The speed of the third piece is about 17 m/s

<h3>Further explanation</h3>
Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !

<u>Given:</u>
mass of object 1 = m₁ = m
mass of object 2 = m₂ = m
velocity of object 1 = v₁ = -24 j
velocity of object 2 = v₂ = -24 i
mass of object 3 = m₃ = 2m
<u>Asked:</u>
velocity of object 3 = v₃ = ?
<u>Solution:</u>
We will use conservation of momentum formula to solve this problem:







<em>We could calculate the magnitude of the vector of velocity of this third piece as follows:</em>





<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant