Explain Kinetic Energy Changes - Middle School Physical Science

1. Identify the slope meaning on a $KE$ vs. mass graph when speed is constant.

   Slope $=\frac{1}{2}v^2$. From $KE = \frac{1}{2}mv^2$, slope equals $\frac{1}{2}$ times speed squared.

2. Which graph shape indicates $KE \propto v^2$ when mass is constant: straight line or curve?

   Curve that gets steeper (quadratic). Squared relationship produces a parabolic curve.

3. What happens to kinetic energy if speed doubles while mass stays the same?

   Kinetic energy becomes $4$ times as large. Since $KE \propto v^2$, doubling $v$ quadruples $KE$.

4. If a $KE$ vs. $m$ graph is a straight line, what does that imply about speed $v$?

   Speed is constant for the data set. Linear $KE$ vs. $m$ means $v^2$ is constant, so $v$ must be constant.

5. Which graph shape best matches $KE$ vs speed $v$ when mass $m$ is constant?

   An upward-curving parabola (quadratic). $KE \propto v^2$ creates a parabolic curve when graphed.

6. Using $KE=\frac{1}{2}mv^2$, what is $KE$ for $m=2$ and $v=3$ (no units)?

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

7. What happens to kinetic energy if speed is cut in half while mass stays the same?

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

8. Which graph becomes linear if you plot $KE$ versus $v^2$ at constant mass?

   $KE$ vs. $v^2$ is linear. Since $KE = \frac{1}{2}mv^2$, KE is linear in $v^2$.

9. Identify which evidence from a graph supports $KE \propto v^2$: doubling $v$ makes $KE$ quadruple or double?

   Doubling $v$ makes $KE$ quadruple. The quadratic relationship means doubling input quadruples output.

10. What happens to kinetic energy if speed triples while mass stays constant?

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

11. Find the kinetic energy ratio if speed triples and mass stays constant: $\frac{KE_2}{KE_1}=?$

    $\frac{KE_2}{KE_1}=9$. Since $KE \propto v^2$, tripling speed gives $3^2 = 9$ times the energy.

12. Identify how kinetic energy changes when mass doubles and speed stays constant.

    Kinetic energy doubles. KE is directly proportional to mass when speed is constant.

13. Which graph is expected to be linear: $KE$ vs. $m$ (constant $v$) or $KE$ vs. $v$ (constant $m$)?

    $KE$ vs. $m$ at constant $v$ is linear. Mass has linear effect while speed has quadratic effect on KE.

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

    Kinetic energy becomes $9$ times as large. Since $KE \propto v^2$, tripling speed multiplies KE by $3^2 = 9$.

15. If two points on a $KE$ vs. mass graph are $(2, 30)$ and $(4, 60)$, what is the relationship shown?

    Direct proportion (doubling mass doubles $KE$). $\frac{60}{30} = 2$ and $\frac{4}{2} = 2$, confirming linear relationship.

16. A $KE$ vs. mass graph passes through $(2,30)$ and $(4,60)$ at constant speed. What does this show about $KE$ and $m$?

    $KE\propto m$. Doubling mass doubles KE, showing direct proportionality.

17. Which option matches a $KE$ vs $v$ graph at constant $m$: constant slope or increasing slope?

    Increasing slope. Quadratic curves steepen as speed increases.

18. Which variable has a squared effect on kinetic energy: mass or speed?

    Speed ($v$) is squared in $KE = \frac{1}{2}mv^2$. The $v^2$ term means speed changes have quadratic effect on KE.

19. Identify the kinetic energy ratio when mass is halved and speed stays constant.

    Kinetic energy becomes $\frac{1}{2}$ as large. KE is directly proportional to mass.

20. Choose the larger kinetic energy: object A has $m=2m_0, v=v_0$; object B has $m=m_0, v=2v_0$.

    Object B. B has $4$ times A's kinetic energy since speed is squared in the formula.

21. Find the kinetic energy ratio if speed halves and mass stays constant: $\frac{KE_2}{KE_1}=?$

    $\frac{KE_2}{KE_1}=\frac{1}{4}$. Since $KE \propto v^2$, halving speed gives $(\frac{1}{2})^2 = \frac{1}{4}$ the energy.

22. What happens to $KE$ when mass triples and speed stays the same?

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

23. If a $KE$ vs. $m$ line goes through $(2,10)$, what is $KE$ at $m=6$ (same $v$)?

    $30$. Linear relationship: if $m$ triples from 2 to 6, $KE$ triples from 10 to 30.

24. Two points on a $KE$ vs. $m$ graph are $(m,KE)=(2,10)$ and $(4,20)$. Identify if $v$ is constant.

    Yes; $KE$ doubles when $m$ doubles. The ratio $\frac{20}{10} = \frac{4}{2}$ confirms direct proportionality.

25. Identify how kinetic energy changes when speed doubles and mass stays constant.

    Kinetic energy becomes $4$ times as large. KE depends on speed squared, so doubling speed quadruples KE.

26. Which change increases kinetic energy more: doubling mass or doubling speed?

    Doubling speed increases kinetic energy more. Speed is squared in the formula, mass is not.

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

    Kinetic energy becomes $9\times$ larger. Since KE ∝ $v^2$, tripling speed gives $3^2 = 9$ times more KE.

28. A graph compares two objects at the same speed; Object A has twice the mass of B. Which has greater $KE$ and by how much?

    Object A; $2$ times the kinetic energy. At same speed, $KE \propto m$, so twice the mass gives twice the $KE$.

29. Identify the proportional relationship between $KE$ and speed when mass is constant.

    $KE \propto v^2$. Quadratic relationship means $KE$ increases with speed squared.

30. Which variable is squared in $KE = \frac{1}{2}mv^2$: mass or speed?

    Speed is squared: $v^2$. The exponent on $v$ is $2$, while mass has exponent $1$.

31. A graph shows $KE$ becomes $4$ times larger when $v$ doubles at constant $m$. Identify the relationship.

    Square relationship: $KE \propto v^2$. Quadrupling $KE$ when doubling $v$ confirms the square relationship.

