Answer to Question #146123 in Mechanics | Relativity for Heather

Question

A 8,100-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the collision and move with speed 7 m/s. Compute how much kinetic energy was "lost" in this inelastic collision.

Answer 1

Mass of the truck is M = 8100 kg.

Mass of the car is m = 1000 kg.

Let speed of the truck before collision be u.

Speed of car before collision is 0.

Then initial momentum of truck is Mu .

And initial momentum of the car is 0.

Then total initial momentum is Mu.

Speed of truck and car after collision is v = 7 m/s

Final momentum of the car and truck is (M+m)v.

Then according to conservation of momentum

Mu = (M+m)v

$u = \frac{(M+m)v}{M}$

$u = \frac{(8100+1000)7}{8100} = 7.86 \;m/s$

Initial kinetic energy before collision is

$K_i = \frac{1}{2}Mu^2$

Final kinetic energy after collision is

$K_f = \frac{1}{2}(M + m)v^2$

Then loss in kinetic energy is

$K_i – K_f = \frac{1}{2}Mu^2 - \frac{1}{2}(M + m)v^2$

$K_i – K_f = \frac{1}{2} \times 8100 \times (7.86)^2 - \frac{1}{2}(8100 + 1000)7^2$

$K_i – K_f = 250207 – 222950 = 27257 \;J$