Question

In Anchorage, collisions of a vehicle with a moose are so commonthat they are referred to with the abbreviation MVC. Suppose a1050 kg car slides into a stationary520 kg moose on a very slippery road,with the moose being thrown through the windshield (a common MVCresult).

(a) What percent of the original kinetic energyis lost in the collision to other forms of energy?
%

(b) A similar danger occurs in Saudi Arabia because ofcamel-vehicle collisions (CVC). What percent of the originalkinetic energy is lost if the car hits a 270 kg camel?
%

(c)Generally, does the percent loss increase or decrease if the animalmass decreases?

increase

decrease

Is this an elastic collision or a completelyinelastic collision? Did you find the kinetic energy of the carbefore and the kinetic energy of the car + moose after? Did youconvert the speed to m/s?

Answer 1

Since momentum is conserved you can write:

   M v_i =   (M +m)v_f         where I used M for the car and m for the moose.

If you square both sides of the equation and multiply by(1/2), then you get

   (1/2) M²v_i²    =   (1/2) (M+ m)²v_f²             

Notice that this is almost the initial and final kineticenergies... but not quite. There is an extra factor of mass on eachside. Since   KE is (1/2) mv²    we can write each side as the KEtimes mass or

     M * K_i   = (M + m ) K_f

Then we can get the ratio of K final to K initial

    K_f / K_i = M / (M+m) = 1050 / (1050 + 520) = 0.6688  

this means 66.88 % of the KE is kept by the car andmoose... so 33.12% is "lost"

Similarly, for the car and camel

      K_f /K_i = M / (M+m) = 1050 / (1050 +270) = 0.7955  

this means 79.55 % of the KE is kept by the car andcamel... so 20.45% is "lost"