32. What happens to kinetic energy if speed triples while mass stays the same?

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

33. A graph shows $KE=50\,\text{J}$ at $v=5\,\text{m/s}$. What is $KE$ at $v=10\,\text{m/s}$ (same $m$)?

    $200\,\text{J}$. Since $KE \propto v^2$, doubling speed quadruples KE: $50 \times 4 = 200$ J.

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

    $KE=\frac{1}{2}mv^2$. Formula shows KE depends on both mass and speed squared.

35. Identify what the slope represents on a graph of $KE$ (y-axis) vs $m$ (x-axis) at constant $v$.

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

36. Identify the relationship between $KE$ and $m$ when $v$ is constant.

    $KE$ is directly proportional to $m$. From $KE = \frac{1}{2}mv^2$, if $v$ is fixed, $KE$ varies linearly with $m$.

37. Identify the kinetic energy ratio when speed triples and mass stays constant.

    Kinetic energy becomes $9$ times as large. Speed is squared, so tripling it multiplies KE by $3^2$.

38. What happens to $KE$ when speed is cut in half and mass stays the same?

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

39. Two objects have the same speed. Object A has twice the mass of B. Which has greater $KE$ and by what factor?

    Object A; $2$ times the kinetic energy. With same $v$, twice the mass gives twice the kinetic energy.

40. A $KE$ vs. speed graph shows speed increases by a factor of $3$. By what factor should $KE$ increase?

    $9$. Since $KE \propto v^2$, factor of $3$ in speed gives $3^2 = 9$ in $KE$.

41. Which change increases kinetic energy more: doubling mass or doubling speed (starting from same $m$ and $v$)?

    Doubling speed (it makes $KE$ $4$ times). Doubling speed quadruples $KE$; doubling mass only doubles $KE$.

42. Identify the meaning of a point at $m=0$ on a $KE$ vs. $m$ graph (with constant $v$).

    $KE=0$ at $m=0$. No mass means no kinetic energy, regardless of speed.

43. Identify the relationship between $KE$ and mass $m$ when speed is constant.

    $KE \propto m$ (linear relationship). From $KE = \frac{1}{2}mv^2$, kinetic energy is directly proportional to mass.

44. What is the relationship between kinetic energy and speed when mass is constant?

    Proportional to speed squared: $KE \propto v^2$. Doubling speed quadruples KE since it's squared in the formula.

45. Identify what the slope represents on a $KE$ vs $m$ graph when speed $v$ is constant.

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

46. Identify the kinetic energy ratio if speed changes from $v$ to $\frac{3}{2}v$ and mass is constant.

    Ratio $KE_2:KE_1 = \frac{9}{4}:1$. $(\frac{3}{2})^2 = \frac{9}{4}$, so KE increases by factor $\frac{9}{4}$.

47. Find the missing point: If $m$ is constant and $KE=20$ at $v=2$, what is $KE$ at $v=4$?

    $80$. Speed doubles: $(4)^2/(2)^2 = 16/4 = 4$, so $KE = 20 \times 4 = 80$.

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

    $KE=\frac{1}{2}mv^2$. Formula shows KE depends on mass and square of speed.

49. If $m$ stays constant and a graph shows $KE$ increases by a factor of $9$, what is the factor change in $v$?

    $v$ increases by a factor of $3$. Since $KE \propto v^2$, if $KE$ increases $9\times$, then $v$ increases $\sqrt{9} = 3\times$.

50. State the formula for kinetic energy using mass $m$ and speed $v$.

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

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

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

52. Which quantity changes faster on a graph: $KE$ vs $v$ or $KE$ vs $m$, and why?

    $KE$ vs $v$, because it depends on $v^2$. The squared relationship causes faster changes than linear.

53. Use the graph: at $m=2\,\text{kg}$, $KE=16\,\text{J}$; at $m=4\,\text{kg}$, $KE=32\,\text{J}$. What is the relationship?

    $KE$ is directly proportional to $m$. Doubling mass from $2$ to $4$ kg doubles $KE$ from $16$ to $32$ J.

54. Which graph shape best represents $KE$ vs. $v$ when $m$ is constant: line or curve?

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

55. Which variable has a squared effect on kinetic energy: mass or speed?

    Speed; $KE\propto v^2$. Speed appears squared in the formula while mass is linear.

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

    Kinetic energy doubles. Since $KE \propto m$, doubling mass doubles KE when speed is constant.

57. A graph shows $KE$ increases from $10\,\text{J}$ to $40\,\text{J}$ while mass stays constant. Identify the speed factor change.

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

58. Find the ratio $\frac{KE_2}{KE_1}$ if mass doubles and speed is cut in half.

    $\frac{1}{2}$. New KE = $\frac{1}{2}(2m)(\frac{v}{2})^2 = \frac{1}{2}(2m)(\frac{v^2}{4}) = \frac{1}{2}$ original.

59. Identify the best graph-based conclusion if $KE$ vs $m$ is a straight line through the origin at fixed $v$.

    $KE$ is directly proportional to $m$. A straight line through origin indicates direct proportionality.

60. A graph shows $KE$ decreases from $50\,\text{J}$ to $25\,\text{J}$ with constant speed. Identify the mass factor change.

    Mass was halved (factor $\frac{1}{2}$). Since $KE \propto m$ at constant speed, halving $KE$ means mass halved.

61. Which graph shape best represents $KE$ vs. $v$ when $m$ is constant?

    An upward-curving parabola-like shape. Quadratic relationship ($v^2$) creates a parabola.

62. What happens to $KE$ when speed doubles and mass stays the same?

    $KE$ becomes $4$ times as large. Since $KE \propto v^2$, doubling speed quadruples kinetic energy.

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

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

64. Which graph shape indicates $KE \propto v^2$ when mass is constant: straight line or curve?

    Upward-curving parabola. Quadratic relationships produce parabolic curves on standard axes.

65. A graph shows $KE$ increases from $30\,J$ to $120\,J$ when speed doubles. What does this imply about $KE$ vs. $v$?

    It implies $KE$ is proportional to $v^2$. KE increased by factor of $4$ when speed doubled, confirming $v^2$ relationship.

66. Identify how kinetic energy changes when mass doubles and speed stays constant.

    Kinetic energy doubles. Since $KE \propto m$, doubling mass doubles kinetic energy.

