# Velocity after 2 cars collide

• Entr0py
In summary, to find the velocity and direction of two cars after a collision where one is traveling north at 60 mph and the other is traveling east at 40 mph and the cars stick together after the collision, the conservation of linear momentum equation is used. The masses of the cars are found to be 62 slugs and 155 slugs, respectively. The initial momentum of car 1 is 3720 slug-m/h and the initial momentum of car 2 is 6200 slug-m/h, resulting in a total initial momentum of 9920 slug-m/h. Using the conservation of linear momentum equation, the final momentum is found to be 217v, where v is the final velocity of the
Entr0py

## Homework Statement

A 2000-lb car traveling north at 60 mph collides with a 5000-lb car traveling east at 40 mph. The cars stick together after the collision. Find the velocity (magnitude and direction) of the cars just after the collision.

## Homework Equations

conservation of linear momentum

## The Attempt at a Solution

Firstly, I need to find the masses of each car. Car 1 has weight of 2000 lbs and has a mass of (2000/32.2) about 62 slugs. Car 2 has weight of 5000 lbs and has mass of (5000/32.2) about 155 slugs. That said, the initial momentum of car 1 equals m(1) times v(1) or (62 slugs times 60 mph) = 3720 slug-m/h. Initial momentum of car 2 equals m(2) times v(2) or (155 times 40 mph) = 6200 slug-m/h. So the total initial momentum is 9920 slug-m/h.

The final momentum of the cars= (total mass of cars since they stick together) times (new velocity) which equals 217v, so v= 9920/217 or 45.7 mph. To find the direction of velocity after collision, wouldn't I do arctan(40/60) = 33.7 degrees NE.

Now I must be doing something wrong because in the back of the book, the answer is 33.3 mph at 31 degrees NE. So i can't figure out what I'm doing wrong.

You cannot add momentum in different directions linearly.

Where is the point in the conversion from one weird unit to another? The masses will cancel anyway, so working with lb everywhere will work. Conversions just introduce rounding errors.

Entr0py said:

## Homework Statement

A 2000-lb car traveling north at 60 mph collides with a 5000-lb car traveling east at 40 mph. The cars stick together after the collision. Find the velocity (magnitude and direction) of the cars just after the collision.

## Homework Equations

conservation of linear momentum

## The Attempt at a Solution

Firstly, I need to find the masses of each car. Car 1 has weight of 2000 lbs and has a mass of (2000/32.2) about 62 slugs. Car 2 has weight of 5000 lbs and has mass of (5000/32.2) about 155 slugs. That said, the initial momentum of car 1 equals m(1) times v(1) or (62 slugs times 60 mph) = 3720 slug-m/h. Initial momentum of car 2 equals m(2) times v(2) or (155 times 40 mph) = 6200 slug-m/h. So the total initial momentum is 9920 slug-m/h.

The final momentum of the cars= (total mass of cars since they stick together) times (new velocity) which equals 217v, so v= 9920/217 or 45.7 mph. To find the direction of velocity after collision, wouldn't I do arctan(40/60) = 33.7 degrees NE.

Now I must be doing something wrong because in the back of the book, the answer is 33.3 mph at 31 degrees NE. So i can't figure out what I'm doing wrong.

(1) Nevermind the units; just use masses of m1 = 2 and m2 = 5. Converting to thousands and slugs serves not purpose in this problem, since you are just multiplying and later dividing by a constant.
(2) The initial momenta of the two cars are VECTORS pointing in different directions, so you need to represent them by their x- and y- coordinates; You cannot simply add up their magnitudes.

mfb said:
You cannot add momentum in different directions linearly.

Where is the point in the conversion from one weird unit to another? The masses will cancel anyway, so working with lb everywhere will work. Conversions just introduce rounding errors.
You're RIGHT, momentum is a vector. Oh shucks

Hello S,

Why you want a mass in slugs is a mystery to me. Never mind.
Momentum is a vector. You can't add the components like this: they are at right angles to each other !

So do a vector addition and use Pythagras to find the total momentum.
The tangent of the direction angle is the y momentum divided by the x momentum

 slow typist: Ray was faster and you already saw what was to be done. All is well.

BvU said:
Hello S,

Why you want a mass in slugs is a mystery to me. Never mind.
Momentum is a vector. You can't add the components like this: they are at right angles to each other !

View attachment 88305

So do a vector addition and use Pythagras to find the total momentum.
The tangent of the direction angle is the y momentum divided by the x momentum

 slow typist: Ray was faster and you already saw what was to be done. All is well.
don't I need their mass in slugs to use in p=mv?

is slugs * mph any better than lb * mph ?

