Momentum

Momentum is the measurement of an object's inertia and is found using the mass and velocity. It is always conserved in a collision, allowing us to find the velocities of objects after a collision.

**Momentum: **Momentum is a vector quantity that measures of an object's inertia when it is in motion. SI units: kg•m/s

**Law of Conservation of Momentum: **Momentum is always conserved during collisions between objects.

**Impulse: **A force exerted over a period of time. SI units: N•s or (kg•m)/s

**Momentum** is the product of mass and velocity (p = mv). We can think of momentum as how hard it is to move/stop a moving object. Thus, between a truck and a car moving at the same speed, the truck has more momentum.

- Newton's second law can be rewritten for momentum. \(\textbf{F}_{net} = m\textbf{a} = m \frac{\Delta \textbf{v}}{\Delta t} = \frac{\Delta \textbf{p}}{\Delta t}\)

Momentum can be transferred from one object to another. The ** law of conservation of momentum** is used to determine the states of objects before or after collisions.

- When dealing with momentum in two dimensions, remember to break down all the vectors into the horizontal and vertical components. Momentum must be conserved in both dimensions.
- There are two main types of collisions:
*Perfectly elastic collisions*occur when energy is conserved in a collision and the colliding objects bounce back perfectly - the total momentum and energy is the same before and after the collision.*Perfectly inelastic collisions are when energy is not conserved in a collision and objects stick together. In this case, only momentum is conserved because some of the energy of the objects goes into producing sound waves and heat.*

Figure: Example of inelastic collision. Initially, only block B is moving. After collision, blocks A and B are stuck together and moving to the right. If the collision is perfectly inelastic, the momentum before and after the collision is the same.

**Impulse **relates force to momentum. By exerting an external force over a certain time, the force is changing the momentum of the object. A large force exerted over a short time will produce the same change in momentum as a small force exerted over a long time.

- If an external force (a force outside the system) is applied to a system, the final momentum is not equal to the initial momentum. The change is equal to the impulse.

Momentum Problem Guide

*p = mv*

*p* - momentum

*m* - mass

*v *- velocity

\(j = \Delta p = F \Delta t\)

*J* - impulse

*F* - force

*t* - time

An object (\(m_1\)) of mass 2kg is moving at speed v1 in the positive x-direction. It collides perfectly elastically with another object (\(m_2\)) of mass 3kg moving at -3m/s. If m1 is moving at +1m/s and \(m_2\) is moving at \(+2m/s\) after the collision, how fast was m1 moving before the collision?

To solve this problem, we will use conservation of momentum. Since the collision was elastic, meaning that the objects did not stick together, we could also use conservation of kinetic energy

\(p_i\)

=

\(p_f\)

start with the conservation of momentum

\(m_1 v_{1i} + m_2 v_{2i}\)

=

\(m_1 v_{1f} + m_2 v_{2f}\)

sum the initial and final momentum of both objects

\(v_{1i}\)

=

\(\frac{m_1 v_{1f} + m_2 v_{2f} - m_2 v_{2i}}{m_1}\)

solve for \(v_{1i}\)

\(v_{1i}\)

=

\(\frac{2 kg . 1 m/s + 3 kg . 2 m/s -3kg . -3m/s}{2 kg}\)

plug in known values

\(v_{1i}\)

=

\(8.5 ms\)

An object (\(m_1\)) of mass 2 kg is moving at 2 m/s in the positive x-direction. It collides perfectly inelastically with another object (\(m_2\)) of mass 3 kg moving at -1 m/s. How fast will the objects be moving after the collision?

This problem is very similar to Example 1, except the objects collide inelastically, which means that they stick together after the collision. In this case, we would not be able to use conservation of kinetic energy.

\(p_i\)

=

\(p_f\)

start with the conservation of momentum

\(m_1 v_{1i} + m_2 v_{2i}\)

=

\((m_1 + m_2)v_f\)

there is only one term for the final momentum because the objects are stuck together

\(v_{f}\)

=

\(\frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}\)

solve for \(v_f\)

\(v_{f}\)

=

\(\frac{2 kg . 2 m/s + 3 kg . -1 m/s}{2 kg + 3 kg}\)

plug in known values

\(v_{f}\)

=

\(.2 m/s\)