67. Identify the kinetic energy ratio if mass doubles and speed triples at the same time.

    $\frac{KE_2}{KE_1} = 18$. Combined effect: $2m \times (3v)^2 = 2m \times 9v^2 = 18(\frac{1}{2}mv^2)$.

68. Which graph shape indicates $KE \propto v^2$ when plotting $KE$ vs. $v$?

    A curve that gets steeper as $v$ increases. Quadratic relationships produce parabolic curves with increasing slope.

69. If $v$ stays constant and a graph shows $KE$ decreases to $\frac{1}{3}$, what happens to $m$?

    $m$ decreases to $\frac{1}{3}$. Since $KE \propto m$, if $KE$ becomes $\frac{1}{3}$, then $m$ becomes $\frac{1}{3}$.

70. Which option matches a linear $KE$ vs $m$ graph at constant $v$: constant slope or changing slope?

    Constant slope. Linear relationships have unchanging slopes.

71. Identify the graph shape expected for $KE$ vs. speed when mass is constant.

    An upward-curving parabola through the origin. Quadratic relationship between $KE$ and speed creates a parabola.

72. A $KE$ vs $m$ graph at fixed $v$ passes through $(2\text{ kg}, 20\text{ J})$ and $(4\text{ kg}, 40\text{ J})$. Identify the relationship.

    Linear; $KE$ is directly proportional to $m$. Doubling mass doubles $KE$, confirming direct proportionality.

73. What happens to kinetic energy if speed doubles while mass stays the same?

    Kinetic energy becomes $4$ times larger. Since $KE \propto v^2$, doubling speed quadruples kinetic energy.

74. Two objects have the same mass. Object A has $v=3\,m/s$ and B has $v=6\,m/s$. What is $\frac{KE_B}{KE_A}$?

    $4$. Same mass means $KE$ ratio equals speed ratio squared: $(\frac{6}{3})^2 = 4$.

75. If $KE$ stays the same but mass increases by a factor of $4$, how must speed change?

    Speed must be cut in half. To keep $KE = \frac{1}{2}mv^2$ constant when $m$ quadruples, $v^2$ must be $\frac{1}{4}$.

76. A graph shows $KE$ doubles when $m$ doubles at constant $v$. Identify the relationship.

    Direct proportion: $KE \propto m$. Doubling one variable doubles the other shows direct proportionality.

77. Identify the relationship between kinetic energy and mass when speed is constant.

    Directly proportional: $KE \propto m$. From $KE = \frac{1}{2}mv^2$, kinetic energy increases linearly with mass.

78. Two lines on a $KE$ vs. $m$ graph have slopes $2$ and $8$. Which object has the greater speed?

    The line with slope $8$ has the greater speed. Higher slope means larger $\frac{1}{2}v^2$, thus greater $v$.

79. If a graph shows $KE$ increases by $30\,\text{J}$ when mass increases by $2\,\text{kg}$ (same $v$), what is the slope?

    $15\,\text{J/kg}$. Slope = $\frac{\Delta KE}{\Delta m} = \frac{30}{2} = 15$ J/kg.

80. Object A has $m=2\,\text{kg}, v=3\,\text{m/s}$; Object B has $m=2\,\text{kg}, v=6\,\text{m/s}$. Which has greater $KE$?

    Object B has greater kinetic energy. B's speed is double A's, so B has $2^2 = 4$ times the kinetic energy.

81. Find the kinetic energy ratio if mass halves and speed stays constant: $\frac{KE_2}{KE_1}=?$

    $\frac{KE_2}{KE_1}=\frac{1}{2}$. Since $KE \propto m$, halving mass halves kinetic energy.

82. A graph shows $KE=72\,\text{J}$ at $v=6\,\text{m/s}$. What is $KE$ at $v=3\,\text{m/s}$ (same $m$)?

    $18\,\text{J}$. Since $KE \propto v^2$, halving speed gives $\frac{1}{4}$ the KE: $72 \times \frac{1}{4} = 18$ J.

83. Identify the kinetic energy ratio when speed is halved and mass stays constant.

    Kinetic energy becomes $\frac{1}{4}$ as large. Halving speed reduces KE by $(\frac{1}{2})^2$.

84. Identify how kinetic energy changes when speed doubles and mass stays constant.

    Kinetic energy becomes $4$ times as large. Since $KE \propto v^2$, doubling speed quadruples kinetic energy.

85. Which graph shape indicates $KE \propto m$ when speed is constant: straight line or curve?

    Straight line (linear). Direct proportionality produces a linear graph through the origin.

86. Which change increases kinetic energy more: doubling mass or doubling speed (same start values)?

    Doubling speed increases kinetic energy more. Speed squared ($2^2=4$) beats mass doubled ($2$).

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

    Kinetic energy becomes $\frac{1}{4}$ as large. Halving speed gives $(\frac{1}{2})^2 = \frac{1}{4}$ the original KE.

88. Find the missing point: If $v$ is constant and $KE=15$ at $m=3$, what is $KE$ at $m=6$?

    $30$. Mass doubles while speed is constant, so KE doubles: $15 \times 2 = 30$.

89. Find the ratio $\frac{KE_2}{KE_1}$ if speed stays constant and mass changes from $m$ to $3m$.

    $3$. KE ratio equals mass ratio when speed is constant.

90. Identify the relationship shown by a straight-line $KE$ vs. mass graph at constant speed.

    Directly proportional: $KE \propto m$. Linear graph shows KE increases at same rate as mass increases.

91. Using $KE=\frac{1}{2}mv^2$, what is the ratio $\frac{KE_2}{KE_1}$ if $m$ is unchanged and $v_2=2v_1$?

    $\frac{KE_2}{KE_1}=4$. Substitute $v_2 = 2v_1$ into ratio: $\frac{(2v_1)^2}{v_1^2} = 4$.

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

    Kinetic energy becomes $4\times$ larger. KE depends on $v^2$, so doubling speed quadruples KE.

93. What is the effect on kinetic energy when mass doubles and speed stays constant?

    Kinetic energy doubles. KE is directly proportional to mass when $v$ is constant.

