Double gravity when standing on two colliders, bug?

Hey,

We are developing a game as part of our semester project. In the game you play as a ball that has to reach the goal. The world itself is contructed from a lot of 1x1 cube prefabs.

The character has a jump ability (done by addforce in y-axis). Our problem is that when the character is in the middle of two adjacent cubes, it will only jump half as high.

Our preliminary guess was that the game somehow applied double gravity(one for each collider). We tested this by making two cubes in the exact same spot, and as assumed, the character jumped half of the intended height.

Just wondering if this is a known bug or if we are doing something wrong? There is a video showcasing our problem: http://www.youtube.com/watch?v=51G2ZozdFZ4

Regards

EDIT: We tried reducing the game to the very basic components and this is the ONLY script running:

 using UnityEngine;
 using System.Collections;
 
 public class VeryBasicMovement : MonoBehaviour {
     Vector3 brake;
     // Use this for initialization
     void Start () {
     
     }
     
     // Update is called once per frame
     void Update () {
     
     }
     
     void FixedUpdate () {
         if(Input.GetKey(KeyCode.LeftArrow)) {
             rigidbody.AddForce(new Vector3(-50, 0, 0));
         }
         
         if(Input.GetKey(KeyCode.RightArrow)) {
             rigidbody.AddForce(new Vector3(50, 0, 0));
         }
         
         if(Input.GetKey(KeyCode.UpArrow)) {
             rigidbody.AddForce(new Vector3(0, 0, 50));
         }
         
         if(Input.GetKey(KeyCode.DownArrow)) {
             rigidbody.AddForce(new Vector3(0, 0, -50));
         }
         
         if(Input.GetKeyDown(KeyCode.Space)) {
             rigidbody.AddForce(new Vector3(0, 400, 0));
         }
         
         ApplyBrakes();
     }
     
     void ApplyBrakes() {
         brake =  -1 * rigidbody.velocity * 400 * Time.deltaTime;
         rigidbody.AddForce(new Vector3(brake.x, 0, brake.z));
     }
 }
 

And it still has the same problem.