When cars are equipped with flexible bumpers, they will bounce off each other du

Question

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 kgcar traveling to the right at 1.50 m/s collides with a 1450 kg car going to the left at 1.10 m/s . Measurements show that the heavier car's speed just after the collision was 0.260 m/s in its original direction. You can ignore any road friction during the collision.

Part A

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

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Part B

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

  J  

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 kgcar traveling to the right at 1.50 m/s collides with a 1450 kg car going to the left at 1.10 m/s . Measurements show that the heavier car's speed just after the collision was 0.260 m/s in its original direction. You can ignore any road friction during the collision.

Part A

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

v =   m/s  

SubmitMy AnswersGive Up

Part B

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

K =

  J  

Explanation / Answer

Let's write the momentum conservation equation:
m1v1 + m2v2 = m1V1 + m2V2
with
m1 = 1850 kg
m2 = 1450 kg
v1 = 1.50 m/s
v2 = - 1.10 m/s
V1 = 0.260 m/s

Then
a) V2 = (m1v1 + m2v2 - m1V1) / m2 =
= (1850*1.50 - 1450*1.10 - 1850*0.260) / 1450 = 0.482 m/s (to the right)

b) Kinetic energy before the collision:
K1 = 1/2 m1v1² + 1/2 m2v2² = 0.5 (1850*1.50² + 1450*1.10²) = 2958.5 J

Kinetic energy after the collision:
K2 = 1/2 m1V1² + 1/2 m2V2² = 0.5 (1850*0.260² + 1450*0.482²) = 230.96 J

Change in kinetic energy:
K = K2 - K1 = 230.96 - 2958.5 = - 2727.56 J

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