94. A graph shows $KE$ vs. $m$ is a straight line through the origin. What does this imply about speed?

    Speed is constant. Linear KE vs. m relationship only occurs when v doesn't change.

95. What happens to kinetic energy if mass doubles while speed stays the same?

    Kinetic energy doubles. Since $KE \propto m$, doubling $m$ doubles $KE$.

96. On a $KE$ vs. $v$ graph, $KE$ is $16\,J$ at $v=4\,m/s$. Predict $KE$ at $v=2\,m/s$ (same $m$).

    $4\,J$. Halving speed reduces $KE$ by factor of $4$: $16 \div 4 = 4\,J$.

97. Identify which object has greater kinetic energy if both have the same speed but one has greater mass.

    The object with greater mass has greater kinetic energy. More mass means more KE at the same speed.

98. A graph shows $KE=40\,\text{J}$ at $m=2\,\text{kg}$. What is $KE$ at $m=6\,\text{kg}$ (same $v$)?

    $120\,\text{J}$. Since $KE \propto m$, tripling mass from 2 to 6 kg triples KE: $40 \times 3 = 120$ J.

99. Choose the correct comparison: which change increases $KE$ more, doubling $m$ or doubling $v$?

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

100. A graph shows $KE$ stays the same while speed increases by a factor of $2$. What happens to mass?

     Mass becomes $\frac{1}{4}$ as large. To keep KE constant when $v^2$ quadruples, mass must quarter.

101. What happens to kinetic energy if speed is cut in half while mass stays the same?

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

102. Two points on a $KE$ vs. $v$ graph are $(v,KE)=(2,8)$ and $(4,32)$. Identify if $m$ is constant.

     Yes; doubling $v$ makes $KE$ $4$ times larger. The ratio $\frac{32}{8} = 4 = (\frac{4}{2})^2$ confirms the $v^2$ relationship.

103. A straight-line graph of $KE$ vs $m$ passes through $(0,0)$ and $(5\,\text{kg},100\,\text{J})$. What is $KE$ at $m=2\,\text{kg}$?

     $40\,\text{J}$. Linear relationship: $\frac{100}{5} = 20$ J/kg, so at 2 kg: $20 \times 2 = 40$ J.

104. Which relationship is linear on a graph: $KE$ vs $m$ (constant $v$) or $KE$ vs $v$ (constant $m$)?

     $KE$ vs $m$ is linear when $v$ is constant. KE is directly proportional to mass, creating a linear relationship.

105. A $KE$ vs. $m$ graph shows $KE=40\,\text{J}$ at $m=2\,\text{kg}$. Find $KE$ at $m=5\,\text{kg}$.

     $100\,\text{J}$. KE is proportional to mass: $\frac{5}{2} \times 40 = 100\,\text{J}$.

106. What happens to $KE$ when speed triples and mass stays the same?

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

107. Identify the relationship shown by a curved upward $KE$ vs. speed graph at constant mass.

     Quadratic: $KE \propto v^2$. Upward curve shows KE increases with square of speed.

108. What happens to kinetic energy if mass doubles while speed stays the same?

     Kinetic energy doubles. Since $KE = \frac{1}{2}mv^2$, doubling $m$ doubles $KE$ when $v$ is constant.

109. Find $\frac{KE_2}{KE_1}$ if mass doubles and speed triples.

     $\frac{KE_2}{KE_1}=18$. Mass doubles ($\times 2$) and speed triples ($\times 3^2 = 9$), so $2 \times 9 = 18$.

110. Which graph is linear: $KE$ vs. $m$ at constant $v$, or $KE$ vs. $v$ at constant $m$?

     $KE$ vs. $m$ at constant $v$. Linear because $KE$ is directly proportional to $m$ when $v$ is fixed.

111. Find the ratio $\frac{KE_2}{KE_1}$ if mass stays constant and speed changes from $v$ to $2v$.

     $4$. Speed ratio squared: $(2)^2 = 4$

112. Which graph shape best matches $KE$ vs. $m$ for constant speed: linear or upward-curving?

     Linear. KE ∝ $m$ creates a straight line through the origin.

113. What happens to kinetic energy if speed doubles while mass stays constant?

     Kinetic energy becomes $4$ times larger. Since $KE \propto v^2$, doubling speed quadruples kinetic energy.

114. Identify how kinetic energy changes when speed doubles and mass stays constant.

     Kinetic energy becomes $4$ times larger. Since $KE \propto v^2$, doubling speed quadruples KE.

115. Identify the kinetic energy ratio $\frac{KE_2}{KE_1}$ when $m_2=m_1$ and $v_2=3v_1$.

     $\frac{KE_2}{KE_1}=9$. Since $KE \propto v^2$, tripling speed gives $3^2 = 9$ times the energy.

116. Which graph should be a straight line through the origin: $KE$ vs $m$ at constant $v$?

     A straight line through the origin. Direct proportionality between KE and mass creates this linear graph.

117. What is the relationship between kinetic energy and mass when speed is constant?

     Directly proportional: $KE \propto m$. When speed is fixed, doubling mass doubles KE.

118. A curved $KE$ vs $v$ graph steepens as $v$ increases. Identify what this indicates about the $KE$-$v$ relationship.

     $KE$ increases with $v^2$ (quadratic growth). The steepening curve shows $KE$ grows faster than linearly with $v$.

119. Which object has greater kinetic energy: same mass, speeds $10\,\text{m/s}$ vs. $20\,\text{m/s}$?

     The $20\,\text{m/s}$ object. Higher speed gives more KE since $KE \propto v^2$.

120. Identify the graph shape expected for $KE$ vs. mass when speed is constant.

     A straight line through the origin. Direct proportionality between $KE$ and mass creates a linear graph.

121. Identify the meaning of a steeper slope on a $KE$ vs. $m$ graph (with constant $v$).

     Greater $v$ (larger proportionality constant). Slope = $\frac{1}{2}v^2$, so steeper means higher $v$.

122. Using $KE=\frac{1}{2}mv^2$, what is $\frac{KE_2}{KE_1}$ if $v$ is unchanged and $m_2=\frac{1}{2}m_1$?

     $\frac{KE_2}{KE_1}=\frac{1}{2}$. Mass halved means KE halved when speed stays constant.

