

96. Identify how $KE$ changes when speed doubles (mass constant) using $KE \propto v^2$.

    $KE$ becomes $4$ times larger. Since $KE \propto v^2$, doubling speed gives $2^2 = 4$ times the energy.

97. State the formula that relates kinetic energy, mass, and speed.

    $KE=\frac{1}{2}mv^2$. Shows kinetic energy equals half the product of mass and velocity squared.

98. Which relationship is correct for constant mass: $KE\propto v$, $KE\propto v^2$, or $KE\propto \frac{1}{v}$?

    $KE\propto v^2$. From $KE = \frac{1}{2}mv^2$, kinetic energy is proportional to speed squared.

99. State the formula for kinetic energy $KE$ in terms of mass $m$ and speed $v$.

    $KE = \frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.

100. On a $KE$ versus $v^2$ graph, which object has greater mass: slope $3$ or slope $5$?

     Slope $5$ corresponds to greater mass. Since slope $= \frac{1}{2}m$, larger slope means larger mass.

101. Identify the best description of the relationship shown by a straight line through the origin on a $KE$ vs $v^2$ graph.

     Direct proportionality: $KE\propto v^2$. A straight line through origin means $y \propto x$.

102. What happens to kinetic energy when speed doubles and mass stays constant?

     It becomes $4\times$ larger. Since $KE \propto v^2$, doubling speed gives $2^2 = 4$ times the energy.

103. A $KE$ vs $v^2$ graph shows two straight lines through the origin; what causes different slopes?

     Different masses. Each mass gives a different slope $\frac{1}{2}m$ on the graph.

104. Which option is a correct unit for kinetic energy in SI units?

     Joules (J). Energy is measured in joules in the SI system.

105. Identify which object has greater mass if its $KE$ versus $v^2$ line is steeper.

     The steeper line has the greater mass. Larger slope means larger $\frac{1}{2}m$, thus larger mass.

106. On a $KE$ vs. $v$ graph for constant mass, what is the $KE$ value when $v = 0$?

     $KE = 0$. No motion means no kinetic energy.

107. Find the mass if a $KE$ versus $v^2$ graph has slope $6\,\text{J}/(\text{m}^2/\text{s}^2)$.

     $m=12\,\text{kg}$. Since slope $= \frac{1}{2}m$, then $m = 2 \times 6 = 12$ kg.

108. What happens to kinetic energy when speed increases by a factor of $5$ and mass stays constant?

     It becomes $25$ times as large. Since $KE \propto v^2$, multiplying $v$ by $5$ gives $5^2 = 25$ times the KE.

109. Identify which graph best matches $KE = \frac{1}{2}mv^2$: a straight line on $KE$ vs. $v$, or a parabola on $KE$ vs. $v$.

     Parabola on $KE$ vs. $v$. The squared relationship creates a parabolic curve.

110. If speed is cut in half from $v$ to $\frac{1}{2}v$, by what factor does kinetic energy change?

     Kinetic energy becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving $v$ gives $(\frac{1}{2})^2 = \frac{1}{4}$ times the energy.

111. On a $KE$ vs. $v^2$ graph, what should the $y$-intercept be if the object starts from rest?

     $0$. When $v = 0$, $KE = 0$ regardless of mass.

112. On a $KE$ vs. $v^2$ graph, what is the slope equal to for an object of mass $m$?

     Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}mv^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.

113. Identify the relationship between kinetic energy and mass for constant speed.

     $KE\propto m$. Kinetic energy is directly proportional to mass.

114. A graph shows $KE=72\,J$ at $v=6\,m/s$. For the same mass, what is $KE$ at $v=3\,m/s$?

     $18\,J$. Halving speed reduces KE by factor of 4: $72J \div 4 = 18J$.

115. Identify the correct pattern: if $v$ increases by $1\ \text{m/s}$ each step, does $KE$ increase by a constant amount?

