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Physics: Collisions

Date: 12/17/98 at 02:04:09
From: Doug Grahame
Subject: Physics: Collisions

Our teacher gave us this problem to puzzle over but I'm completely 
confused as to how to solve it! Here it is:

   Before Collision:
   Two masses M1: 94 kg M2: 114 kg 
   Two velocities V1: 5.6 m/s  V2: 5.6 m/s
   Conditions of Collision: M2 is heading due south, while M1 is     
   heading due North. After the collision M2 veers off at due East.  
   M1 veers off at angle pheta.

   After Collison

   Two masses: M1: 94 kg M2: 114 kg 
   Two velocities: V1: x V2: 3.24  

   Using the formula x^2= V1^2 + V2^2 - 2(V1)(V2)cos(pheta) 

   This still leaves us with two variables. Considering it is a head-on 
   collision I thought computing the Forces of the two objects would 
   give me some useful information. M2 has enough of a force to 
   cancel out M1's force, plus still have an acting force of 291 
   newtons due south. I wasn't sure how to draw a vector diagram, as 
   the 291N is straight down. That's all that I could think of to try 
   to figure out this question. Hope you can help!

Date: 12/17/98 at 12:58:37
From: Doctor Rick
Subject: Re: Physics: Collisions

Hi, Doug. Welcome to Ask Dr. Math.

First of all, Doug, is the angle theta or phi? It looks like you tried 
for something halfway in between! Theta is the O with a horizontal 
line in the middle; phi is the o with a vertical | or / through it.

I can't figure out why you tried to apply the Law of Cosines with 
theta as the angle between V1 and V2. You can't apply formulas blindly! 
Draw a diagram:

                  | 5.6m/s
          theta      3.24
                   --------> M2
           x    /
             /    ^
          /       |
       M1         |
                  | 5.6m/s

So now, what can you use to figure out the direction and magnitude 
of M1's final velocity? Force is not going to be very helpful, because 
the forces in a collision depend strongly on the details of the 
interaction. (Force is proportional to acceleration, and acceleration 
depends on how much time it takes for the velocities to change.)

Try working with momentum. In Newton's laws, the one thing that is 
always conserved in a collision is the total momentum. I will just get 
you started.

Momentum is a vector, and it's best to think about the x (E-W) 
component and the y (N-S) component separately. 

What is the total E-W component of momentum initially? It is 0. 
Neither mass is moving horizontally. 

What is the total E-W component of momentum after the collision? M2 is
moving east at 3.24 m/s, and its mass is 114 kg, so its momentum (all 
in the x direction) is

  P_2x = (3.24)(114) kg-m/s.

M1 is moving at an angle. The E-W component of its velocity is -x sin 
theta. Its mass is (still) 94 kg, so the x component of momentus is 

  P_1x = (-x sin theta)(94) kg-m/s

Equating the initial and final totals gives you one equation to solve. 
Do the same with the N-S component of momentum and you will have a 
second equation with which to solve for your two variables.

See if you can make this work.

- Doctor Rick, The Math Forum   
Associated Topics:
High School Physics/Chemistry

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