Calculating Force in Car-Railroad Collision

[remaan]

Homework Statement

A 1250 kg car moving with a speed of 28.1 m/s collides head on with a railroad car of mass 2.5 * 10^4 kg moving with a speed of 3.00 m/s The car becomes attached to the railroad car in the collision IF the collison takes 0.952 s what was the average force on the car in N ?

Homework Equations

F avr. = pf - Pi / t

The Attempt at a Solution

I am fine with every thing, but my problem is calculation ??

I did that and got 211.2 N

Is that Right ?

 

[alphysicist]

Hi remaan,

Homework Statement

A 1250 kg car moving with a speed of 28.1 m/s collides head on with a railroad car of mass 2.5 * 10^4 kg moving with a speed of 3.00 m/s The car becomes attached to the railroad car in the collision IF the collison takes 0.952 s what was the average force on the car in N ?

Homework Equations

F avr. = pf - Pi / t

The Attempt at a Solution

I am fine with every thing, but my problem is calculation ??

I did that and got 211.2 N

Is that Right ?

That force seems rather small for the 1250kg car in this case. Can you show what numbers you used, and how you got that result?

 

[remaan]

In fact, I can say that I used the same num. in the ques.

and the law of cons. of Momentom.

First, I found the velocity after the collison which equals to 4. 195

and then I subtract Pf - Pi and / .592

and then I got 57.16 * 10 ^ 3 What do you think ??

 

[remaan]

Or, I am not sure of using the numbers. I mean I used the whole number without rounding ??

 

[djeitnstine]

\Delta p = m \Delta v_f - m \Delta v_i then the resulting momentum equation is m_1v_{i1} + m_2v_{i2} = m_1v_{f1} + m_2v_{f2}, and since the collision is inelastic the equation reduces to m_1v_{i1} + m_2v_{i2} = (m_1+ m_2)v was this your approach?

 

[alphysicist]

In fact, I can say that I used the same num. in the ques.

Sure, but how did you use those numbers? What equation did you set up, and how did you put the numbers in the equation? It appears to me that you used the numbers in the wrong way, because the velocity after the collision is not 4.195m/s. (The train car here should not speed up due to the collision.) It seems to me that you are perhaps using speed instead of velocity, but without knowing what you did I can't be sure.

and the law of cons. of Momentom.

First, I found the velocity after the collison which equals to 4. 195

and then I subtract Pf - Pi and / .592

The time given in the original post was 0.952; was this just a typo? (Also, what were you using for Pf and Pi?)

and then I got 57.16 * 10 ^ 3 What do you think ??

 

[remaan]

\Delta p = m \Delta v_f - m \Delta v_i then the resulting momentum equation is m_1v_{i1} + m_2v_{i2} = m_1v_{f1} + m_2v_{f2}, and since the collision is inelastic the equation reduces to m_1v_{i1} + m_2v_{i2} = (m_1+ m_2)v was this your approach?

Yes, I used this equation.

 

[remaan]

ok, let me show you what I did.
m1v1+ m2v2 = V ( m1 + m2)
I subs. the masses

(1250 * 28.1 + 2.5 * 10 ^4 * 3 ) = V ( 1250+ 2.5 * 10^4 )

And V = 4.195 ( after the collision )

After that, I found the initial and final momentom.

Pi = ( 1250 * 28.1 + 2.5 * 10^ 4 ) = 60125

And Pf = (4.195 ) ( 1250 + 3 * 2.5 * 10 ^ 4 )
And when I subtract them and divide by .592 I got 52.5 * 10 ^ 3

I don't know why each time I do it I get different answer ??

 

[Redbelly98]

Since the collisions is "head on", does that mean the velocities are in the same direction or opposite directions before colliding?

Can you use + or - to indicate the direction of a velocity?