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1 | initial version |

As @Peter Mitrano suggests, the **force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default maxtime step is 0.001 *(1 ms)*

F = m*a
v = a*t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F*t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + (100*0.001)/5 v(0.001) = 0.02 m/s *or*

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

2 | No.2 Revision |

As @Peter Mitrano suggests, the **force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default maxtime step is 0.001 *(1 ms)*

F = m*a
v = a*t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F*t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + (100*0.001)/5 v(0.001) = 0.02 m/s *or*

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

3 | No.3 Revision |

As @Peter Mitrano suggests, the **force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default maxtime step is 0.001 *(1 ms)*

F = m*a
v = a*t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F*t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + ~~(100~~*0.001)/5 **(100*0.001)/5 v(0.001) = 0.02 m/s *or* or

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

4 | No.4 Revision |

**force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default maxtime step is 0.001 *(1 ms)*

F = m*a
v = a*t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F*t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + (100*0.001)/5

**v(0.001) = 0.02 m/s** or **2 cm/s**

Throwing in the effects of gravity, we can determine *(roughly)* that the 5 kg sphere in your example should - in theory - reach a maximum vertical *(+z)* height of 0.01 m *(1 cm)* and return to a height of 0 m after 0.02 s *(20 ms)*.

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

5 | No.5 Revision |

**force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default maxtime step is 0.001 *(1 ms)*

F = m*a
v = a*t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F*t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + (100*0.001)/5

**v(0.001) = 0.02 m/s** or **2 cm/s**

Throwing in the effects of gravity, we can determine *(roughly)* that the 5 kg sphere in your example should - in theory - reach a maximum vertical *(+z)* height of 0.01 m *(1 cm)* and return to a height of 0 m after 0.02 s *(20 ms)*.

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

6 | No.6 Revision |

**force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default maxtime step is 0.001 *(1 ms)*

F = m*a
v = a*t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F*t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + (100*0.001)/5

**v(0.001) = 0.02 m/s** or **2 cm/s**

Throwing in the effects of gravity, we can determine *(roughly)* that the 5 kg sphere in your example should - in theory - reach a maximum vertical *(+z)* height of 0.01 m *(1 cm)* and return to a height of 0 m after 0.02 s *(20 ms)*.

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

7 | No.7 Revision |

**force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default maxtime step is 0.001 *(1 ms)*

F = ~~m~~*m * a
v = a*a * t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F*t)/m(F * t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + ~~(100*0.001)/5 ~~(100 * 0.001)/5

**v(0.001) = 0.02 m/s** or **2 cm/s**

*(roughly)* that the 5 kg sphere in your example should - in theory - reach a maximum vertical *(+z)* height of 0.01 m *(1 cm)* and return to a height of 0 m after 0.02 s *(20 ms)*.

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

8 | No.8 Revision |

**force is applied over only 1 simulation time step**, resulting in a **small impulse** and therefore a **very small change in velocity**.

When I run Gazebo simulator, the default ~~maxtime step ~~max time (step) size is 0.001 *(1 ms)*

F = m * a

v = a * t

v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then

**v(t) = v(t=0) + (F * t)/m**

*Given:*

v(t=0) = 0

F = 100 N *(F > Fgravity)*

t = 0.001 s *(1 simulation time step)*

m = 5 kg

v(0.001) = 0 + (100 * 0.001)/5

**v(0.001) = 0.02 m/s** or **2 cm/s**

*(roughly)* that the 5 kg sphere in your example should - in theory - reach a maximum vertical *(+z)* height of 0.01 m *(1 cm)* and return to a height of 0 m after 0.02 s *(20 ms)*.

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

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