CHAPTER 6
Momentum and Collisions
Momentum: The linear momentum of an object of mass m
moving with velocity v is defined as the product
of its mass and its velocity
p = mv
Momentum is a vector.
It can be broken down into components.
px = mvx
py = mvy
Newton’s Second Law can be re-written in terms of momentum
F = ma
F ∆t = m∆v
F = m∆v
F ∆t = ∆p
∆t
Impulse = change in Momentum
Impulse – Momentum Theorem
F ∆t = ∆p = mv – mvo
Applications
Batter Striking a Ball
Automobile Bumper Design
Objective
Maximize ∆v of the ball
Maximize ∆t so the F is
minimized
Figure 6.1(a) could be an illustration of the force involved
when a tennis ball hits a racket.
Figure 6.1(b) illustrates how to estimate the impulse
involved in the collision.
Example Problem (Baseball)
A baseball thrown at 45m/s (100mph) is struck by a bat and
moves toward centerfield at 85m/s. The contact with the bat
last for only .0015 seconds. The ball has a mass of .35kg.
Calculate the average force caused by the bat and the
change in momentum of the ball.
Problem Solving Strategy
Identify the variables
F=?
∆t = .0015sec
m = .35kg
∆v = 85m/s – -45m/s (moving towards centerfield was
= 130m/s
chosen as the positive direction)
Solve the Problem
∆p = mv
∆p = 46kg·m/s
∆p = .35kg · 130m/s
F ∆t = ∆p
F = ∆p = 46kg·m/s
∆t
.0015s
F = 31,000N
Conservation of Momentum
Any collision: Momentum is conserved
m 1v 1 + m 2v 2 = m 1v 1’ + m 2v 2’
Total momentum
before collision
Total momentum
after collision
Linear Momentum Problem
A cue with a mass of . 25kg ball rolls along a pool table with a
velocity of .85m/s and strikes the eight ball head-on which
has a mass of .28kg and is at rest. After the collision, the cue
ball continues to roll forward with a velocity of .15m/s. What
is the velocity of the eight ball after the collision?
Problem Solving Strategy:
1. Extract the data.
2. Plug data into Conservation
m1 = .25kg
of Momentum equation and
m2 = .28kg
solve for the unknown variable.
v1 = .85m/s
m 1v 1 + m 2v 2 = m 1v 1’ + m 2v 2’
v2 = 0m/s
v1’ = .15m/s
v2’ = .63m/s
v 2’ = ?
The total momentum before and after the collision
remains constant at .21kg m/s
Example Problem (Railroad Car Coupling)
A 91,000kg railroad car is moving at .35m/s toward a
45,000kg railroad currently at rest. What will be the
velocity of the two cars after they collide and couple up?
Identify the variables
m1 = 91,000kg
m2 = 45,000kg
v1 = .35m/s
v2 = 0m/s
v 1’ = ?
V2’ = ?
v1’ = v2’ = vT = Total velocity
Solve the Problem
m 1v 1 + m 2v 2 = m 1v 1’ + m 2v 2’
= m1VT + m2vT
= (m1+m2) · vT
All variables known except vT
vT = .23m/s
Isolated Systems
Momentum is conserved regardless of the nature of the
forces between the objects.
Applications:
Objects Involved:
Rocketry
Rocket and Propellant
Ice Skating
Two Skaters pushing away
from each other
Guns
Gun and Bullet
Example Problem (Two Figure Skaters)
An 80.kg male skater pushed away from his 40.kg female
partner and achieves a velocity (relative to the ice) of
2.5m/s. What will be the velocity of the female partner?
Identify the variables
m1 = 80.kg
m2 = 40.kg
v1 = 0m/s
v2 = 0m/s
v1’ = 2.5m/s
v 2’ = ?
Solve the Problem
m 1v 1 + m 2v 2 = m 1v 1’ + m 2v 2’
v2’ = -5.0m/s
Plug and solve for v2’
Elastic Collisions
Momentum is always conserved. In “elastic” collisions total
kinetic energy is also conserved. This allows us to write a
second equation (in addition to the conservation of
momentum equation) and solve two equations for two
unknowns. Thankfully, this is not required in this class.
Second Equation (Conservation of Kinetic Energy):
m 1v 12 + m 2 v 22 = m 1v 1’ 2 + m 2v 2’ 2
Elastic Collisions occur between atoms, molecules, and
sub-atomic particles. Some larger collisions may be
approximated as elastic collisions.
Inelastic Collisions:
Momentum is conserved. Total kinetic energy is not
conserved. Kinetic energy is dissipated as heat.