Question

A car of mass 2500 kg collides with a truck of mass 4400 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car's velocity just before the collision was <31, 0, 0> m/s, and the truck's velocity just before the collision was <-13, 0, 24> m/s.

b)what is the increase in internal energy of the car and truck (thermal energy and deformation)?

Answer #1

Given,

m1 = 2500 kg ; m2 = 4400 kg ; u1 <31,0,0> and u2 = <-13,0,24>

from conservation of momentum

along x

vfx (2500 + 4400) = 2500 x 31 - 4400 x 13 = 2.94

vfy = 0

vfz (2500 + 4400) = 2500 x 0 + 4400 x 24 => vfz = 16.3 m/s

vf = sqrt (2.94^2 + 0^2 + 16.3^2) = 15.58 m/s

KEf = 1/2 x 6900 x 15.58^2 = 837890 J

KE1 = 0.5 x 2500 x 31^2 = 1201250 J

KE2 = 0.5 x 2500 x (-13^2 + 24^2) = 931250 J

KEi = 931250 + 1201250 = 2132500 J

dec in KE = 2132500 - 837440.58 = -1295059.42 J

Hencem KE = -1.29 x 10^6 J

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