Chapter 9: Linear Momentum and Collisions
Rachel Min
Background: From Newton’s Third Law, we know that any force applied to an object has an equal and
opposite force pushing back on it, so what happens when two objects moving with different velocities collide?
This unit covers linear momentum and collisions in both one and two dimensions in isolated and non-isolated
systems.
Relevant Formulae:
Key Points:
Momentum: is the product of the velocity and the mass of a
moving object, with the unit 𝑘𝑔 ∙ 𝑚/𝑠
Conservation of Momentum: the momentum in an isolated
system always remains constant
Impulse: change in momentum, denoted with a 𝐽⃗
Relation to Force: the integral of force over time is equal to
the impulse on an object
Elastic Collisions: systems in which the total kinetic energy
at the beginning and end of the collision is the same
Inelastic Collisions: systems in which the energy at the
beginning and at the end of the collision is not the same,
though momentum is still conserved
Perfectly Inelastic: when the objects colliding stick together
to form one object
Center of Mass: the point representing the average position
of matter in a system
𝐽⃗ = ∫ 𝐹⃗ 𝑑𝑡 = ∆𝑝⃗ (Impulse)
!!⃗
𝐹⃗ = !" (Force in relation to momentum)
𝑥!" =
∑ !! !!
∑ !!
(Center of Mass)
!
𝑟!" = ! ∫ 𝑟⃗𝑑𝑚 (Vector position of center of
mass of an extended object)
!
𝑣⃗!" = ! ∑! 𝑚! 𝑣⃗! (Velocity of center of mass)
!
𝑎⃗ !" = ! ∑! 𝑚! 𝑎⃗ ! (Acceleration of center of
mass)
1-D versus 2-D Collisions:
1-D Collisions occur on a single, straight-line
Diagrams:
axis and are the least complicated problems to
Elastic Collision:
Before
𝑝⃗ = 𝑚𝑣⃗ (Base momentum formula)
After
solve.
2-D Collisions occur on two axis, x and y, and
require the use of angles and separate
equations for each component. These problems
Inelastic Collision:
Before
After
Perfectly Inelastic Collision:
Before
After
are generally more complicated.
Helpful Strategy:
1. Draw a diagram
2. Analyze the problem (what are you
trying to find?)
3. Write out your relevant momentum
formula
4. Write out your momentum equation to
fit the problem
5. Solve
Practice Problem 1:
A 10.0-g bullet is fired into a stationary block of wood having mass m=5.00 kg. The bullet imbeds into the block.
The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original
speed of the bullet?
Practice Problem 2:
A 1 200-kg car traveling initially at vCi =25.0 m/s in an easterly direction crashes into the back of a 9,000-kg truck
moving in the same direction at vTi =20.0 m/s. The velocity of the car immediately after the collision is vCf =18.0 m/s
to the east. What is the velocity of the truck immediately after the collision?
Practice Problem 3:
An object of mass 3.00 kg, moving with an initial velocity of 5.00i m/s, collides with and sticks to an object of mass
2.00 kg with an initial velocity of -3.00j m/s. Find the final velocity of the composite object.
Practice Problem 1:
We first need to draw a diagram:
Before
After
What are we trying to find? Final velocity of the truck
Relevant momentum formula: 𝑝⃗ = 𝑚𝑣⃗
Now we can solve:
Equation to fit problem 𝑚! 𝑣! + 𝑚! 𝑣! = 𝑚! 𝑣′! + 𝑚! 𝑣′!
Entered numbers
(. 01𝑘𝑔)(𝑣! ) + 0 = (5.0𝑘𝑔 + .01𝑘𝑔)(.600𝑚/𝑠)
𝑣! = 301𝑚/𝑠
Practice Problem 2:
We first need to draw a diagram:
Before
After
What are we trying to find? Initial velocity
Relevant momentum formula: 𝑝⃗ = 𝑚𝑣⃗
Now we can solve:
Equation to fit problem 𝑚! 𝑣! + 𝑚! 𝑣! = 𝑚! 𝑣′! + 𝑚! 𝑣′!
Entered numbers (1200𝑘𝑔)(25𝑚/𝑠) + (9000𝑘𝑔)(20𝑚/𝑠) = (1200𝑘𝑔)(18𝑚/𝑠) + (9000)(𝑣 ! ! )
𝑣′! = 20.9 𝑚/𝑠
Practice Problem 3:
We first need to draw a diagram:
Before
What are we trying to find? Final velocity of the combined object
Relevant momentum formula: 𝑝⃗ = 𝑚𝑣⃗
Now we can solve:
Equation to fit problem 𝑚! 𝑣! + 𝑚! 𝑣! = (𝑚! + 𝑚! )𝑣′
Entered numbers
𝑣! =
!(3𝑘𝑔)(5𝑖)𝑚/𝑠 + (2𝑘𝑔)(−3𝑗!𝑚/𝑠 = (3 + 2𝑘𝑔)(𝑣′)
After
−1.2
= −28.1°
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We need 𝜃 because
direction matters
𝜃 = tan!!
15𝑖 − 6𝑗
= (3𝑖 − 1.2𝑗)𝑚/𝑠 ⇒ 𝑣 ! = !3! + 1.2! = 3.23𝑚/𝑠@ − 28.1°
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