     No; the increases get larger each step. Since $KE \propto v^2$, equal increases in $v$ produce increasingly larger $KE$ changes.

116. If speed doubles and mass stays constant, by what factor does kinetic energy change?

     It becomes $4$ times larger. Since $KE \propto v^2$, doubling $v$ gives $(2)^2 = 4$ times the energy.

117. What happens to $KE$ when speed triples from $v$ to $3v$ for the same mass?

     $KE$ becomes $9$ times as large. Since $KE \propto v^2$, tripling $v$ gives $(3v)^2 = 9v^2$.

118. For constant mass, what is the slope of a straight-line graph of $KE$ (y) vs $v^2$ (x)?

     Slope $=\frac{1}{2}m$. From $KE = \frac{1}{2}m \cdot v^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.

119. What pattern between kinetic energy and speed is shown by $KE = \frac{1}{2}mv^2$ when $m$ is constant?

     Kinetic energy is proportional to $v^2$. Since $m$ is constant, $KE$ changes with the square of speed.

120. Data show $KE=8\ \text{J}$ at $v=2\ \text{m/s}$. For the same mass, what is $KE$ at $v=5\ \text{m/s}$?

     $50\ \text{J}$. Find $\frac{1}{2}m = \frac{8}{4} = 2$, then $KE = 2 \times 5^2 = 50$ J.

121. Two objects have the same speed. Which has greater $KE$: the one with larger $m$ or smaller $m$?

     The one with larger $m$. From $KE = \frac{1}{2}mv^2$, larger $m$ gives larger $KE$ when $v$ is constant.

122. Which type of graph should $KE$ vs. $v$ form for constant mass: linear or curved?

     Curved (parabolic). The $v^2$ relationship creates a parabola when graphing $KE$ vs. $v$.

123. What is the pattern between kinetic energy and speed when mass stays constant?

     $KE\propto v^2$ (quadratic increase). KE increases with the square of velocity when mass is constant.

124. If speed is cut in half and mass stays constant, by what factor does kinetic energy change?

     It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving $v$ gives $(\frac{1}{2})^2 = \frac{1}{4}$ times the energy.

125. On a $KE$ vs $v$ graph for constant mass, what is the kinetic energy when $v=0$?

     $0\ \text{J}$. When velocity is zero, there's no motion and no kinetic energy.

126. Identify the graph shape of $KE$ versus $v$ for a constant mass object.

     An upward-opening curve (parabola-like). The $v^2$ relationship creates a parabolic curve.

127. A $KE$ vs. $v^2$ line has slope $3$. What is the mass of the object?

     $6\,kg$. Slope equals $\frac{1}{2}m$, so $m = 2 \times 3 = 6\,kg$.

128. A graph shows $KE$ rising faster as $v$ increases; which relationship best fits: linear or quadratic?

     Quadratic ($KE \propto v^2$). Rising faster with increasing $v$ indicates a power relationship.

129. A graph shows $KE$ rises faster as $v$ increases. Identify the best description: constant rate or increasing rate of change.

     Increasing rate of change. The parabolic curve shows $KE$ increases faster at higher speeds.

130. If the slope of a $KE$ vs $v^2$ graph is $2.5$, what is the mass $m$?

     $m=5$. Since slope $= \frac{1}{2}m$, then $2.5 = \frac{1}{2}m$, so $m = 5$.

131. On a $KE$ vs $v$ graph for one object, where is kinetic energy $0$?

     At $v=0$. When speed is zero, $KE = \frac{1}{2}m(0)^2 = 0$.

132. If speed triples and mass stays constant, what happens to kinetic energy?

     It becomes $9\times$ as large. Since $KE \propto v^2$, tripling $v$ gives $3^2 = 9$ times the energy.

133. State the formula for kinetic energy $KE$ in terms of mass $m$ and speed $v$.

     $KE = \frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.

