Momentum and Collision
Momentum
Momentum: A vector quantity defined as
the product of an object’s mass and
velocity.
p = momentum
momentum = mass x velocity
p = mv
Units of momentum: kg m/s
Sample Problem
A 2250 kg truck has a velocity of 25 m/s to the
east. What is the momentum of the truck?
p = mv
p = (2250 kg)(25 m/s)
p = 5.6 x 104 kg m/s to the east
Impulse
From Newton’s Second Law:
F = ma
F = mΔv
Δt
F = Δp
Δt
FΔt = Δp or FΔt = mvf – mvi
Impulse
Impulse: A force applied over a period of time.
Impulse is measured in Newton seconds, Ns.
force
Impulse = FΔt = Δp
Area under curve is
the impulse
time
Impulse-Momentum Theorem
The impulse on an object is equal to the object’s
final momentum minus the object’s initial
momentum.
FΔt = pf – pi
FΔt = mvf – mvi
Impulse Problem (FΔt = Δp)
A driver accelerates a 240 kg snowmobile, which
results in a force being exerted that speeds up
the snowmobile from 6.0 m/s to 28.0 m/s over
a time interval of 60 s.
 What is the change in momentum
 What is the impulse on the snowmobile
 What is the average force that is exerted on the
snowmobile
Impulse Problem (FΔt = Δp)
A driver accelerates a 240 kg snowmobile, which
results in a force being exerted that speeds up
the snowmobile from 6.0 m/s to 28.0 m/s over
a time interval of 60 s.
 What is the change in momentum
Δp = mvf – mvi
= 240 kg(28 m/s) – 240 kg(6 m/s)
= 5280 kg m/s
Impulse Problem (FΔt = Δp)
A driver accelerates a 240 kg snowmobile, which
results in a force being exerted that speeds up
the snowmobile from 6.0 m/s to 28.0 m/s over
a time interval of 60 s.
 What is the impulse on the snowmobile
FΔt = Δp Δp =5280 kg m/s
FΔt = impulse
Impulse = 5280 kg m/s
Impulse Problem (FΔt = Δp)
A driver accelerates a 240 kg snowmobile, which
results in a force being exerted that speeds up
the snowmobile from 6.0 m/s to 28.0 m/s over
a time interval of 60 s.
What is the average force that is exerted on the
snowmobile
FΔt = Δp
F(60s) = 5280 kg m/s
F = 88 N
Impulse Problem (FΔt = Δp)
Suppose a 60 kg person is traveling in a vehicle moving at
32 m/s. The car hits a concrete wall and the car comes
to a stop in 0.2 s. What is the force exerted on the
person during the crash?
FΔt = Δp
F(.20s) = mvf – mvi = 60kg(0m/s) – 60kg(32m/s)
F(.20s) = -1920 kg m/s
F = -9600 N
Impulse Problem (FΔt = Δp)
Suppose that this person is now in a specially engineered
car that crumples upon impact and takes 0.8 s to come
to a complete stop. What is the force on the passenger
now?
FΔt = Δp
= mvf – mvi = 60kg(0m/s) – 60kg(32m/s)
= -1920 kg m/s
FΔt = Δp
F(.80s) = -1920 kg m/s
F = -2400 N
Collisions
Collisions: When two or more objects come into
contact with each other and exchange energy
and momentum.
Elastic Collision: total kinetic energy of the
system before the collision equals the total
kinetic energy after the collision. Momentum is
also conserved.
Elastic Collisions
• No energy is dissipated as heat energy
• No energy is spent on deforming the bodies
• Occurs when two objects “bounce” apart after
they collide
• Close examples:
– Bumper cars
– Pool table balls
– Rubber balls colliding
Elastic Collision
m1v1,f + m2v2,f = m1v1,i + m2v2,i
m1
m2
m1
m2
½ m1v1,f2 + ½ m2v2,f 2= ½ m1v1,i2 + ½ m2v2,i2
Elastic Collision Problem
Ball A (0.355 kg) moves to the right on a
frictionless surface at 0.095 m/s and hits ball B
(0.710 kg) moving in the same direction at a
speed of 0.045 m/s. After the collision the
speed of ball A is 0.035 m/s. What is the speed
of ball B?
mava,f + mbvb,f = mava,i + mbvb,i
.355 kg(.035 m/s) + .71 kg(vb,f) = 0.355 kg(.095m/s) + .71kg(.045m/s)
Vb,f = +0.075 m/s
Elastic Collision Problem 2
Two balls are moving towards each other. The
blue ball is moving to the right with a mass of
0.2 kg and a velocity of 1 m/s. The green ball,
0.5 kg, is moving to the left with a velocity of 3
m/s. If the green ball after the collision is
moving to the left with a velocity of 2 m/s, what
is the velocity of the blue ball?
0.2 kg
1 m/s
3 m/s
0.5 kg
Problem 2
0.2 kg
1 m/s
3 m/s
0.5 kg
mbluevf + mgreenvf = mbluevi + mgreenvi
0.2kg(vf) +0.5kg(-2 m/s) = 0.2 kg(1 m/s) + .5 kg( -3 m/s)
0.2vf – 1 kg m/s = -1.3 kg m/s
0.2vf = -0.3 kg m/s
Vf(blue) = -1.5 m/s
Inelastic Collisions
Inelastic Collision: A collision between bodies where
momentum is conserved but the total kinetic energy is
not. Part of the kinetic energy is transformed into
another energy form. Usually colliding objects are
distorted in some way.
Examples:
- car collisions
- two balls of clay colliding and sticking together
- asteroid colliding with a planet
- bullet embedding itself in a block of wood
Inelastic Collision
m1v1,f + m2v2,f = m1v1,i + m2v2,i
m1
m2
m1
m2
If objects stick together after collision v1,f = v2,f
vf(m1 + m2) = m1v1,i + m2v2,i
Inelastic Collision Problem 1
A 35 g bullet strikes a 5.0 kg stationary wooden
block and embeds itself in the block. The block
and the bullet fly off together at 8.6 m/s. What
was the original velocity of the bullet?
0.035 kg
8.6 m/s
5.0 kg
? m/s
Inelastic Collision Problem 1
0.035 kg
8.6 m/s
5.0 kg
? m/s
vf(m1 + m2) = m1v1,i + m2v2,i
8.6 m/s(5.0kg + 0.035kg) = 5.0kg(0) + 0.035kg(v)
43.301 kg m/s = 0.035kg(v)
v = 1237 m/s
Inelastic Collision Problem 2
A 0.105 kg hockey puck moving at 48 m/s is
caught by a 75 kg goalie at rest. With what
speed does the goalie slide on the ice?
vf(m1 + m2) = m1v1,i + m2v2,i
vf(0.105kg+ 75 kg) = 75kg(0) + 0.105kg(48m/s)
vf(75.105kg) = 5.04 kg m/s
vf = 0.067 m/s