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

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

124. Identify the kinetic energy ratio if speed changes from $v$ to $2v$ and mass is constant.

     Ratio $KE_2:KE_1 = 4:1$. Doubling speed quadruples KE when mass is constant.

125. Find the kinetic energy ratio if mass triples and speed stays constant: $\frac{KE_2}{KE_1}=?$

     $\frac{KE_2}{KE_1}=3$. Since $KE \propto m$, tripling mass triples kinetic energy.

126. If mass is halved and speed is tripled, by what factor does kinetic energy change?

     $\frac{9}{2}$ times. Factor = $(\frac{1}{2})(3^2) = \frac{1}{2} \times 9 = \frac{9}{2}$.

127. What happens to $KE$ when speed is cut in half and mass stays the same?

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

128. Which graph shape best represents $KE$ vs. $m$ when $v$ is constant: line or curve?

     Line (linear relationship). Direct proportionality between $KE$ and $m$ creates a straight line.

129. Identify the kinetic energy ratio $\frac{KE_2}{KE_1}$ when $v_2=v_1$ and $m_2=\frac{3}{2}m_1$.

     $\frac{KE_2}{KE_1}=\frac{3}{2}$. Since $KE \propto m$, multiplying mass by $\frac{3}{2}$ multiplies $KE$ by $\frac{3}{2}$.

130. State the formula for kinetic energy in terms of mass and speed.

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

131. What does the slope of a $KE$ vs. $v^2$ graph represent when mass is constant?

     $\frac{1}{2}m$. From $KE = \frac{1}{2}mv^2$, slope equals $\frac{1}{2}m$.

132. Choose the correct statement: On a $KE$ vs $v^2$ graph (constant $m$), the line is straight or curved?

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

133. Identify the relationship shown by a curved upward $KE$ vs. speed graph at constant mass.

     Quadratic: $KE \propto v^2$. Upward curve shows KE increases faster as speed increases.

134. A $KE$ vs. mass graph shows $KE$ doubles when mass doubles. What does this imply about $KE$ vs. $m$?

     It implies $KE$ is directly proportional to $m$. Equal increases show linear relationship between KE and mass.

135. What happens to kinetic energy if speed triples while mass stays the same?

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

136. A graph shows $KE$ increases from $12\,\text{J}$ to $24\,\text{J}$ with constant speed. Identify the mass factor change.

     Mass doubled (factor $2$). Since $KE \propto m$ at constant speed, doubling $KE$ means mass doubled.

137. Identify the kinetic energy ratio $\frac{KE_2}{KE_1}$ when $m_2=2m_1$ and $v_2=\frac{1}{2}v_1$.

     $\frac{KE_2}{KE_1}=\frac{1}{2}$. Doubling mass ($\times 2$) and halving speed ($\times \frac{1}{4}$) gives $2 \times \frac{1}{4} = \frac{1}{2}$.

138. Use the graph: at $v=2\,\text{m/s}$, $KE=8\,\text{J}$; at $v=4\,\text{m/s}$, $KE=32\,\text{J}$. What changed?

     Speed doubled, and $KE$ increased by a factor of $4$. $KE$ increased by factor of $\frac{32}{8} = 4 = 2^2$, so speed doubled.

139. What is the effect on kinetic energy when speed triples and mass stays constant?

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

140. What happens to kinetic energy if mass doubles while speed stays the same?

     Kinetic energy doubles. Since $KE \propto m$, doubling mass doubles kinetic energy.

141. Identify the correct conclusion from a $KE$ vs. $v$ graph that curves upward as $v$ increases.

     $KE$ increases faster than speed because $KE \propto v^2$. Parabolic growth shows $KE$ increases with square of speed.

142. Identify how kinetic energy changes when speed is halved and mass stays constant.

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

143. Identify what the slope represents on a $KE$ vs $v^2$ graph when mass $m$ is constant.

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

144. Identify how kinetic energy changes when speed triples and mass stays constant.

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

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

     $KE \propto v^2$ (quadratic relationship). From $KE = \frac{1}{2}mv^2$, kinetic energy varies with the square of speed.

146. Identify how kinetic energy changes when mass doubles and speed stays constant.

     Kinetic energy doubles. Since $KE \propto m$, doubling mass doubles KE when speed is fixed.

147. Object A has $m=2\,\text{kg}, v=4\,\text{m/s}$; Object B has $m=4\,\text{kg}, v=2\,\text{m/s}$. Which has greater $KE$?

     Object A has greater kinetic energy. A: $KE = \frac{1}{2}(2)(16) = 16$ J; B: $KE = \frac{1}{2}(4)(4) = 8$ J.

148. Identify how kinetic energy changes when mass is halved and speed stays constant.

     Kinetic energy is halved. Since $KE \propto m$, halving mass halves kinetic energy.

149. Two objects have the same mass. Object A has $3$ times the speed of B. Which has greater $KE$ and by what factor?

     Object A; $9$ times the kinetic energy. With same $m$, triple speed gives $3^2 = 9$ times the energy.

150. Identify the relationship shown by a straight-line $KE$ vs. mass graph at constant speed.

     Directly proportional: $KE \propto m$. Linear graph shows KE increases proportionally with mass.

151. A graph shows $KE$ decreases from $36\,\text{J}$ to $9\,\text{J}$ with constant mass. Identify the speed factor change.

     Speed was halved (factor $\frac{1}{2}$). $KE$ decreased by factor of $\frac{1}{4}$, so speed decreased by $\sqrt{\frac{1}{4}} = \frac{1}{2}$.

152. If $KE$ stays the same but speed increases by a factor of $3$, how must mass change?

     Mass must become $\frac{1}{9}$ as large. To keep $KE = \frac{1}{2}mv^2$ constant when $v$ triples, $m$ must be $\frac{1}{9}$.

153. What happens to kinetic energy if mass doubles while speed stays constant?

     Kinetic energy doubles. Since $KE \propto m$, doubling mass doubles kinetic energy.

154. Identify the relationship between kinetic energy and mass when speed is constant.

     Directly proportional: $KE \propto m$. From $KE = \frac{1}{2}mv^2$, kinetic energy varies linearly with mass.

