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 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 0.260 m/s in its original direction. You can ignore any road friction during the collision. Calculate the change in the combined kinetic energy of the two-car system during this collision.

Explanation / Answer

m1 = 1850 kg         m2 = 1350 kg

speeds before collision

u1 = 1.6 m/s        u2 = -1.1 m/s

speeds after collision

v1 = 0.26        v2 = ?

initial momentum before collision

Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from moentum conservation

total momentum is conserved

Pf = Pi

m1*u1 + m2*u2 = m1*v1 + m2*v2

(1850*1.1) - (1350*1.1) = (1850*0.26) + (1350*v2).....(1)

v2 = 0.05 m/s

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEi = 0.5*1850*1.6^2 + 0.5*1350*1.1^2 = 3184.75

KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2

KEf = 0.5*1850*0.26^2 + 0.5*1350*0.05^2 ........(2)

KEf = 64.22 J

change = KEi - KEf = -3120.5 J

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