[Physics] How to calculate the time until collision

collisionhomework-and-exercisesnewtonian-mechanics

I'm programming a 1 dimensional physics simulator in python. How do I calculate the amount of time until two particle points that are traveling towards eachother collide? I am using the equation:

time = (m2.pos - m1.pos) / (m1.velocity - m2.velocity).

You can assume m1 is always on the left, and m2 is always on the right. This equation works fine in my program if both points are traveling in the same direction at slightly different speeds, but when m1 has a positive velocity and m2 has a negative velocity, that gives the denominator of my equation a negative number which produces a negative amount of time until the next collision.

Best Answer

If m1 has a positive velocity and m2 has a negative velocity, you get (m1.velocity - m2.velocity) = (positive - negative) = positive. On the other hand if the signs are switched, you'll get a negative result because the particles are moving away from each other.

Related Solutions

Astrophysics – Help with Equations for Deflecting an Asteroid

A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass $m$ at distance $D$, heading for earth of radius $R$ with velocity $v$.

The equations you need are conservation of angular momentum and energy. If you give the asteroid a lateral kick $F\Delta t$ (impulse = change in momentum $m\Delta v$), then the angular momentum of the new orbit is

$$L = m \Delta v \cdot r$$

And this will still be the angular momentum when you reach earth. So the question then becomes, what velocity will it have when it gets to earth? The answer - at a distance of $3R$ it will have increased its velocity (because of the gravitational potential energy it had). Assuming you started very far away ($r>>R$) you find this from

$$\Delta E = \frac{GmM_e}{3R} = \frac{gmR}{3}$$

since we know that the gravitational acceleration of the earth at the surface, $g$, is given by

$$g = \frac{GM_e}{R^2}$$

Finally, kinetic energy is $\frac12 m v^2$ as you may remember. So velocity will increase when energy increases.

I'm going to leave you the fun of combining these equations to solve your problem. I must admit I am a little bit unsure of how you plan to model the explosion - you might want to elaborate a bit on that if you need help figuring out the impulse $F\Delta t$.

Newtonian Mechanics – Analyzing Multiple Colliding Balls Using Computational Physics

You need to setup a system of $K$ equations with $K$ impulse unknowns if derived from the $N$ balls whereas only $K$ of them are colliding at some time.

Here is how to do it:

Create a (block) vector with all velocity vectors in sequence. The size is $2N\times1$ $$ \mathbf{v} = \begin{pmatrix} \vec{v}_1 \\ \vec{v}_2 \\ \vdots \\ \vec{v}_N \end{pmatrix} = \begin{pmatrix} \dot{x}_1 \\ \dot{y}_1 \\ \dot{x}_2 \\ \dot{y}_2 \\ \vdots \\ \dot{x}_N \\ \dot{y}_N \end{pmatrix} $$
Create a (block) matrix with all the masses, but also of $2N$ rows $$\mathbf{M} = \begin{bmatrix} m_1 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & m_1 & 0 & 0 & & 0 & 0 \\ 0 & 0 & m_2 & 0 & & 0 & 0 \\ 0 & 0 & 0 & m_2 & & 0 & 0 \\ & & & & \ddots & & \\ 0 & 0 & 0 & 0 & & m_N & 0 \\0 & 0 & 0 & 0 & & 0 & m_N \end{bmatrix}$$ such that $\mathbf{M} \mathbf{v}$ would be all the momentum componnets.
You are looking for the change in velocity $\Delta \mathbf{v}$ due to all the impulses from the collisions. If you knew the sum of all impulses $\boldsymbol{\ell}$ (again in $2N\times1$ vector) then you would write the change in velocity as $$\Delta \mathbf{v} = \mathbf{M}^{-1} \boldsymbol{\ell}$$
Create a $2N \times K$ contact pair block matrix $\mathbf{Z}$. Each column represents a contact between two bodies ($i \rightarrow j$) by containing the contact normal vector $\hat{n}_{ij}$ in the $j$-th block position, and the negative normal vector $-\hat{n}_{ij}$ in the $i$-th block position. For example: $$\mathbf{Z} = \begin{bmatrix} -\hat{n}_{13} & 0 & -\hat{n}_{14} \\ 0 & -\hat{n}_{24} & 0 \\ \hat{n}_{13} & 0 & 0 \\ 0 & \hat{n}_{24} & \hat{n}_{14} \end{bmatrix}$$ represents the contact pair matrix when bodies ($1 \rightarrow 3$, $2 \rightarrow 3$ and $1 \rightarrow 4$) are colliding. The arrow indicates the direction at which the normal vector points, and hence the direction where the positive impulse is applied. Remember that each contact normal vector has two components $\hat{n} = ( \cos \psi, \sin \psi )$.
Construct a $K \times 1$ vector of unknown impulse magnitudes for each contact pair. $$ \mathbf{J} = \begin{pmatrix} J_{13} \\ J_{24} \\ J_{14} \end{pmatrix}$$ Now the sum of all impulses are computed by $$\boldsymbol{\ell} = \mathbf{Z} \,\mathbf{J}$$
Calculate the relative speed of each contact pair ($K \times 1$ vector) by projecting the speeds along the contact normal $$ \mathbf{u}_{imp} = \mathbf{Z}^\top \mathbf{v}$$
Establish the law of collision in terms of the relative speeds before and after the collision. $$ \mathbf{Z}^\top \left( \mathbf{v} + \Delta \mathbf{v} \right) = -\epsilon \mathbf{Z}^\top \mathbf{v}$$ where $\epsilon$ is the coefficient of restitution. Move the knowns to the right hand side and find the change in relative speeds by $$ \mathbf{Z}^\top \Delta \mathbf{v} = -(1+\epsilon) \mathbf{Z}^\top \mathbf{v} = -(1+\epsilon) \mathbf{u}_{imp}$$
Put 3. 5. and 7. together to make $$ \mathbf{Z}^\top \left(\mathbf{M}^{-1} \left( \mathbf{Z} \,\mathbf{J} \right) \right) = -(1+\epsilon) \mathbf{u}_{imp} $$ and solve for the unknown impulses $$ \mathbf{J} = -\left(\mathbf{Z}^\top \mathbf{M}^{-1} \mathbf{Z} \right)^{-1} (1+\epsilon) \mathbf{u}_{imp} $$ Note that $\mathbf{Z}^\top \mathbf{M}^{-1} \mathbf{Z}$ is a $K \times K$ matrix that needs inversion.
When all the impulses are calculated, each body velocity vector changes by $$ \Delta \mathbf{v} = \mathbf{M}^{-1} \mathbf{Z} \,\mathbf{J} $$

