- #1

Eclair_de_XII

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## Homework Statement

"Suppose a ##1000 kg## car slides into a stationary ##500 kg## moose on a very slippery road, with the moose being thrown through the windshield. (a) What percent of the original kinetic energy is lost in the collision to other forms of energy? (b) What percent of the original kinetic energy is lost if the car hits a ##300 kg## camel? (c) Generally, does the percentage loss increase or decrease if the animal mass decreases?"

## Homework Equations

##m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}##

##K=\frac{1}{2}mv^2##

##m_0=1000kg##

##m_1=500kg##

##m_2=300kg##

##K_0=\frac{1}{2}m_0v_0^2##

## The Attempt at a Solution

Okay, so I figured out pretty much (a) and (b), but will need someone to check how I got the values for those problems. I also made a derivative proof for (c) that I would like someone to go over. Let ##m_0## be the car, ##m_1## be the moose, and ##m_2## be the camel.

(a)

##(m_0)(v_0)=(m_0+m_1)(v_1)##

##v_1=(\frac{m_0}{m_0+m_1})(v_0)=\frac{2}{3}v_0##

I set this equal to the new velocity for the kinetic energy calculation needed to calculate the loss in ##K_0##.

##K_0'=\frac{1}{2}m_0v_1^2=\frac{1}{2}m_0(\frac{2}{3}v_0)^2=\frac{2}{9}m_0v_0^2##

##K_1'=\frac{2}{9}m_1v_0^2##

Now here, I took the sum of all the kinetic energy remaining (in the moose and in the car).

##∑K'=\frac{2}{9}v_0^2(m_1+m_0)##

Then I divided it by ##K_0## to yield the kinetic energy remaining out of kinetic energy before the collision.

##\frac{∑K'}{K_0}=(\frac{\frac{2}{9}v_0^2(m_1+m_0)}{\frac{1}{2}m_0v_0^2})=\frac{4}{9}(\frac{m_1+m_0}{m_0})=\frac{2}{3}##

The total percentage of energy lost is given by: ##1-\frac{∑K'}{K_0}=\frac{1}{3}##

(b)

##(m_0)(v_0)=(m_0+m_2)(v_2)##

##v_2=(\frac{m_0}{m_0+m_2})(v_0)=\frac{10}{13}v_0##

##K_0'=\frac{1}{2}m_0v_2^2=\frac{1}{2}m_0(\frac{10}{13}v_0)^2=\frac{50}{169}m_0v_0^2##

##K_2'=\frac{50}{169}m_2v_0^2##

##∑K''=\frac{50}{169}v_0^2(m_2+m_0)##

##\frac{∑K''}{K_0}=\frac{\frac{50}{169}v_0^2(m_2+m_0)}{\frac{1}{2}m_0v_0^2}=\frac{100}{169}(\frac{m_2+m_0}{m_0})=(\frac{100}{169})(\frac{1300}{1000})=\frac{130}{169}##

The percentage of energy lost is: ##1-\frac{∑K''}{K_0}=\frac{39}{169}##

(c) Here is the part that I wish for someone to check.

I'm looking for the percentage of kinetic energy loss when an object of fixed mass is forced into an inelastic collision with a body of variable mass. This value is given by:

##g(m)=1-\frac{∑K}{K_0}=1-2(\frac{m_0}{m+m_0})^2(\frac{m_0+m}{m})=1-2(\frac{m_0^2}{m+m_0})(\frac{1}{m})##

So: ##g(m)=1-2(\frac{m_0^2}{m^2+m_0m})##

I differentiated this with respect to ##m## and got:

##g'(m)=\frac{2(m_0^2)(2m+m_0)}{(m^2+m_0m)^2} ≥ 0## for all ##m##, since mass can never be negative or zero.

Therefore, as mass increases, the percentage of energy loss will increase. Conversely, as mass decreases (as is asked by the problem), the percentage of energy loss will decrease. Can anyone check my derivation of part (c)? Additionally, can anyone check if my process in (a) and (b) are correct? I'm particularly worried about if ##m## goes to zero, in part (c).