One Each Day Till They Go Away (Part 4)

Classical Mechanics Level pending

A particle of mass M = 1 kg M = 1 \text{kg} is launched from the origin of the x y xy plane with speed v 0 = 50 m/s v_0 = 50 \text{m/s} at an angle θ = π / 3 \theta = \pi/3 with respect to the positive x x axis. The ambient gravitational acceleration g = 10 m/s 2 g = 10 \text{m/s}^2 is in the negative y y direction.

There is a force field in the region ( x − 190 ) 2 + ( y − 50 ) 2 ≤ 400 (x - 190)^2 + (y - 50)^2 \leq 400 . In this region, the field supplies a force F ⃗ = ( F x , F y ) = ( − 10 N , 50 N ) \vec{F} = (F_x, F_y) = (-10 \, \text{N}, 50 \, \text{N}) .

When the particle lands again at y = 0 y = 0 , what is its x x coordinate?

Note: Gravity is active within the force field region


The answer is 283.2.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Jun 7, 2021

Simulation code attached below. I think that this problem can be managed analytically, although might be a bit tedious. Attached is also the particle's trajectory. 

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% OCTAVE
clear all
clc

% Parameters:
theta    = 60*(pi/180);
g        = 10;
vo       = 50;
M        = 1;

% Initial conditions:
dx(1)    = vo*cos(theta);
dy(1)    = vo*sin(theta);
x(1)     = 0;
y(1)     = 0;

% Numerical resoloution and time initialisation:
dt       = 1e-5;
t(1)     = 0;
k        = 1;

% Simulation loop till y coordinate becomes zero again:
while y(k) >= 0

    % Force field components zeroed by default:
    Fx         = 0;
    Fy         = 0;

    % Force field circle:
    R_sq       = (x(k) - 190)^2 + (y(k) - 50)^2;

    % Within circle force field is non zero:
    if R_sq <= 400
        Fx     = -10;
        Fy     = 50;
    end

    % Newton's second law:
    ddr        = ([0;-M*g] + [Fx;Fy])/M;

    % Acceleration components:
    ddx        = ddr(1);
    ddy        = ddr(2);

    % Numerical integration: semi implicit Euler:
    dx(k+1)    = dx(k) + ddx*dt;
    dy(k+1)    = dy(k) + ddy*dt;

    x(k+1)     = x(k) + dx(k+1)*dt;
    y(k+1)     = y(k) + dy(k+1)*dt;

    % Time increment:
    t(k+1)     = t(k) + dt;

    % Index increment:
    k          = k + 1;
end


ANSWER     = x(end);
% ANSWER   = 283.1864

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