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Conservation of Linear Momentum

As the particle interact within a system their individual momentum will transfer to the particle that they come in contact with.  However, if after analyzing the system you notice that there is no external forces acting on the system than you can use the conservation of linear momentum.

(Eq 1) $∑m_i(v_i)_1=∑m_i(v_i)_2$

The above equation represents the conservation of linear momentum, where it states that the total linear momentum will remain the same within the system of particles within a time period from $t_1$ to $t_2$.  Due to this, a system of particle will have a constant velocity around its center of mass.  This is regardless of how the particle within in the system interact with each other.

(Eq 2)  $(v_G)_1=(v_G)_2$

As previously mentioned, particle within the system will transfer there individual momentum to each other.  Because of this particle, when to particle of different mass contact each the respective velocity of one particle will be different than the other.

(Eq 3)  $m_1v_1 = m_2v_2$

$m_1$ = mass of particle 1

$m_2$ = mass of particle 2

$v_1$= velocity of particle 1

$v_2$ = velocity of particle 2

As seen in the above equation for a particle of a greater mass to have the same linear momentum as a particle of a smaller mass it will have a slower velocity.  On the other hand the particle with the smaller mass will have a greater velocity than the particle with the larger mass.  As a result heavier object are harder to stop than a lighter object that is going the same speed.

Example 

A cannon with a mass of 1200 kg shoots a cannon ball that has a mass of 12 kg at a velocity of 74$\frac{m}{s}$.  What is the velocity of the cannon After the cannon ball is fired.

Solution 

Step 1:  Determine the conservation of momentum equation.

$v_{c_1}m_c+v_{b_1}m_b=v_{c_2}m_c+v_{b_2}m_b$

where $v_{c_1}=0$, $v_{b_1}=0$, $v_{b_2}=75\frac{m}{s}$

$v_{c_2}m_c=v_{b_2}m_b$

Step 2: Determine the velocity of the cannon.

$v_{c_2}=\frac{12~kg(75~m/s)}{1200~kg}=.75~m/s$

 

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