But as Ray says, lb, slugs or stones, or the usual kg: you are multiplying them in and then dividing them out again later on

BvU said:
is slugs * mph any better than lb * mph ?

But as Ray says, lb, slugs or stones, or the usual kg: you are multiplying them in and then dividing them out again later on
Oh ok. So i calculated the net momentum by pnet= p1 + p2 and that gave me 233. Then to find the resultant velocity I I did v=(233)/7 = 33.3 mph. Now I can't seem to find the direction. I thought it would be arctan (3) but that's not correct.

I figured the answer out. Now I need to spend quite some time figuring out and learning THE PROCESS to get the answers. Knowing that momentum IS a vector it has vertical and horizontal components. So initial momentum^2= P(ix)^2 + p(iy)^2 and so p(i)=sqrt of all of that. Then to find p(f) i need to do the same thing but note that mass=M=combined mass of the two cars. From momentum conservation I know that the COMPONENTS of initial and final momentum equal one another. Doing this I can solve for final vertical velocity and final horizontal velocity. Knowing these I add the sum of their squares and then sqrt that to find final velocity vector. To find its direction I just do arctan (final vertical velocity/final horizontal velocity). HOWEVER, I'm still confused why I can just use pounds instead of slugs in this question. Since i need to find the momentum of each car I need to use their mass (but pound is a measurement of weight and slugs is a measurement of mass). That's why I converted pounds into slugs. because isn't momentum measured in kgm/s in SI units? That is mass times velocity, so how would (pounds miles an hour) be just as useful as (slugs miles and hour) if slugs measure mass and pounds measure weight?

Entr0py said:
Oh ok. So i calculated the net momentum by pnet= p1 + p2 and that gave me 233. Then to find the resultant velocity I I did v=(233)/7 = 33.3 mph. Now I can't seem to find the direction. I thought it would be arctan (3) but that's not correct.
Why 3? It's the momentum ratio you want, not the velocity ratio.

Entr0py said:
HOWEVER, I'm still confused why I can just use pounds instead of slugs in this question. Since i need to find the momentum of each car I need to use their mass (but pound is a measurement of weight and slugs is a measurement of mass). That's why I converted pounds into slugs. because isn't momentum measured in kgm/s in SI units? That is mass times velocity, so how would (pounds miles an hour) be just as useful as (slugs miles and hour) if slugs measure mass and pounds measure weight?
If each car were exactly twice the mass given, would it change the answers?

haruspex said:
Why 3? It's the momentum ratio you want, not the velocity ratio.
I figured out my mistake.

haruspex said:
If each car were exactly twice the mass given, would it change the answers?
No it wouldn't. Does that mean i say p=(lb)(mph)?

p=mv is independent of units. You can use lb*mph, you can use kg*m/s, you can use (multiples of your body weight)*(average speed of an unladen European swallow) - it does not matter.
You just have to be consistent in the calculation.

mfb said:
p=mv is independent of units. You can use lb*mph, you can use kg*m/s, you can use (multiples of your body weight)*(average speed of an unladen European swallow) - it does not matter.
You just have to be consistent in the calculation.
I'm sorry, I understand that you and the other mentors are trying to help me understand that momentum is IND of units used. it's just isn't a pound a measurement of weight which measures force? Then how can i use p=(measurement of force)(rate). I understand that rate is just velocity which is change in displacement. I get that. But I'm frankly stuck on why I can use a measurement of force in place for a measurement of mass.

pound is a unit of mass.
pound-force (sometimes called "pound" as well which is wrong) is a unit of force.

Entr0py
mfb said:
pound is a unit of mass.
pound-force (sometimes called "pound" as well which is wrong) is a unit of force.
THANKS for the clarification. So when the author says a car of 2000 pounds he is just saying that this car has a MASS of 2000 pounds not that the force of gravity acting on it is 2000 FORCE-POUNDS?

Entr0py said:
THANKS for the clarification. So when the author says a car of 2000 pounds he is just saying that this car has a MASS of 2000 pounds not that the force of gravity acting on it is 2000 FORCE-POUNDS?
The author should have been more precise. Usually we designate ##lb_f and lb_m## to distinguish between the two.

Remember : ##1 lb_f = 1 lb_m ## so lbs could be either units of force or units of mass.

Sometimes we use this equation ## F = \frac{m * a}{g_c} ## where the ##G_c ##term is used to convert us from slugs to## lb_m##

NickAtNight said:
The author should have been more precise. Usually we designate ##lb_f and lb_m## to distinguish between the two.

Remember : ##1 lb_f = 1 lb_m ## so lbs could be either units of force or units of mass.

Sometimes we use this equation ## F = \frac{m * a}{g_c} ## where the ##G_c ##term is used to convert us from slugs to## lb_m##
where gc is 32.2 ft/s^2?