155. What does the slope of a $KE$ vs. $m$ graph equal when speed $v$ is constant?

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

156. Identify the relationship between $KE$ and $v$ when $m$ is constant.

     $KE$ is proportional to $v^2$. From $KE = \frac{1}{2}mv^2$, if $m$ is fixed, $KE$ varies with square of $v$.

157. Identify how kinetic energy changes when speed is cut in half and mass stays constant.

     Kinetic energy becomes $\frac{1}{4}$ as large. Halving speed gives $(\frac{1}{2})^2 = \frac{1}{4}$ of original KE.

158. Identify the kinetic energy ratio if mass stays constant and speed changes from $v$ to $2v$.

     $\frac{KE_2}{KE_1} = 4$. Speed squared: $(2v)^2 = 4v^2$, so KE increases by factor of 4.

159. Identify the relationship shown by a curved upward graph of $KE$ vs. $v$ when $m$ is constant.

     $KE$ is proportional to $v^2$. Parabolic curve shows quadratic relationship between KE and speed.

160. A $KE$ vs. speed graph shows $KE$ is $10\,J$ at $2\,m/s$ and $40\,J$ at $4\,m/s$. What is the pattern?

     When $v$ doubles, $KE$ becomes $4$ times as large. KE increased by factor of $4$ when speed doubled ($40÷10=4$).

161. What does the slope represent on a $KE$ vs. $m$ graph when speed is constant?

     Slope $= \frac{1}{2}v^2$. For $KE = \frac{1}{2}mv^2$, the coefficient of $m$ gives the slope.

162. Identify what it means if a $KE$ vs $m$ graph has slope $8$ in SI units.

     The constant factor is $\frac{1}{2}v^2 = 8$. Speed must be $4$ m/s since $\frac{1}{2}(4)^2 = 8$.

163. Identify which change increases kinetic energy more: doubling mass or doubling speed (starting from same $m$ and $v$).

     Doubling speed. Speed appears squared in $KE$, so its effect is greater than mass.

164. On a $KE$ vs. $m$ graph, $KE$ is $30\,J$ at $m=3\,kg$. Predict $KE$ at $m=5\,kg$ (same $v$).

     $50\,J$. Since $KE \propto m$, use ratio: $\frac{30}{3} = \frac{KE}{5}$, so $KE = 50\,J$.

165. Which graph shape indicates $KE \propto m$ when plotting $KE$ vs. $m$?

     A straight line through the origin. Linear relationships produce straight lines passing through $(0,0)$.

166. If two points on a $KE$ vs. speed graph are $(2, 20)$ and $(4, 80)$, what relationship is shown?

     Quadratic (doubling speed makes $KE$ $4$ times). $\frac{80}{20} = 4$ and $\frac{4}{2} = 2$, so $KE$ increases as speed squared.

167. What happens to $KE$ when speed triples and mass stays the same?

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

168. What does the slope represent on a graph of $KE$ (y-axis) versus $v^2$ (x-axis) at fixed $m$?

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

169. Using $KE=\frac{1}{2}mv^2$, what is $KE$ for $m=4$ and $v=2$ (no units)?

     $8$. $KE = \frac{1}{2}(4)(2)^2 = \frac{1}{2}(4)(4) = 8$

170. A $KE$ vs. mass graph shows mass decreases by a factor of $5$. By what factor should $KE$ change?

     $\frac{1}{5}$. Since $KE \propto m$, factor of $\frac{1}{5}$ in mass gives same factor in $KE$.

171. A graph shows $KE=50$ J at $v=5$ m/s for an object. What is $KE$ at $v=10$ m/s (same $m$)?

     $200$ J. Doubling speed quadruples $KE$: $50 \times 4 = 200$ J.

172. What happens to kinetic energy if speed is cut in half while mass stays constant?

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

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

     $KE = \frac{1}{2}mv^2$. Formula shows KE depends on mass and speed squared.

174. Which graph shape best matches $KE$ versus speed $v$ when mass is constant?

     An upward-curving parabola (quadratic increase). The $v^2$ relationship creates a parabolic curve.

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

     $KE = \frac{1}{2}mv^2$. This fundamental formula shows KE depends on both mass and speed squared.

176. If a $KE$ vs. speed graph curves upward, what does that imply about the effect of increasing speed?

     Each increase in $v$ adds more $KE$ than the previous increase. Squared relationship means each increment has greater effect.

177. A $KE$ vs. speed graph shows $KE=25$ J at $v=5$ and $KE=100$ J at $v=10$. What does this show?

     $KE\propto v^2$. Quadrupling KE when speed doubles confirms squared relationship.

178. A graph shows $KE$ changes from $20\,J$ to $60\,J$ at constant speed. What happened to mass?

     Mass tripled. $\frac{60}{20} = 3$, so mass increased by factor of $3$.

179. A graph shows $KE$ changes from $36\,J$ to $4\,J$ at constant mass. What happened to speed?

     Speed became $\frac{1}{3}$ as large. $\frac{4}{36} = \frac{1}{9} = (\frac{1}{3})^2$, so speed decreased by factor of $3$.

180. Identify the relationship between kinetic energy and speed when mass is constant.

     Quadratic: $KE \propto v^2$. From $KE = \frac{1}{2}mv^2$, kinetic energy varies with speed squared.

181. What happens to kinetic energy if speed is cut in half while mass stays the same?

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

182. If a $KE$ vs. $m$ line passes through the origin, what does that indicate about the relationship?

     Proportional relationship: $KE$ is directly proportional to $m$. A line through the origin indicates direct proportionality with no offset.

183. A $KE$ vs. $m$ graph shows $KE=12$ J at $m=3$ kg. What is $KE$ at $m=1$ kg (same $v$)?

     $4$ J. Direct proportion: $\frac{KE_1}{m_1} = \frac{KE_2}{m_2}$, so $\frac{12}{3} = \frac{KE_2}{1}$.

184. What happens to kinetic energy if speed doubles while mass stays constant?

     Kinetic energy becomes $4$ times as large. Since $KE \propto v^2$, doubling speed quadruples kinetic energy.

185. A graph shows $KE$ doubles while mass stays constant. What is $\frac{v_2}{v_1}$?

     $\sqrt{2}$. KE doubled means $v^2$ doubled, so $v$ increased by $\sqrt{2}$.

