Force exerted in inelastic collision

[chrisbowe82]

Hi, this is my first post, I've had a look at some examples to try and use them to help with my answer, but I've had no luck so far. I've just started doing physics and not spent much time on this board, so if my formatting/layout anything is a bit dodgy, then please tell me.
I Would really appreciate it if somebody could tell me what I'm doing wrong here, as the answer doesn't seem right to me, so here goes...

Homework Statement

A 7000kg wagon traveling at 8m/s runs into the back of a 5000kg stationary van.
a)Calculate their combined speed immediately after the collision
b)What is the force exerted on the wagon if the time taken to reach their combined speed after the collision is 0.3s?

Homework Equations

(m1 u1) + (m2 u2) = (m1+m2)v

F=m(v-u)/t

The Attempt at a Solution

a)Momentum of the truck =56,000kg m/s
I've worked out the combined speed
(m1 u1) + (m2 u2) = (m1+m2)v
v=61,000/12,000
v=5.08m/s (Combined speed after the collision)

b)The best I can work out is F=m(v-u)/t (I'm guessing the masses of the 2 vehicles are added together?)
so f=m(5.08-8)/t
F=(mv-mu)/t
F=60,960-96,000/0.3
F=-116,800 N

Am I right here, or way off?
Any help here will be much appreciated.

 

[Doc Al]

The Attempt at a Solution

a)Momentum of the truck =56,000kg m/s
I've worked out the combined speed
(m1 u1) + (m2 u2) = (m1+m2)v
v=61,000/12,000
v=5.08m/s (Combined speed after the collision)

Where did the 61,000 come from? (What's the initial speed of the van? What's its initial momentum?)

b)The best I can work out is F=m(v-u)/t (I'm guessing the masses of the 2 vehicles are added together?)

That equation is fine, but why add the masses? Since you want the force on the wagon, use the change in momentum of the wagon. (Of course, you may also take advantage of Newton's 3rd law and look at the van.)

 

[chrisbowe82]

Well, the initial velocity of the van is 0m/s before the collision
the 61,000 came from
7000kgx8m/s(the wagon) + 5000kgx0m/s(the truck) =61,000

So, I'm thinking now...
F=mass of the wagon x final velocity (7000 x 5.08) - mass x initial velocity (7000x8)
giving me f=35,560-56,000 /0.3

F=18760 N

or does m(v-u)/0.3 mean 'final mass - initial mass'/0.3? (I'm sorry for the daft question -it's 10 years since I last did physics...)

 

[Doc Al]

Well, the initial velocity of the van is 0m/s before the collision

So, I'm thinking now...
F=mass of the wagon x final velocity (7000 x 5.08) - mass x initial velocity (7000x8)
giving me f=35,560-56,000 /0.3

Good, except that your answer to part a for the final speed was wrong. (Redo it--see my earlier comment.)

F=18760 N

How did you get this number?

or does m(v-u)/0.3 mean 'final mass - initial mass'/0.3?

No, you had it right. v is the final speed; u is the initial speed.

 

[chrisbowe82]

The way I worked out part a) was this
(m1 u1) + (m2 u2) = (m1+m2)v
(7000x8) + (5000x0) = (7000+5000) ie, the 7000kg truck moving at 8m/s and 5000kg van at 0m/s
so 56,000 + 5000 = 12,000v
v=61,000/12,000
v=5.08m/s after the collision

so it gives me 5.08m/s as the velocity after the collision, which is where i put it into
the equation for the final velocity for part b)
I got F=18,760 by doing the calculation wrong. (not sure what happened there...)
F=(7000x5.08) - (7000x8)
F=35,560 -56,000 /0.3
F= -20440/0.3
F=-68133.33N

 

[Doc Al]

The way I worked out part a) was this
(m1 u1) + (m2 u2) = (m1+m2)v
(7000x8) + (5000x0) = (7000+5000) ie, the 7000kg truck moving at 8m/s and 5000kg van at 0m/s
so 56,000 + 5000 = 12,000v
v=61,000/12,000
v=5.08m/s after the collision

What's 5000x0?