Example

Imagine three spheres impacting at the same time:

Use this MATLAB script to find their final velocties:

%% Example triple impact (with 3D vectors)
% Three spheres impact at the same time. See Figure for their
% masses, positions and velocities before impact

% define unit vectors
o_ = [0;0;0];
i_ = [1;0;0];
j_ = [0;1;0];
k_ = [0;0;1];

R = 20;
r_1 = R*i_ 
r_2 = -R*i_
r_3 = R*tand(60)*j_

% Three possible contacts: 1*2, 2*3, 1*3
n_12 = (r_2-r_1)/norm(r_2-r_1)
n_23 = (r_3-r_2)/norm(r_3-r_2)
n_13 = (r_3-r_1)/norm(r_3-r_1)

% velocity vectors before impact
v_1 = o_;
v_2 = -1.0*n_12
v_3 = -2.0*n_13

v = [v_1;v_2;v_3]

% mass matrix
M = blkdiag(5.0*eye(3),4.0*eye(3),3.0*eye(3))

% contact pair matrix
Z = [-n_12, o_, -n_13; n_12, -n_23, o_; o_, n_23, n_13]
% Z =
% 
%     1.0000         0    0.5000
%          0         0   -0.8660
%          0         0         0
%    -1.0000   -0.5000         0
%          0   -0.8660         0
%          0         0         0
%          0    0.5000   -0.5000
%          0    0.8660    0.8660
%          0         0         0

% coefficient of restitution
eps = 0.5

% impact speeds
u_imp = Z.'*v

% impulses
J = -(1+eps)*inv(Z.'*inv(M)*Z)*u_imp
% J =
% 
%     1.7021
%     2.1702
%     4.6277

% velocity vector change
Dv = inv(M)*Z*J

% final velocity
v_final = v+Dv
% v_final =
% 
%     0.8032
%    -0.8015
%          0
%     0.3032
%    -0.4699
%          0
%     0.5904
%     0.2303
%          0


% check for error
actual = Z.'*(v+Dv);        %(relative speeds after impact) =
expected = -eps * u_imp;    % -eps*(relative speeds before impact)
err = actual-expected

% err =
% 
%   1.0e-015 *
% 
%    -0.1110
%    -0.6661
%    -0.6661

In the last part, I check the results against the law of contacts and the error is ~1e-15.

Best Answer

Related Solutions

Astrophysics – Help with Equations for Deflecting an Asteroid

Newtonian Mechanics – Analyzing Multiple Colliding Balls Using Computational Physics

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