Did This AP Physics Problem Analysis Get It Right?

[WoodenFire]

Here is the actual physics problem. Below it I will try to explain exactly how I thought to figure it out. I would appreciate it if you corrected me If I made any errors. I am very sketchy on this one after being out for a week from class.

http://img74.imageshack.us/img74/2937/img012brz6.th.jpg
For some reason the picture isn't showing to most people... here is the imageshack direct link. http://img74.imageshack.us/my.php?image=img012brz6.jpg
**Note : the friction value is for the stretch of 9.6m in the picture.

u = .89, u = (Ff/Fn), 1714kg + 750kg = (2464kg)(9.8m/s/s) = 24147.2N, .89 = (Ff)/(24147.2), Ff = 21491N

F = ma, 21491 = 2464a, a = 8.71m/s/s

d = (1/2)(a)(t^2), 9.6m = 4.36(t^2), t = 1.48s

P = Momentum, (P)/(Ff) = t **I made this equation up because of the fact that momentum is Newtons x Seconds and dividing it by Newtons will give me time., (P)/(21491)=(1.48), P = 31806.7NS

Plugging into a triangle the 31806.7NS as the hypotinuse and 48.8degrees as the angle, I used cosine and sine to get the following information.
1) North = 20950.7NS
2) East = 23931.8NS

P = (M)(V)

Car : 20950.7 = (750)(V), V = 27.9m/s ~ 56mph
Truck : 23931.8 = (1714)(V), V = 13.96m/s ~ 28mph

My conclusion is that the car obviously didn't stop at the stopsign and was speeding while the truck kept just under the speed limit. The blame would entirely be on the driver of the car.

What do you think?
Due tomarrow!

 

[radou]

Hello WoodenFire, and welcome to PF!

It would be better you used the homework thread. Anyway, I can't see at which problem you actually stated what exactly the problem is (i.e. the text of the problem). Without it, I'm afraid no one will be able to help you.

 

[WoodenFire]

What are you talking about? The question in the problem clearly states "Who was lieing in the accident report?". The car obviously didn't stop at the stop sign and was going over the speed limit while the truck was ~2mph below the 30mph speed limit. The blame would entirely be placed on the car.

I showed all of my work and how I did it.

I am sorry I thought this was the homework thread. :O
But I am just wondering, at the velocities I found, and the coefficent of friction slowing the inelastic truck+car bundle down, does it all work out that they travel exactly 9.6m? If it does, doesn't that mean I did it correctly? I really am not an expert on physics. This is the first physics course I have ever taken.

Have you tried solving the problem from the picture given and masses of the car and truck? It really doesn't take more than 10-15minutes if you can spare it! :P

Thankyou again for your help and welcoming me.

 

[OlderDan]

Here is the actual physics problem. Below it I will try to explain exactly how I thought to figure it out. I would appreciate it if you corrected me If I made any errors. I am very sketchy on this one after being out for a week from class.

http://img74.imageshack.us/img74/2937/img012brz6.th.jpg
For some reason the picture isn't showing to most people... here is the imageshack direct link. http://img74.imageshack.us/my.php?image=img012brz6.jpg
**Note : the friction value is for the stretch of 9.6m in the picture.

u = .89, u = (Ff/Fn), 1714kg + 750kg = (2464kg)(9.8m/s/s) = 24147.2N, .89 = (Ff)/(24147.2), Ff = 21491N

F = ma, 21491 = 2464a, a = 8.71m/s/s

d = (1/2)(a)(t^2), 9.6m = 4.36(t^2), t = 1.48s

P = Momentum, (P)/(Ff) = t **I made this equation up because of the fact that momentum is Newtons x Seconds and dividing it by Newtons will give me time., (P)/(21491)=(1.48), P = 31806.7NS

Plugging into a triangle the 31806.7NS as the hypotinuse and 48.8degrees as the angle, I used cosine and sine to get the following information.
1) North = 20950.7NS
2) East = 23931.8NS

P = (M)(V)

Car : 20950.7 = (750)(V), V = 27.9m/s ~ 56mph
Truck : 23931.8 = (1714)(V), V = 13.96m/s ~ 28mph

My conclusion is that the car obviously didn't stop at the stopsign and was speeding while the truck kept just under the speed limit. The blame would entirely be on the driver of the car.

What do you think?
Due tomarrow!

Looks about right. You could have found the deceleration from friction by simply using

F = Ma = μMg
a = μg

You would do well to change your style a bit from what I call "stream of consciousness computing" to writing actual equations. Statements like

1714kg + 750kg = (2464kg)(9.8m/s/s) = 24147.2N

are pure nonsense. Kilograms do not equal Newtons. Get in the habit of writing equal signs only when relating quantities that are equal, such as

M = 1714kg + 750kg = (2464kg)
W = Mg = (2464kg)(9.8m/s/s) = 24147.2N