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.
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.
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.
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
\(v_{f}\)
=
\(.2 m/s\)