Entr0py said:
where gc is 32.2 ft/s^2?
Correct

Entr0py
Ray Vickson said:
(1) Nevermind the units; just use masses of m1 = 2 and m2 = 5. Converting to thousands and slugs serves not purpose in this problem, since you are just multiplying and later dividing by a constant.
(2) The initial momenta of the two cars are VECTORS pointing in different directions, so you need to represent them by their x- and y- coordinates; You cannot simply add up their magnitudes.
This was before I knew there was such a thing as pound being used to measure mass but ALSO being used to measure force. The author didn't clarify, and hence my conversion to slugs (I did NOT know that pounds were ALSO a unit of mass).

NickAtNight said:
Correct
Thank you. Your clarification helped out a lot.

NickAtNight said:
Correct
So if I say I weigh 150 pounds, that is my weight not my mass. So would it be more accurate to say I weigh 150 force-pounds not 150 mass-pounds?

Entr0py said:
Thank you. Your clarification helped out a lot.
Lbs were around before ##F= m*a## was discovered. The so the system had to be adjusted to fix the discovery.

I have never used "slugs" in my career... Not sure they ever enjoyed much use.

The businessmen and farmer would continue to use their practical measure of lb_m.

The scientists would have used the lb_f.

If you are going to convert to other units, go to the modern SI system.

Entr0py said:
So if I say I weigh 150 pounds, that is my weight not my mass. So would it be more accurate to say I weigh 150 force-pounds not 150 mass-pounds?
It is both. We are on earth. The a is 32.2 ft/s^2 which cancels the G_c. So you could say either.

You step on the scale and it says 150 lbs.

That is 150 lbs_f.

If you convert it to mass, it is also 150 lb_m.

But if you were on the moon, where the gravity is less, the scale would read 12 lbs_f.

And when you convert it, you would get 150 lbs_m

Wiki says. "
The slug is part of a subset of units known as the gravitational FPS system, one of several such specialized systems of mechanical units developed in the late 19th and the 20th century. Geepound was another name for this unit in early literature.[4]

The name "slug" was coined before 1900 by British physicist Arthur Mason Worthington,[5] but it did not see any significant use until decades later. A 1928 textbook says:

No name has yet been given to the unit of mass and, in fact, as we have developed the theory of dynamics no name is necessary. Whenever the mass, m, appears in our formulae, we substitute the ratio of the convenient force-acceleration pair (w/g), and measure the mass in lbs. per ft./sec.2 or in grams per cm./sec.2.

—Noel Charlton Little, College Physics, Charles Scribner's Sons, 1928, p. 165."

NickAtNight said:
Lbs were around before ##F= m*a## was discovered. The so the system had to be adjusted to fix the discovery.

I have never used "slugs" in my career... Not sure they ever enjoyed much use.

The businessmen and farmer would continue to use their practical measure of lb_m.

The scientists would have used the lb_f.

If you are going to convert to other units, go to the modern SI system.
thanks

"...So what the heck is a slug?

Well in the British and US system of weights in which a pound is the unit of weight/force under the acceleration of gravity, which in English Units is 32 feet/sec2, the unit of mass is the slug.

The word slug seems to have been created around the beginning of the 20th century by A. M. Worthington [from the OED]"1902, A. M. WORTHINGTON Dynamics of Rotation (ed.4) p. viii, I have ventured to give the name of a ‘slug’ to the British Engineer's Unit of Mass, i.e., to the mass in which an acceleration of one foot-per-sec.-per-sec. is produced by a force of one pound."

The OED also gives clues to why the term is so little known to students;

"1936, F. W. LANCHESTER Theory of Dimensions & its Application for Engineers v. 37

Even amongst the advocates of Perry's system.., the slug has never taken shape except on paper; it has, and has had no real material existence."

And if it ever did live, it may have been short lived,

[OED]"1973, Nature 20 July 184/3 The statement that the unit of mass in the British system is the slug is several years out of date." https://apps.spokane.edu/InternetContent/AutoWebs/markg/Slugs,etc.docx

Entr0py
Entr0py said:
thanks

You are welcome.

Oh, the SI system does not get off quite so easy either.

If you weigh yourself on the bathroom scales and get 68 kg (150 lbs).

