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4vir4ik [10]
4 years ago
14

Two observers times the motion of a car from one place to another. The first observer's clock read 262 seconds at the start and

375 seconds at the end of the car's motion. The second observer's clock read -86 seconds at the start. What did it read at the end?
Physics
1 answer:
icang [17]4 years ago
4 0

Answer:

The time of the second observer at the end of the car motion is 27 s

Explanation:

Initial time of the first observer, t₁ = 262 s

final time of the first observer, t₂ = 375 s

The time of the car motion, t = t₂ - t₁

t = 375 s - 262 s

t = 113 s

Initial time of the second observer, t₁ = -86 s

final time of the second observer, t₂ = ?

The time of the car motion, t = t₂ - t₁

113 = t₂ - t₁

113 = t₂ - (-86)

113 = t₂ + 86

t₂ = 113 - 86

t₂ = 27 s

Therefore, the time of the second observer at the end of the car motion is 27 s

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v_a=0.8176 m/s

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Explanation:

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Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

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After the collision and being v'_a and v'_b the respective velocities, the total momentum is

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Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

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v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

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The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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