Was the Second Driver Speeding in the Collision?

[Rijad Hadzic]

Homework Statement

A police officer is attempting to reconstruct an accident in which a car traveling southward with a speed of 23 mph colided with another car of equal mass traveling eastward at an uknown speed. After the collision, the two cars coupled and slid at an angle of 60 degrees south of east. If the speed limit in that neighborhood is 25 mph, should the officer cite the second driver for speeding?

Homework Equations

The Attempt at a Solution

https://imgur.com/a/FbBgU
https://imgur.com/a/N3gk2

This is the solution manual answer.

in the third box when they write:

m(10.3 m/s) = 2m vcos(60)

This does not make sense to me. The car is traveling south at 23 mph, which = 10.3 m/s, but that is the Y COMPONENT, not the x component,

Wouldn't it make sense to write:

m(x) = 2m vcos(60)
and
m(10.3 m/s) = 2m vsin(60)

But doing it that way I get the wrong answer. Is anyone able to explain this to me please?

 

[QuantumQuest]

I would advise you to try to solve the problem yourself. Have you understood the conservation of momentum?

 

[Rijad Hadzic]

I would advise you to try to solve the problem yourself. Have you understood the conservation of momentum?

I did try the problem myself but got the wrong answer.

The way I set it up was:

(m)(x)= (2m) (ycos(300))
(m)(23) = (2m) (ysin(300))

which makes sense to me because on the first line, car 1 doesn't have an x component of velocity so the equation checks out
and on line two car 2 doesn't have a y velocity so the equation checks out

conservation of momentum means Pinitial = P final right?

 

[collinsmark]

Homework Statement

A police officer is attempting to reconstruct an accident in which a car traveling southward with a speed of 23 mph colided with another car of equal mass traveling eastward at an uknown speed. After the collision, the two cars coupled and slid at an angle of 60 degrees south of east. If the speed limit in that neighborhood is 25 mph, should the officer cite the second driver for speeding?

Homework Equations

The Attempt at a Solution

https://imgur.com/a/FbBgU
https://imgur.com/a/N3gk2

This is the solution manual answer.

in the third box when they write:

m(10.3 m/s) = 2m vcos(60)

This does not make sense to me. The car is traveling south at 23 mph, which = 10.3 m/s, but that is the Y COMPONENT, not the x component,

Wouldn't it make sense to write:

m(x) = 2m vcos(60)
and
m(10.3 m/s) = 2m vsin(60)

But doing it that way I get the wrong answer. Is anyone able to explain this to me please?

Well, there's obviously something wrong with the either the problem statement or the given solution.

The problem statement, as you've written it above, makes the claim "... a car traveling southward with a speed of 23 mph..." (There is no screenshot of the problem statement provided.)

Yet in the solution screenshot (second page) the claim is "... the initial speed of the southbound car was significantly greater than 25 mi/hr..."

Something isn't consistent there. Are you sure the you are not supposed to calculate the speed of the southbound car?

 

[Rijad Hadzic]

Well, there's obviously something wrong with the either the problem statement or the given solution.

The problem statement, as you've written it above, makes the claim "... a car traveling southward with a speed of 23 mph..." (There is no screenshot of the problem statement provided.)

Yet in the solution screenshot (second page) the claim is "... the initial speed of the southbound car was significantly greater than 25 mi/hr..."

Something isn't consistent there. Are you sure the you are not supposed to calculate the speed of the southbound car?

I am not sure if I even want to finish this problem anymore...

The question in my book came with no diagram/picture, and I wrote the question out word for word...

 

[QuantumQuest]

I did try the problem myself but got the wrong answer.

The way I set it up was:

(m)(x)= (2m) (ycos(300))
(m)(23) = (2m) (ysin(300))

which makes sense to me because on the first line, car 1 doesn't have an x component of velocity so the equation checks out
and on line two car 2 doesn't have a y velocity so the equation checks out

conservation of momentum means Pinitial = P final right?

