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 1800 kg car traveling to the right at 1.40 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.

A-What was the speed of the lighter car just after the collision? v=

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

Answer #1

Since there is no external force applied, So using momentum conservation

Pi = Pf

m1*V1i + m2*V2i = m1*V1f + m2*V2f

given that m1 = mass of car 1 traveling to right = 1800 kg

m2 = mass of car 2= 1450 kg

V1i = 1.40 m/s

V2i = -1.10 m/s

V1f = final speed of car m1 = 0.260 m/s

V2f = final speed of car 2 = ?

Using above given values:

1800*1.40 + 1450*(-1.10) = 1800*0.260 + 1450*V2f

V2f = (1800*1.40 - 1450*1.10 - 1800*0.260)/1450

**V2f = 0.315 m/s**

**B)**

now change in kinetic energy is given by,

dKE = KEf - KEi

dKE = 0.5*m1*V1f^2 + 0.5*m2*V2f^2 - 0.5*m1*V1i^2 - 0.5*m2*V2i^2

dKE = 0.5*1800*0.260^2 + 0.5*1450*0.315^2 - 0.5*1800*1.40^2 - 0.5*1450*(-1.10)^2

**dKE = -2508.5 J = change in the combined kinetic energy
of the two-car system during this collision**

**Let me know if you've any query.**

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