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A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial velocity of 3.33 m/s. The car then runs up the frictionless slope, gaining 0.108 m in altitude before leveling out to another horizontal segment at the higher level. What is the final velocity of the car if we neglect air resistance

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boffeemadrid

Answer:

2.994 m/s

Explanation:

m = Mass of the car = 100 g

[tex]v_1[/tex] = Initial velocity = 3.33 m/s

h = Height = 0.108 m

g = Acceleration due to gravity = 9.81 m/s²

We know that energy in the system is conserved so

[tex]\dfrac{1}{2}mv_1^2=mgh+\dfrac{1}{2}mv_f^2\\\Rightarrow v_f=\sqrt{2\left({\dfrac{1}{2}v_1^2-gh}\right)}\\\Rightarrow v_f=\sqrt{2\left(\dfrac{1}{2}3.33^2-9.81\times 0.108\right)}\\\Rightarrow v_f=2.994\ m/s[/tex]

The final velocity of the car is 2.994 m/s

onyebuchinnaji

The final velocity of the car along the horizontal segment is 3.63 m/s.

The given parameters;

  • mass of the toy, m = 100 g
  • initial velocity of the car, u = 3.33 m/s
  • increase in height of the car, Δh = 0.108 m

The final velocity of the car is determined by applying the principle of conservation of mechanical energy;

[tex]\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2 - u^2) = mg(\Delta h)\\\\v^2 - u^2 = 2g\Delta h\\\\v^2 = 2g\Delta h + u^2\\\\v = \sqrt{2g\Delta h + u^2} \\\\v = \sqrt{(2 \times 9.8 \times 0.108) \ + \ (3.33)^2} \\\\v = 3.63 \ m/s[/tex]

Thus, the final velocity of the car along the horizontal segment is 3.63 m/s.

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