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Sophie [7]
3 years ago
11

At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 55 kg, 90 kg, and 42.5 kg respecti

vely. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vC = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming car A hits car B before car C does. Assume positive sign denoting forward motion and negative sign denoting backward motion.
Physics
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

Vb = 0.334 m/s

Va = -1.265 m/s

Vc = 1.424 m/s

Explanation:

Favorite Answer

Initial momentum = 255(2) – 242.5(1.5) = 146.25

Final momentum = 255Va + 290Vb + 242.5 Vc = 146.25

Vb - Va = 0.8(2) = 1.6

Vc - Vb = 0.8(1.5) = 1.2

Va = Vb -1.6

Vc = Vb + 1.2

255(Vb -1.6) + 290Vb + 242.5(Vb + 1.2) = 146.25

255 Vb – 408 + 290 Vb + 242.5 Vb + 291 = 146.25

787.5 Vb = 263.25

Vb = 0.334 m/s

Va = Vb -1.6 = 0.334 – 1.6 = -1.265 m/s

Vc = Vb + 1.2 = 0.224 + 1.2 = 1.424 m/s

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The shape is connected in parallel so;

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\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

5.2) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

I hope I helped you^_^

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2 years ago
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Answer:

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Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

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v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

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