186. Which change increases kinetic energy more: doubling mass or doubling speed?

     Doubling speed increases kinetic energy more. Speed squared means $2^2 = 4$ times increase vs only $2$ times for mass.

187. What happens to $KE$ when mass doubles and speed stays the same?

     $KE$ doubles. Since $KE \propto m$, doubling mass doubles kinetic energy.

188. Identify which object has greater kinetic energy if both have the same mass but one has greater speed.

     The object with greater speed has greater kinetic energy. Higher speed means more KE at the same mass.

189. Which graph shape best represents $KE$ vs. $m$ when $v$ is constant?

     A straight line through the origin. Direct proportionality creates a linear relationship.

190. Which graph shape indicates $KE \propto m$ when speed is constant: straight line or curve?

     Straight line (linear). Direct proportionality creates a linear relationship.

191. What happens to $KE$ when mass doubles and speed stays the same?

     $KE$ doubles. Since $KE \propto m$, doubling mass doubles kinetic energy.

192. Which graph shape best matches $KE$ versus mass $m$ when speed is constant?

     A straight line through the origin (linear increase). Direct proportionality to mass creates a linear relationship.

193. What happens to kinetic energy if speed doubles while mass stays the same?

     Kinetic energy becomes $4$ times larger. Since $KE \propto v^2$, doubling speed quadruples kinetic energy.

194. Which object has greater kinetic energy: same speed, masses $2\,\text{kg}$ vs. $6\,\text{kg}$?

     The $6\,\text{kg}$ object. Higher mass gives more KE since $KE \propto m$.

195. Identify the kinetic energy ratio when mass triples and speed stays constant.

     Kinetic energy becomes $3$ times as large. KE scales linearly with mass at constant speed.

196. A graph shows $KE$ increases from $18$ to $72$ when only mass changes. What is $\frac{m_2}{m_1}$?

     $4$. KE increased 4×, so mass increased 4× at constant speed.

197. If a $KE$ vs. mass graph is a straight line through the origin, what does the origin crossing mean physically?

     If $m=0$, then $KE=0$ (for constant $v$). No mass means no matter to have kinetic energy.

198. Identify the relationship shown by a straight-line graph of $KE$ vs. $m$ when $v$ is constant.

     $KE$ is directly proportional to $m$. Linear relationship means KE changes by same factor as mass.

199. Identify the best x-axis choice to make a linear graph for speed effects: $v$ or $v^2$ (with $KE$ on y-axis).

     Use $v^2$. Plotting $KE$ vs $v^2$ linearizes the quadratic relationship.

200. From a $KE$ vs. $v$ graph, $KE$ at $3\,m/s$ is $45\,J$. Predict $KE$ at $6\,m/s$ for the same mass.

     $180\,J$. Speed doubles so KE quadruples: $45 × 4 = 180\,J$.

201. Identify the proportional relationship between $KE$ and mass $m$ when speed is constant.

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

202. Identify the kinetic energy ratio if mass changes from $m$ to $3m$ and speed is constant.

     Ratio $KE_2:KE_1 = 3:1$. Tripling mass triples KE when speed is constant.

203. What happens to kinetic energy if mass doubles while speed stays constant?

     Kinetic energy doubles. Since $KE \propto m$, doubling mass doubles kinetic energy.

204. Identify the kinetic energy ratio if mass doubles and speed triples at the same time.

     Ratio $KE_2:KE_1 = 18:1$. Mass doubles ($\times 2$) and speed triples ($\times 3^2 = 9$): $2 \times 9 = 18$.

205. On a $KE$ vs. $v$ graph, $KE=16$ J at $v=4$ m/s. What is $KE$ at $v=2$ m/s (same $m$)?

     $4$ J. Halving speed: $KE' = KE(\frac{1}{2})^2 = \frac{1}{4} \times 16 = 4$ J.

206. Which graph is linear: $KE$ vs. $m$ (constant $v$) or $KE$ vs. $v$ (constant $m$)?

     $KE$ vs. $m$ at constant $v$. KE is linear in mass but quadratic in speed.

207. A graph shows $KE$ rising from $5\,J$ to $20\,J$ when $v$ rises from $2$ to $4\,m/s$. What is the relationship?

     $KE \propto v^2$ (doubling $v$ makes $KE$ $4\times$). Speed doubled while $KE$ quadrupled, confirming the $v^2$ relationship.

208. On a $KE$ vs $m$ graph at constant $v$, how does the line compare for a larger constant speed?

     Steeper slope because slope $= \frac{1}{2}v^2$. Higher speed increases the slope proportionally to $v^2$.

209. What is the effect on kinetic energy when speed is cut in half and mass stays constant?

     Kinetic energy becomes $\frac{1}{4}$ as large. Halving speed gives $(\frac{1}{2})^2 = \frac{1}{4}$ of original KE.

210. A graph shows $KE$ rising from $10\,J$ to $20\,J$ when $m$ rises from $2$ to $4\,kg$. What is the relationship?

     $KE \propto m$ (doubling $m$ doubles $KE$). Both values doubled, confirming direct proportionality to mass.

211. Which graph should be curved upward: $KE$ vs $v$ at constant $m$?

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

212. Which change produces a larger $KE$ increase: doubling mass or doubling speed (starting from same $m$ and $v$)?

     Doubling speed produces the larger increase. Doubling speed quadruples $KE$; doubling mass only doubles $KE$.

213. Two objects have the same speed; one has $3$ times the mass. What is the kinetic energy comparison?

     The heavier object has $3$ times the $KE$. At same speed, KE is directly proportional to mass.

214. Identify how kinetic energy changes when speed triples and mass stays constant.

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

215. Using $KE = \frac{1}{2}mv^2$, find the factor change in $KE$ if $m$ stays constant and $v$ increases by a factor of $5$.

     Factor change is $5^2 = 25$. Only speed changes, so factor is just $5^2 = 25$.

216. Which graph shape best matches $KE$ vs. $v$ for constant mass: linear or upward-curving?

     Upward-curving (quadratic). KE ∝ $v^2$ creates a parabola, not a straight line.

