U3AOS1 Topic 9: Momentum

Momentum is a vector that can be defined as "a mass in motion" and is described by the equation:

\[ \displaystyle \Large \rho = mv\]


The Law of Conservation states that all objects within an isolated system will have a constant momentum such that the initial and final momentum will be equal.


In other words, the momentum will be constant in an isolated system.

Adding all of the initial and final momentums will give the same value:

\[ \displaystyle \Large \sum \rho_{\text{initial}}= \sum \rho_{\text{final}}\]


An isolated system is where only specific components are considered and focused on. It can be seen in the visual representation below




Scenarios


Momentum questions generally involve one or two objects and require you to calculate the momentum before and after a potential collision. There are two situations that can occur when this happens.


1. They move at the same speed and you need to find the velocity

2. They are moving at different speeds and you need to find one velocity

It is unlikely you will need to calculate the mass but if you do the process is just rearranging the equations.



For the first scenario, the formula below must be applied:

\[ \displaystyle \Large m_{1}u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v_{1} \]


For the second scenario, the formula below must be applied:

\[ \displaystyle \Large m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \]


The equations look similar because they are the same equation but the first one has just been factorised. 




Process of solving:


1. Label the left direction as negative and the right as positive

2. Identify and label all masses and velocities

3. Write the appropriate equation depending on the scenario

4. Substitute your variables

5. Solve methodically and carefully

There is always an urge to do it quickly but it's more important to be well practiced and correct first.

6. Identify the direction (if solving for the velocity)



Simulation:


Use the simulation below to assist you in your working. Note that an e value of 1 is elastic and any other value is inelastic (this will be covered in the collision topic)



Example 1



A \( 2 kg \) cart moving rightward at \(4 m/s \) collides with a \(3 kg \) cart moving leftward at \(2 m/s \). After the collision, the \(2 kg \) cart is observed moving leftward at \(2 m/s\). What is the velocity of the \(3 kg\) cart after the collision?

Step 1: Label that the left is negative and the right direction is positive

−←→+- \leftarrow \rightarrow+

Step 2: Identify and label all masses and velocities
\( m_{1}=2\), \( u_{1}=4 \), \( v_{1} = -2 \)
\( m_{2} = 3 \), \(u_{2}=-3 \), \(v_{2}=\text{?}\)
m1=2,u1=4,v1=−2m_1 = 2, u_1 = 4, v_1 = -2
m2=3,u2=−3,v2=?m_2 = 3, u_2 = -3, v_2 = ?

Step 3: Choose the appropriate equation
we know the velocities are not the same so the second scenario is more appropriate
\[\displaystyle \Large m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}\]
m1u1+m2u2=m1v1+m2v2m_1u_1+m_2u_2 = m_1v_1 + m_2v_2

Step 4: Substitute your variables
\[ \displaystyle \Large (2)(4) + (3)(-3)= (2)(-2)+ (3) v_{2}\]
(2)(4)+(3)(−3)=(2)(−2)+(3)v2(2)(4)+(3)(-3) = (2)(-2) + (3)v_2

Step 5: Solve each momentum and sum carefully.
\[ 8+ (-9) = -4 + 3v_{2}\]
\[ -1 = -4 + 3v_{2}\]
\[-1 +4 = 3v_{2}\]
\[ 3= 3v_{2}\]
\[1=v_{2}\]
8+−9=−4+3v28+-9 = -4+ 3v_2
−1=−4+3v2-1 = -4+ 3v_2
−1+4=3v2-1 +4= 3v_2
3=3v23= 3v_2
1=v21= v_2

Step 6: Identify the direction
The value is positive so it's moving to the right
 →

Therefore the final velocity for the \(3kg \) mass is \(1 m/s\) to the right.

Example 2



A \( 2 kg \) cart moving rightward at \(4 m/s \) collides with a \(3 kg \) cart moving leftward at \(2 m/s \). After the collision, the \(2 kg \) cart is observed moving leftward at \(2 m/s\). What is the velocity of the \(3 kg\) cart after the collision?

Step 1: Label that the left is negative and the right direction is positive

−←→+- \leftarrow \rightarrow+

Step 2: Identify and label all masses and velocities
\( m_{1}=2\), \( u_{1}=4 \), \( v_{1} = -2 \)
\( m_{2} = 3 \), \(u_{2}=-3 \), \(v_{2}=\text{?}\)
m1=2,u1=4,v1=−2m_1 = 2, u_1 = 4, v_1 = -2
m2=3,u2=−3,v2=?m_2 = 3, u_2 = -3, v_2 = ?

Step 3: Choose the appropriate equation
we know the velocities are not the same so the second scenario is more appropriate
\[\displaystyle \Large m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}\]
m1u1+m2u2=m1v1+m2v2m_1u_1+m_2u_2 = m_1v_1 + m_2v_2

Step 4: Substitute your variables
\[ \displaystyle \Large (2)(4) + (3)(-3)= (2)(-2)+ (3) v_{2}\]
(2)(4)+(3)(−3)=(2)(−2)+(3)v2(2)(4)+(3)(-3) = (2)(-2) + (3)v_2

Step 5: Solve each momentum and sum carefully.
\[ 8+ (-9) = -4 + 3v_{2}\]
\[ -1 = -4 + 3v_{2}\]
\[-1 +4 = 3v_{2}\]
\[ 3= 3v_{2}\]
\[1=v_{2}\]
8+−9=−4+3v28+-9 = -4+ 3v_2
−1=−4+3v2-1 = -4+ 3v_2
−1+4=3v2-1 +4= 3v_2
3=3v23= 3v_2
1=v21= v_2

Step 6: Identify the direction
The value is positive so it's moving to the right
 →

Therefore the final velocity for the \(3kg \) mass is \(1 m/s\) to the right.

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