Newton’s Laws of Motion
Newton’s First Law of Motion
http://www.wisc-online.com/objects/tp1202/tp1202.swf
Newton’s First Law of Motion
Demo using the Barbie car with Barbie in the front seat and a baby in
the back seat without child restraint.
Newton’s Second Law of Motion
The acceleration of an object is directly proportional to the force
applied to it and inversely proportional to the mass of the object.
a=F
m
F = ma
Note: both acceleration and force have magnitude and direction, so
both are vectors!
The newton is the unit of force N = kg*m/s2
Newton’s Second Law of Motion
Two horses are pulling a
barge with a mass of 2.00 x
103 kg along a canal. The
cable connecting to the first
horse makes an angle of
30.0 with respect to the
direction of the canal, while
the cable connected to the
second horse makes an
angle of 45.0. Find the
initial acceleration of the
barge, starting at rest, if
each horse exerts a force of
magnitude 6.00 x 102 N on
the barge. Ignore forces of
resistance on the barge.

F1
Barge = 2.00 x 103 kg
q1
q2

F2
Problem modified from Serway and Vuille’s College Physics 8th Edition
Newton’s Third Law of Motion
http://www.cpo.com/home/portals/2/Media/post_sale_content/newton
s%20third%20law.swf
Newton’s Laws of Motion
Here is a neat quiz that you can use with your students!
http://www.softschools.com/quizzes/science/newtons_laws/quiz384.ht
ml
Newton’s Laws of Motion
Let’s do some problems.
Momentum
The linear momentum of an object is dependent upon the mass and
the velocity of the object.
p = m*v
Notice that both momentum and velocity are vectors and, thus, have
both magnitude and direction!
Changing the momentum of an object requires that a force be applied.
Dp = Dt*F
Momentum
The impulse is defined as the force applied to an object over a given
period of time.
I = Dt*F
Substituting in the equation from the previous slide, we can get
I = DP = mvf - mvi
Momentum
Let’s try a problem!
In a crash test, a car of mass 1.50 x 103 kg collides with a wall
and rebounds. The initial and final velocities of the car are vi = 15.0 m/s and vf = 2.60 m/s, respectively.
If the collision lasts for 0.15 s, what is the impulse delivered
to the car due to the collision?
What is the size and direction of the average force exerted
on the car?
Conservation of Momentum
In a closed system, momentum must be conserved.
If 2 objects in motion collide, there can be a shift in momentum, but
the total sum must remain the same.
http://www.youtube.com/watch?v=mFNe_pFZrsA
Conservation of Momentum
If 2 objects in motion collide, there can be a shift in momentum, but
the total sum must remain the same.
V1i
V1f
V2i
V2f
F21Dt = - F12Dt
V1fm1 – V1im1 = V2fm2 – V2im2
V1im1 + V2im2 = V2fm2 + V1fm1
Inelastic Collisions
Inelastic collisions – when momentum is conserved but kinetic energy
is not. Some energy lost when the one object deforms to “stick” to the
other.
KE = p2
2m
V1im1 + V2im2 = Vf(m2 + m1)
V1im1 + V2im2 = Vf
(m2 + m1)
Elastic Collisions
Elastic collisions – when momentum is conserved and kinetic energy is
also conserved.
V1im1 + V2im2 = V2fm2 + V1fm1
½ m1V1i2 + ½ m2V2i2 = ½ m1V1f2 + ½ m2V2f2
Which can simplify to
V1i - V2i = - (V1f - V2f)
Collisions
Let’s do some problems!
• An SUV with a mass of 1.80 x 103 kg is traveling eastbound at 15.0
m/s, while a compact car with mass 9.00 x 103 kg is traveling
westbound at – 15.0 m/s. The cars collide head-on, becoming
entangled.
– Find the speed of the entangled cars after the collision.
•
– Find the change in the velocity of each car.
•
– Find the change in the kinetic energy of the system consisting of
both cars.
Collisions
Let’s do some problems!
Two billiard balls of identical mass move toward each other. Assume
that the collision between them is perfectly elastic. If the initial
velocities of the balls are + 30 cm/s and – 20 cm/s, respectively, what
is the velocity of each ball after the collision? Assume that friction and
rotation are unimportant.