217. Identify which graph would show constant $KE$ while $m$ changes: $KE$ vs. $m$ is horizontal or slanted?

     Horizontal line. Constant $KE$ despite changing $m$ produces a horizontal line.

218. Which relationship is linear: $KE$ vs $m$ (constant $v$) or $KE$ vs $v$ (constant $m$)?

     $KE$ vs $m$ is linear (when $v$ is constant). $KE = \frac{1}{2}mv^2$ shows direct proportionality to $m$ when $v$ is fixed.

219. Which graph should be linear if you plot $KE$ on the $y$-axis: vs. $m$, $v$, or $v^2$?

     Linear for $KE$ vs. $m$ (constant $v$) or $KE$ vs. $v^2$ (constant $m$). Both relationships are linear when the other variable is held constant.

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

     $KE = \frac{1}{2}mv^2$. Shows kinetic energy depends on mass and the square of velocity.

221. A graph shows $KE$ increases from $10$ to $40$ when only speed changes. What is $\frac{v_2}{v_1}$?

     $2$. KE increased 4×, so speed doubled since $v^2$ ratio is 4.

222. If a $KE$ vs. $m$ graph becomes steeper while remaining a line through the origin, what changed about $v$?

     Speed increased (since slope $= \frac{1}{2}v^2$). A steeper slope means larger $\frac{1}{2}v^2$, so speed increased.

223. What does the slope represent on a graph of $KE$ (y-axis) versus mass $m$ (x-axis) at fixed $v$?

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

224. On a $KE$ vs. $v$ graph, $KE=9$ J at $v=3$ m/s. What is $KE$ at $v=6$ m/s (same $m$)?

     $36$ J. Since $KE \propto v^2$: $\frac{9}{3^2} = \frac{KE_2}{6^2}$ gives $KE_2 = 36$ J.

225. From a $KE$ vs. $m$ graph, $KE$ at $2\,kg$ is $50\,J$. Predict $KE$ at $6\,kg$ for the same speed.

     $150\,J$. Mass triples so KE triples: $50 × 3 = 150\,J$.

226. Identify the proportional relationship between $KE$ and mass when speed is constant.

     $KE \propto m$. Direct proportionality means $KE$ increases linearly with mass.

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

     Kinetic energy doubles. KE is directly proportional to mass when speed is constant.

228. What is the effect on kinetic energy when speed doubles and mass stays constant?

     Kinetic energy becomes $4$ times as large. KE depends on $v^2$, so doubling speed quadruples KE.

229. If mass triples and speed doubles, by what factor does kinetic energy change?

     $12$ times. Factor = $(3)(2^2) = 3 \times 4 = 12$.

230. Which statement matches a curved upward $KE$ vs. speed graph when mass is constant?

     $KE\propto v^2$ (quadratic relationship). Quadratic relationships produce upward-curving parabolas.

231. Using $KE = \frac{1}{2}mv^2$, find the factor change in $KE$ if $m$ triples and $v$ doubles.

     Factor change is $3 \times 4 = 12$. Mass triples ($3×$) and speed effect is $2^2 = 4×$, so total is $3 × 4 = 12×$.

232. Identify which change increases kinetic energy more: doubling mass or doubling speed (starting from same $m$ and $v$).

     Doubling speed increases kinetic energy more. Speed affects KE quadratically ($v^2$) while mass affects it linearly.

233. Identify the kinetic energy ratio if speed stays constant and mass changes from $m$ to $3m$.

     $\frac{KE_2}{KE_1} = 3$. Mass triples while speed stays constant, so KE triples.

234. Identify the proportional relationship between $KE$ and speed $v$ when mass is constant.

     $KE\propto v^2$. Kinetic energy varies with the square of speed when mass is fixed.

235. Which statement matches a straight-line $KE$ vs. mass graph when speed is constant?

     $KE\propto m$ (linear relationship). Direct proportionality creates a straight line through origin.

236. Identify the relationship between kinetic energy and speed when mass is constant.

     Proportional to $v^2$: $KE \propto v^2$. From $KE = \frac{1}{2}mv^2$, kinetic energy varies with the square of speed.

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

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

238. A $KE$ vs. $v$ graph shows $KE=50$ at $v=5$ and $KE=200$ at $v=10$. Identify the speed effect.

     Doubling $v$ increases $KE$ by a factor of $4$. From $(5,50)$ to $(10,200)$: speed doubles, $KE$ quadruples.

239. A graph shows $KE$ changes from $25\,J$ to $100\,J$ at constant mass. What happened to speed?

     Speed doubled. $\frac{100}{25} = 4 = 2^2$, so speed increased by factor of $2$.

240. Which change has the larger effect on kinetic energy: doubling mass or doubling speed?

     Doubling speed. Speed affects KE quadratically ($v^2$) while mass affects it linearly.

241. A curved graph shows $KE=20\,\text{J}$ at $v=2\,\text{m/s}$ (same $m$). What is $KE$ at $v=4\,\text{m/s}$?

     $80\,\text{J}$. Since $KE \propto v^2$, doubling speed quadruples KE: $20 \times 4 = 80$ J.

242. What happens to $KE$ when speed doubles and mass stays the same?

     $KE$ becomes $4$ times larger. Since $KE \propto v^2$, doubling speed quadruples kinetic energy.

243. A $KE$ vs $v$ graph at fixed $m$ shows $v$ doubles while $KE$ quadruples. Identify the relationship.

     Quadratic; $KE \propto v^2$. When speed doubles, $KE$ increases by factor of $2^2 = 4$.

244. Two objects have the same speed. Object A has $m=2\,kg$ and B has $m=6\,kg$. What is $\frac{KE_B}{KE_A}$?

     $3$. Same speed means $KE$ ratio equals mass ratio: $\frac{6}{2} = 3$.

245. A graph shows $KE$ increases from $20$ J to $80$ J when speed doubles. What relationship does this support?

     $KE\propto v^2$. Quadrupling KE when speed doubles confirms squared relationship.

246. Identify what it means if a $KE$ vs $m$ graph line passes through the origin.

     At $m=0$, $KE=0$ (direct proportion). Zero mass means zero kinetic energy, showing direct proportionality.

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