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23 January, 03:46

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750kg car traveling to the right at 1.60m/s collides with a 1450kg car going to the left at 1.10m/s. Measurements show that the heavier car's speed just after the collision was 0.260m/s in its original direction. You can ignore any road friction during the collision.

A:What was the speed of the lighter car just after the collision?

B:Calculate the change in the combined kinetic energy of the two-car system during this collision.

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  1. 23 January, 04:29
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    We shall apply law of conservation of momentum.

    m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, m₁, m₂ are masses of two bodies colliding with velocities u₁ and u₂ respectively. v₁ and v₂ are their velocities after collision.

    m₁ = 1750, m₂ = 1450, u₁ = 1.6 m / s, u₂ = - 1.1 m / s, v₁ =.26 m / s

    substituting the values

    1750 x 1.6 + 1450 x - 1.1 = 1750 x. 26 + 1450 v₂

    2800 - 1595 = 455 + 1450v₂

    1450v₂ = 750

    v₂ =.517 m / s

    B) initial kinetic energy

    = 1/2 x 1750 x 1.6² + 1/2 x 1450 x - 1.1²

    = 2240 + 877.25

    = 3117.25 J

    final kinetic energy

    = 1/2 x 1750 x. 26² + 1/2 x 1450 x. 517²

    = 59.15 + 193.78

    = 252.93

    loss of kinetic energy

    = 3117.25 - 252.93

    = 2864.32 J
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