Conservation of momentum means that the total momentum of a system is conserved. Now, the car going eastward has only $x$ component of velocity and the car going southwards has only $y$ component of velocity. That's how the things are before collision. After the collision, the system of the two cars ($2m$) goes in the direction shown. The car that was moving southwards gets an $x$ component of velocity and the car moving eastwards gets a $y$ component. So, if you take conservation of momentum for each axis separately, you end up with the equations shown in the image. Be careful in the right triangle that is formed from the final velocity of the system of the two collided cars and the axes. For an horizontal contributed component you must take $\cos60^0$ and $\sin60^0$ for a vertically contributed component.

 

[Rijad Hadzic]

Conservation of momentum means that the total momentum of a system is conserved. Now, the car going eastward has only $x$ component of velocity and the car going southwards has only $y$ component of velocity. That's how the things are before collision. After the collision, the system of the two cars ($2m$) goes in the direction shown. The car that was moving southwards gets an $x$ component and the car moving eastwards gets a $y$ component. So, if you take conservation of momentum for each axis separately, you end up with the equations shown in the image. Be careful in the right triangle that is formed from the final velocity of the system of the two collided cars and the axes. For an horizontal contributed component you must take $\cos60^0$ and $\sin60^0$ for a vertically contributed component.

When I did the problem, I set my equations up like:m(x) = 2m vcos(60)
and
m(10.3 m/s) = 2m vsin(60)

Which (I think?) agrees with what you said.

But the book set up:

m(10.3) = (2)vcos(60)

why are they setting up a y component with an x component?

 

[scottdave]

Well, there's obviously something wrong with the either the problem statement or the given solution...
"... a car traveling southward with a speed of 23 mph..." (There is no screenshot of the problem statement provided.)

Yet in the solution screenshot (second page) the claim is "... the initial speed of the southbound car was significantly greater than 25 mi/hr..."

Something isn't consistent there. Are you sure the you are not supposed to calculate the speed of the southbound car?

I agree with inconsistency between the problem statement (as presented to us) and the solution manual's outcome. But look at the "check and think" section of the solution manual: It is obvious that the southbound car was traveling faster than the eastbound car, since they are the same mass and the angle is closer to south than east. So if it was the southbound car which was 23 miles per hour, the eastbound car was well below the speed limit (no calculating necessary). If instead the eastbound car was the 23 mph, then you have do to the calculations. Also note that the step of converting to meters per second is not necessary to solve this problem. We are not using any momentum units (they don't even tell you the mass) - Keeping all of the speed units consistent, would be all that is necessary to calculate the answer.

 

[Rijad Hadzic]

I agree with inconsistency between the problem statement (as presented to us) and the solution manual's outcome. But look at the "check and think" section of the solution manual: It is obvious that the southbound car was traveling faster than the eastbound car, since they are the same mass and the angle is closer to south than east. So if it was the southbound car which was 23 miles per hour, the eastbound car was well below the speed limit (no calculating necessary). If instead the eastbound car was the 23 mph, then you have do to the calculations. Also note that the step of converting to meters per second is not necessary to solve this problem. We are not using any momentum units (they don't even tell you the mass) - Keeping all of the speed units consistent, would be all that is necessary to calculate the answer.

Exactly. In my answer that I did before viewing this solotion I didn't change to meters per second and instead kept it at mph I was wondering why they changed it..

 

[collinsmark]

Try re-doing the problem this modified problem statement and I think the solution works out:

A police officer is attempting to reconstruct an accident in which a car traveling eastward with a speed of 23 mph collided with another car of equal mass traveling southward at an unknown speed. After the collision, the two cars coupled and slid at an angle of 60 degrees south of east. If the speed limit in that neighborhood is 25 mph, should the officer cite the second driver for speeding?​

 

[scottdave]

I am not sure if I even want to finish this problem anymore...

The question in my book came with no diagram/picture, and I wrote the question out word for word...

Sounds like it is time to email your teacher, with your concerns about this problem.