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 kg car traveling to the right at 1.60 m/s collides with a 1350 kg car going to the left at 1.10 m/s . measurements show that the heavier car's speed just after the collision was 0.270 m/s in its original direction. you can ignore any road friction during the collision?

ΔK=- 2765.75 J

Explanation:

write the momentum conservation equation:

m1v1 + m2v2 = m1V1 + m2V2

with  given data

m1 = 1850 kg

m2 = 1350 kg

v1 = 1.60 m/s

v2 = - 1.10 m/s

V1 = 0.270 m/s

Then

V2 = (m1v1 + m2v2 - m1V1) / m2  

V2= (1850*1.60 - 1350*1.10 - 1850*0.270) / 1350 = 0.722 m/s (to the right)

Kinetic energy before the collision:

K1 = 1/2 m1v1² + 1/2 m2v2² = 0.5 (1850*1.60² + 1350*1.10²) = 3184.75 J

Kinetic energy after the collision:

K2 = 1/2 m1V1² + 1/2 m2V2² = 0.5 (1850*0.270² + 1350*0.722²) = 419 J

Change in kinetic energy:

ΔK = K2 - K1 = 419 - 3184.75 = - 2765.75 J

the answer is 12 as i may remember

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