In Anchorage, collisions of a vehicle with a moose are so common that they are r
ID: 2105200 • Letter: I
Question
In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviation MVC. Suppose a 1150 kg car slides into a stationary 550 kg moose on a very slippery road, with the moose being thrown through the windshield (a common MVC result). (a) What percent of the original kinetic energy is lost in the collision to other forms of energy? A similar danger occurs in Saudi Arabia because of camel–vehicle collisions (CVC). (b) What percent of the original kinetic energy is lost if the car hits a 310 kg camel? (c) Generally, does the percent loss increase (if so type 1) or decrease (if so type 2) if the animal mass decreases?
Explanation / Answer
m(c) * v(c) + m(m)* 0 = [m(c)+m(m)] V
m(c) * v(c) = [m(c)+m(m)] V ------- (1)
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V = joint or combined speed of both together > ineastic
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energy conservation> before =after
0.5[m(c)*v^2(c) +0] = 0.5[m(c)+m(m)] V^2 + E(lost)
E(lost) = 0.5 {m(c)*v^2(c) - [m(c)+m(m)] V^2}
eliminate V from (1)
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[m(c)+m(m)] V^2 = m^2(c) * v^2(c) / [m(c)+m(m)]
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E(lost) = 0.5 {m(c)*v^2(c) - m^2(c) * v^2(c) / [m(c)+m(m)] }
E(lost) = 0.5 m(c)*v^2(c) {1 - m(c) / [m(c)+m(m)] }
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ratio of original kinetic energy is lost = E(lost)/0.5 m(c)*v^2(c)
% kinetic energy lost = {1 - m(c) / [m(c)+m(m)] }*100
% kinetic energy lost = {1 - 1050 / [1050+540] }*100
= 0.3396*100 = 33.96%
= 34% approx
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b) % kinetic energy lost = {1 - 1150 / [1150+310] }*100
= 21.232%