[Python] How to make turtle detect which side it collided on

Here is a MRE of my code:

from turtle import Turtle, Screen
import time
import random

# --------------- GLOBAL VARIABLES --------------------
speed_x = 20
speed_y = 20
bricks = []


def check_collision():
global speed_x, speed_y, bricks

if ball.xcor() > (window.window_width() / 2) - 20 or ball.xcor() < -(window.window_width() / 2) + 20:
speed_x *= -1
elif ball.ycor() > (window.window_height() / 2) - 20 or ball.ycor() < -(window.window_height() / 2) + 20:
speed_y *= -1
elif abs(paddle.xcor() - ball.xcor()) < 20/2 + 150/2 and abs(paddle.ycor() - ball.ycor()) < 20/2 + 60/2:
if abs(paddle.xcor() - ball.xcor()) == 20/2 + 150/2:
speed_x *= -1
else:
speed_y *= -1


def game_loop():
ball.goto(ball.xcor() + speed_x, ball.ycor() + speed_y)
check_collision()

window.update()

window.ontimer(game_loop, 1)

def move_right():
paddle.setx(paddle.xcor() + 25)


def move_left():
paddle.setx(paddle.xcor() - 25)


window = Screen()

time.sleep(2)

ball = Turtle(shape='circle')

ball.pu()
ball.speed(0)
ball.goto(0, -300)

ball.speed(0)
ball.left(random.randint(45, 135))

paddle = Turtle(shape='square')
paddle.speed(0)
paddle.goto(0, -400)
paddle.shapesize(1, 7.5, 3)
paddle.up()


window.listen()
window.onkeypress(move_right, 'Right')
window.onkeypress(move_left, 'Left')
window.onkeypress(move_right, 'd')
window.onkeypress(move_left, 'a')


game_loop()

window.mainloop()


Is it possible to detect if the ball hit the side (the short one) of the paddle? The reason I need this is since if It collides on the left side of the paddle, I want to ball x to change and not the y. In my code, the if abs(paddle.xcor() - ball.xcor()) == 20/2 + 150/2: is my attempt to recognize the side and the top (big side).

Continue reading...