(a) 33 %
In order to find the fraction of the original kinetic energy lost in the collision, we have to use the conservation of momentum first. We can write:
Mu = (M+m)v
where
Mu is the momentum before the collision, with
M = 1150 kg being the mass of the car
u is the initial velocity of the car
(M+m)v is the momentum after the collision, with
m = 500 kg being the mass of the moose
v the final velocity of the car+moose sticking together
Solving the equation, we find an expression for v:
[tex]v=\frac{Mu}{M+m}[/tex]
Now we can calculate the fraction of kinetic energy lost as:
[tex]\frac{\Delta K}{K_i}=\frac{K_i-K_f}{K_i}=1-\frac{K_f}{K_i}[/tex] (1)
where Ki is the total kinetic energy before the collision and Kf the total kinetic energy after the collision. Using the formula for the kinetic energy:
[tex]K_i = \frac{1}{2}Mu^2[/tex]
[tex]K_f = \frac{1}{2}(M+m)v^2=\frac{u^2}{2(M+m)}[/tex]
And substituting into (1), we find:
[tex]\frac{\Delta K}{K_i} = 1-\frac{M}{M+m}[/tex]
And using M = 1150 kg and m = 500 kg, we find
[tex]\frac{\Delta K}{K_i}=1-\frac{1150}{1150+500}=0.33[/tex]
which means that 33% of the original kinetic energy is lost.
(b) 22%
We can redo the exercise exactly in the same way, although this time we have to consider the mass of the camel instead, which is:
m = 330 kg
And replace the mass of the moose with this. If we do so, we obtain:
[tex]\frac{\Delta K}{K_i}=1-\frac{1150}{1150+330}=0.22[/tex]
So, only 22% of the total kinetic energy is lost in this case.
(c) Decreases
We can analyze again the formula for the fraction of kinetic energy lost:
[tex]\frac{\Delta K}{K_i} = 1-\frac{M}{M+m}[/tex]
In the formula, M (the mass of the car is constant). We notice that if the mass of the animal (m) decreases, the denominator decreases, so the value of the fraction increases, and therefore the whole term [tex]\frac{\Delta K}{K_i}[/tex] will decrease.
