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 1850 car traveling to the right at 1.40 collides with a 1350 car going to the left at 1.10 . Measurements show that the heavier car's speed just after the collision was 0.250 in its original direction. You can ignore any road friction during the collision.

A.) What is the speed of the lighter car just after collision?

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

Question

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akindelemf akindelemf · Answer 1 · 2020-05-04T13:29:26+02:00

Answer:

a) 0.4759 m/s

b) -754.1 J

Explanation:

a) From conservation of momentum, the final velocity of the light car is,

m₁v₁(f) + m₂v₂(i) = m₁v₁(f) + m₂v₂(f)

v₂(f) = m₁v₁(f) + m₂v₂(i) - m₁v₁(f) / m₂

= (1850) (1.40) + (1350) (-1.10) - (1850) (0.250) / 1350

= 2590 - 1485 - 462.5 / 1350

= 642.5 / 1350

= 0.4759 m/s

b) Change in K.E = After collision K.E - initial K.E

= 1/2 m₁v₁² + 1/2 m₂v₂² - 1/2 m₁v₁² + 1/2 m₂v₂²

= 0.5{ (1850*0.250² + 1350* 0.4759²) - (1850*0.40² + 1350* 1.10²)}

= 0.5{ (115.6 + 305.7) - (296 + 1633.5) }

= 0.5 {421.3 - 1929.5}

= 0.5 ( -1508.2)

= -754.1 J