134. Identify the expected shape of a scatter plot of $KE$ versus $v$ for constant mass and increasing speed.

     An upward-curving (concave up) trend. The quadratic relationship $KE \propto v^2$ creates an upward-curving parabola.

135. What pattern should a $KE$ vs. $v^2$ graph show for constant mass: line or curve?

     Straight line. Plotting $KE$ against $v^2$ linearizes the quadratic relationship.

136. Which object has greater mass if both lines are on a $KE$ vs. $v^2$ graph?

     The object with the steeper slope. Steeper slope means larger $\frac{1}{2}m$, therefore larger mass.

137. State the formula for kinetic energy of an object with mass $m$ and speed $v$.

     $KE=\frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.

138. Calculate $KE$ for $m=2\,\text{kg}$ and $v=3\,\text{m/s}$.

     $9\,\text{J}$. Using $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9$ J.

139. A graph shows $KE = 50\,\text{J}$ at $v = 10\,\text{m/s}$. Find $KE$ at $v = 5\,\text{m/s}$ for the same mass.

     $12.5\,\text{J}$. Halving speed gives $\frac{1}{4}$ the $KE$: $50 \times \frac{1}{4} = 12.5\,\text{J}$.

140. Identify the slope of a $KE$ vs. $v^2$ graph for an object of mass $m$.

     Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}m \cdot v^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.

141. If speed triples and mass stays constant, by what factor does kinetic energy change?

     It becomes $9$ times larger. Since $KE \propto v^2$, tripling $v$ gives $(3)^2 = 9$ times the energy.

142. What happens to kinetic energy when speed doubles and mass stays constant?

     It becomes $4$ times as large. Since $KE \propto v^2$, doubling $v$ gives $2^2 = 4$ times the KE.

143. If a $KE$ vs $v^2$ line has slope $2.5$, what mass $m$ does it represent?

     $m=5\ \text{kg}$. Since slope $= \frac{1}{2}m$, then $m = 2 \times \text{slope} = 2 \times 2.5 = 5$ kg.

144. If speed doubles from $v$ to $2v$, by what factor does kinetic energy change?

     Kinetic energy becomes $4$ times as large. Since $KE \propto v^2$, doubling $v$ gives $(2)^2 = 4$ times the energy.

145. A graph shows $KE$ increases by a factor of $9$ when $v$ increases by a factor of $3$. Identify the relationship.

     $KE \propto v^2$. Factor of 9 increase for factor of 3 in speed indicates $v^2$ relationship.

146. Which type of graph should $KE$ vs. $v^2$ form for constant mass: linear or curved?

     Linear. Since $KE = \frac{1}{2}mv^2$, $KE$ is directly proportional to $v^2$.

147. Choose the correct comparison for same mass: $KE(2v)$ equals what multiple of $KE(v)$?

     $4\times$. Doubling speed quadruples kinetic energy since $KE \propto v^2$.

148. A graph shows $KE$ quadruples when $v$ doubles. What relationship does this indicate?

     $KE \propto v^2$. Quadrupling when doubling confirms the squared relationship.

149. If speed doubles and mass stays constant, what happens to kinetic energy?

     It becomes $4\times$ as large. Since $KE \propto v^2$, doubling $v$ gives $2^2 = 4$ times the energy.

150. What happens to kinetic energy when mass doubles and speed stays constant?

     It doubles. Since $KE \propto m$ at constant $v$, doubling mass doubles KE.

151. Find $v$ if $m=2\ \text{kg}$ and $KE=9\ \text{J}$ using $KE=\frac{1}{2}mv^2$.

     $3\ \text{m/s}$. $9 = \frac{1}{2}(2)v^2$ gives $v^2 = 9$, so $v = 3$ m/s.

152. Which variable must be held constant to isolate the pattern between $KE$ and $v$ on a graph?

     Mass $m$ must be constant. Isolating the $v$ effect requires keeping other variables fixed.