This is also Force units. So ##kg_f##

And the ##G_c## for the SI system is ##9.8 kg*m/kg_f/s^2 ##

So ##kg_f = kg_m##

If you step on the same scale on the moon, you would only weigh 11.6 kg_f. But when you do the conversion, you will still be 68 kg mass.From that other source.
...
Students who learn physics using the SI system are often confused when they encounter kilograms as weight or what the Marks Standard Handbook for Mechanical Engineers calls kilogram force. This term comes as a surprise to students who have been taught that mass in the metric system was kilograms and force was Newtons. They are further bewildered when presented with a spring scale calibrated in kilograms (a force/weight), use a Rockwell hardness tester with loads(force/weight) measured in kilograms or find the term kilogram force in a bearing catalog. They are then more totally confused when they take a dynamics course and they are told to divide the weight of an object in pounds by g=32 ft/sec2 to obtain its mass in slugs but that the weight in kilograms is not divided by the metric g=9.8 m/sec2 to get kilogram mass.

The Kilogram is the base unit of mass in the SI and MKS versions of the metric system. The kilogram is defined as the mass of the standard kilogram, a platinum-iridium bar in the custody of the International Bureau of Weights and Measures (BIPM) near Paris, France. Copies of this bar are kept by the standards agencies of all the major industrial nations, including the U.S. National Institute of Standards and Technology (NIST). It is unique because it is the only physical object used to define an international standard.

The weight of one kilogram of mass near the surface of the Earth is called a Kilogram Force, and is about 2.204,622 pounds. This is NOT the same as a Newton, in fact it is about 9.8 Newtons. A Newton is the amount of force(weight) created by a mass of 1 Kg under an acceleration of 1 meter/sec2. A Kgf (kilogram force) is the force/weight created by a mass of 1 Kg under the acceleration of gravity on the earth, about 9.8 meters/sec2.
[\Quote]

BvU and Entr0py
NickAtNight said:
You are welcome.

Oh, the SI system does not get off quite so easy either.

If you weigh yourself on the bathroom scales and get 68 kg (150 lbs).

This is also Force units. So ##kg_f##

And the ##G_c## for the SI system is ##9.8 kg*m/kg_f/s^2 ##

So ##kg_f = kg_m##

If you step on the same scale on the moon, you would only weigh 11.6 kg_f. But when you do the conversion, you will still be 68 kg mass.From that other source.
You sure do know a lot about weights and masses.

Boy, if I have ever seen a plea for metrication, this is the one !

On the other hand it's good to realize -- as entro did -- that momentum really has the dimension mass times length divided by time, in whatever units are in fashion !

NickAtNight said:
You are welcome.

Oh, the SI system does not get off quite so easy either.

If you weigh yourself on the bathroom scales and get 68 kg (150 lbs).

This is also Force units. So ##kg_f##

And the ##G_c## for the SI system is ##9.8 kg*m/kg_f/s^2 ##

So ##kg_f = kg_m##

If you step on the same scale on the moon, you would only weigh 11.6 kg_f. But when you do the conversion, you will still be 68 kg mass.From that other source.
No, the SI doesn't have that mess. There is no such thing as "kg_f".
The bathroom scale measures a force of 667 N and converts it for you to m=F/g = 68 kg.
On the Moon the bathroom scale measures ~110 N and converts it (wrongly) to about 11 kg because it is not calibrated for use on Moon. A properly calibrated scale there would convert 110 N to 68 kg as well. If scales would be used in different environments frequently, one would add a test mass to the scale to make it self-calibrating.

Entr0py

## 1. What factors affect the velocity of two cars after a collision?

The velocity of two cars after a collision is affected by several factors, including the mass and speed of each car, the angle of impact, and the coefficient of restitution (a measure of how much energy is lost during the collision).

## 2. How is velocity calculated after a collision?

The velocity of two cars after a collision can be calculated using the conservation of momentum principle, which states that the total momentum before the collision is equal to the total momentum after the collision. This can be expressed as m1v1 + m2v2 = m1v1' + m2v2', where m is the mass of each car and v is the velocity before and after the collision.

## 3. Can the velocity of two cars after a collision be greater than the sum of their initial velocities?

No, according to the conservation of momentum principle, the total momentum after a collision must be equal to the total momentum before the collision. Therefore, the velocity of two cars after a collision cannot be greater than the sum of their initial velocities.

## 4. How does the angle of impact affect the velocity of two cars after a collision?

The angle of impact can greatly affect the velocity of two cars after a collision. If the cars collide head-on (at a 180-degree angle), the velocity of both cars will be greatly reduced. However, if the cars collide at an angle, the velocity of one car may be significantly reduced while the other car may experience a change in direction but maintain a similar velocity.

## 5. What is the coefficient of restitution and how does it affect the velocity of two cars after a collision?

The coefficient of restitution is a measure of how much energy is lost during a collision. It is represented by the symbol e and ranges from 0 to 1, with 0 representing a perfectly inelastic collision (where the two cars stick together after the collision) and 1 representing a perfectly elastic collision (where no energy is lost). The higher the coefficient of restitution, the less energy is lost during the collision, resulting in a higher velocity for both cars after the collision.

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