# How much energy is lost in an inelastic collision?

• Eclair_de_XII
In summary: If m=0, then the derivative is undefined. The percentage of kinetic energy lost as the animal mass decreases is undefined. How do I interpret this?It means that as the mass of the animal decreases, the percentage of kinetic energy lost becomes larger and larger, approaching infinity as the mass approaches zero. This may seem counterintuitive, but it makes sense if you think about it. As the animal mass decreases, the car has less and less mass to transfer its kinetic energy to, resulting in a larger percentage of energy loss.
Eclair_de_XII

## 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).

If you take the equation you start with in c) and plug in the values for the masses you do not get the answers you found in a) and b).
The answer to b) can be simplified.

You are free to pick variables to be anything you like - and you have been clear. It is best practice to make the variables easy to read though, so m_m and m_c for the mass of the moose and the camel respectively and, maybe M for the car since it is a big mass? You can see how it makes the equations later easier to read and thus troubleshoot if needed? (You can also use u and v for initial and final velocities etc.) It's a niggle only because different people feel differently about this, and you were clear enough in your definitions.

You appear to have assumed that the animal sticks to the car in each case - reasonable but, ewww.

loss of KE would be ##-\Delta K## (check: because a negative loss would be a "gain".)
The percentage loss would be ##-100\Delta K/K## right?
If ##\Delta K = K'-K## (final minus initial) and we take ##g## as the "percentage loss", then:
##g = 100(K-K')/K = 100K(1-K'/K)/K = 100(1-K'/K)## ... aaand this is where you lost me, because my equation now looks different from yours.

... I am not following the reasoning, so I cannot show you where it went wrong.
Suggest recheck.

Simon Bridge said:
You appear to have assumed that the animal sticks to the car in each case - reasonable but, ewww.
It says the animal joins the driver in the vehicle. Passenger seat, one hopes.

Yeah, "through the windsheild"... ikk. And the driver keeps going straight through the whole thing.

haruspex said:
If you take the equation you start with in c) and plug in the values for the masses you do not get the answers you found in a) and b).

Let me try again: ##g(m)=1-\frac{m_0}{m+m_0}##. Differentiate to get: ##g'(m)=\frac{m_0}{(m+m_0)^2}≥0## for all ##m##. So as ##m## increases, percentage loss in ##K_0## increases; conversely, as ##m## decreases, percentage loss in ##K_0## decreases. My only gripe with this, now, is that if ##m=0##, ##g'(0)=\frac{1}{m_0}##. How would I interpret this if it were correct?

haruspex said:
The answer to b) can be simplified.

##g(m_1)=\frac{3}{13}##

Simon Bridge said:
this is where you lost me, because my equation now looks different from yours.

I'm sorry. I should have labelled my variables better. I also took into account the kinetic energy that was transferred to the stationary animal when calculating percentage of kinetic energy loss. That was what the ##K_2'## and ##K_1'## were supposed to represent. I added the kinetic energy of the animals and the car, and divided them over the initial kinetic energy of the car before the collision.

Eclair_de_XII said:
My only gripe with this, now, is that if ##m=0##, ##g'(0)=\frac{1}{m_0}##.
Why is that a problem? g(0)=0.

You're right. I'm just wondering about how I should interpret ##g'(0)=\frac{1}{m_0}##.

Eclair_de_XII said:
You're right. I'm just wondering about how I should interpret ##g'(0)=\frac{1}{m_0}##.
It tells you how the energy loss grows, from zero, as the mass of the victim increases, from zero. Hit a fly mass mf and the fraction of energy lost is mf/m0.

Mm, okay then. Thanks for your help, everyone.

Simon Bridge

## 1. How is energy lost in an inelastic collision?

In an inelastic collision, energy is lost through the conversion of kinetic energy into other forms, such as heat or sound. This can occur due to deformation of the objects involved or friction between them.

## 2. What factors determine the amount of energy lost in an inelastic collision?

The amount of energy lost in an inelastic collision depends on the materials and structures of the objects involved, the speed and angle of the collision, and external factors such as air resistance or surface conditions.

## 3. Is it possible for no energy to be lost in an inelastic collision?

No, it is not possible for no energy to be lost in an inelastic collision. Some energy will always be converted into other forms, even if it is a small amount.

## 4. How can we calculate the amount of energy lost in an inelastic collision?

The amount of energy lost in an inelastic collision can be calculated by comparing the kinetic energy of the objects before and after the collision. The difference between the two values represents the energy that was lost.

## 5. Are all collisions in the real world inelastic?

No, not all collisions in the real world are inelastic. Some collisions, such as those between perfectly elastic objects, do not result in any energy loss. However, inelastic collisions are more common as most real-world objects are not perfectly elastic.

• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
38
Views
2K
• Introductory Physics Homework Help
Replies
26
Views
364
• Introductory Physics Homework Help
Replies
15
Views
453
• Introductory Physics Homework Help
Replies
4
Views
205
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
206
• Introductory Physics Homework Help
Replies
19
Views
1K