153. If two objects have the same kinetic energy, which has greater speed: larger mass or smaller mass?

     The object with smaller mass has greater speed. For same $KE$, $v \propto \frac{1}{\sqrt{m}}$, so smaller mass needs higher speed.

154. State the formula that relates kinetic energy, mass, and speed.

     $KE=\frac{1}{2}mv^2$. Shows kinetic energy depends on mass and speed squared.

155. Find $KE$ for $m = 2\ \text{kg}$ and $v = 3\ \text{m/s}$.

     $KE = 9\ \text{J}$. Using $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9$ J.

156. What happens to $KE$ when speed doubles from $v$ to $2v$ for the same mass?

     $KE$ becomes $4$ times as large. Since $KE \propto v^2$, doubling $v$ gives $(2v)^2 = 4v^2$.

157. At $v=3\ \text{m/s}$, object A has $KE=18\ \text{J}$ and B has $KE=36\ \text{J}$. Which has larger mass?

     Object B. At same speed, higher $KE$ means larger mass since $KE \propto m$.

158. Find $v$ if $m = 2\ \text{kg}$ and $KE = 16\ \text{J}$.

     $v = 4\ \text{m/s}$. From $16 = \frac{1}{2}(2)v^2$, solving gives $v^2 = 16$, so $v = 4$ m/s.

159. If mass is constant, which ratio should stay constant across data: $\frac{KE}{v}$ or $\frac{KE}{v^2}$?

     $\frac{KE}{v^2}$ stays constant. Since $KE = \frac{1}{2}mv^2$, the ratio $\frac{KE}{v^2} = \frac{1}{2}m$ is constant.

160. If a graph shows $KE$ proportional to $v$ (a straight line), what is the correct conclusion about the model?

     It is inconsistent with $KE=\frac{1}{2}mv^2$. The formula requires $KE \propto v^2$, not $KE \propto v$.

161. Find the speed for $m=4\ \text{kg}$ and $KE=50\ \text{J}$.

     $v=5\ \text{m/s}$. $50 = \frac{1}{2}(4)v^2$, so $v^2 = 25$, thus $v = 5$ m/s.

162. Identify the relationship between kinetic energy and speed for constant mass.

     $KE\propto v^2$. Kinetic energy is proportional to the square of speed.

163. A $KE$ vs. $v^2$ line has slope $0.5$. What is the mass of the object?

     $1\,kg$. Slope equals $\frac{1}{2}m$, so $m = 2 \times 0.5 = 1\,kg$.

164. A graph shows $KE=80\ \text{J}$ at $v=4\ \text{m/s}$. What is the mass $m$?

     $m=10\ \text{kg}$. $80 = \frac{1}{2}m(4)^2 = 8m$, so $m = 10$ kg.

165. Two objects have the same speed. Object A has mass $2m$ and object B has mass $m$. What is $\frac{KE_A}{KE_B}$?

     $2$. KE ratio equals mass ratio at same speed: $\frac{2m}{m} = 2$.

166. Find $KE$ for $m=2\ \text{kg}$ and $v=3\ \text{m/s}$ using $KE=\frac{1}{2}mv^2$.

     $9\ \text{J}$. $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9$ J.

167. Which graph shape best represents $KE$ vs. $v$ for a constant mass: line or curve?

     Curve (upward-opening parabola). The $v^2$ relationship creates a parabolic curve, not a straight line.

168. On a $KE$ vs. $v^2$ graph, what physical quantity equals the slope?

     $\frac{1}{2}m$. From $KE = \frac{1}{2}mv^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.

169. If a $KE$ vs $v^2$ line has slope $6\ \text{J}/(\text{m}^2/\text{s}^2)$, what is the mass $m$?

     $m=12\ \text{kg}$. Since slope $= \frac{1}{2}m$, then $m = 2 \times \text{slope} = 2 \times 6 = 12$ kg.

Study these flashcards