Conservation of momentum
When is momentum conserved?
Outside force... What does that mean?
FΔt = Δ(mv)  m(Δv)
(m1v1 +m2v2 +...)before = (m1v’1 +m2v’2 +...)after
Is momentum conserved in both elastic and
inelastic collisions?
Can you predict?
A big fish swallows a little fish. (The big
fish is 4 times the mass of the little fish).
The little fish is at rest, while the big fish
is moving at 5 km/hr.
What kind of collision is this?
How fast does the fish move after
swallowing the little fish?
Can you predict?
A big fish swallows a little fish. (The big
fish is 4 times the mass of the little fish).
This time the big fish is at rest, while the
little fish is moving at 5 km/hr.
What kind of collision is this?
How fast does the fish move (backward)
after swallowing the little fish?
Uh, is that a collision?
What kind of collision is this?
Student mass = 40kg Ball mass = 20kg
He throws the ball which causes him to glide
back at 4.0 m/s. Why does this happen?
How fast must he have thrown the ball?
Uh, is that a collision?
What kind of collision is this? Elastic inelastic
Rifle mass = M = 4.0kg Bullet mass = m = 0.010kg
If the velocity of the bullet is v = 300 m/s, what is
the recoil speed of the rifle? (V)
MVb + mvb = MVa + mva
What is the recoil speed of the rifle? (V)
M = 4 kg m = 0.010 kg v = 300 m/s V = ?
(4)(0) + (0.010)(0) = 4Va + (0.010)(300)
0 = 4Va + 3  4Va = -3  Va = -3/4 = -0.75 m/s
I’m losing my marbles
What kind of collision is this… elastic or inelastic?
The mass of the red marble = 5 kg.
The mass of the blue marble = 10 kg
Vel. of 1st marble (red) to right is 40 m/s.
Vel. of 2nd marble (blue) to left is -50 m/s.
After the collision the vel. of the 2nd (blue) = 10 m/s to right.
What’s the vel. of the 1st marble (red) after the collision?
(Use m1v1 + m2v2 = m1v3 + m2v4)
(5)(40) + (10)(-50) = 5V + (10)(10)  Solve for V
Solving the equation
(5)(40) + (10)(-50) = 5V + (10)(10)  Solve for V
200 – 500 = 5V + 100
-300 = 5V + 100
5V = -300 – 100 = -400 
V = -400/5 = -80 m/s
An inelastic auto collision
A car rear-ends a truck.
mcar = 1000kg, while mtruck = 3000kg.
The vel. of the car was 20.0 m/s to the right.
The truck is at rest.
What’s the velocity of the car and truck (stuck
together) after the collision?
An elastic auto collision
A car rear-ends a truck, again.
mcar = 1000kg, while mtruck = 3000kg.
The velocity of the car was 20.0 m/s the right.
The truck is initially at rest.
What’s the velocity of the car after the
collision, if the truck moves forward at 10 m/s?
Another elastic auto collision
This time, a truck rear-ends a car.
mcar = 1000kg, while mtruck = 3000kg.
The velocity of the truck was 10.0 m/s to the right.
The car is initially at rest.
What’s the velocity of the truck after the collision, if the
car moves forward at 15 m/s after being struck?
First, calculate the total momentum before the collision,
and assume that momentum is conserved in all collisions.
You calculate the total momentum. Work on next slide…
Another elastic auto collision
The momentum of car before the collision is 0 since it is not moving.
So pCar = 0. Hence all of the momentum before the collision is in the
truck:
pTruck = mTruckvTruck = (3000kg)(10 m/s) = 30,000 kg m/sec.
After the collision we are told that the car moves forward at 15 m/s.
Hence the momentum of the car after the collision is:
pCar = mCarvCar = (1000kg)(15 m/s) = 15,000 kg m/sec.
So that the total momentum after the collision remains at 30,000 the
momentum of the truck must be 15,000 kg m/sec. (30,000 – 15,000)
pTruck = mTruckvTruck = 15,000 kg m/sec  3000vTruck = 15,000 Hence…
vTruck = 15,000/3,000 = 5 m/s.
Another take on it
Let’s do this again, using the conservation of
momentum formula:
mCarvCar-Before + mTruckvTruck-Before = mCarvCar-After + mTruckvTruck-After
You do it:
(1000)(0) + (3000)(10) = (1000)(15) + (3000)(vTruck-After)
30,000 = 15,000 + (3000)(vTruck-After)
So, subtracting 15,000 from both sides:
15,000 = (3000)(vTruck-After)
vTruck-After = 15,000/3,000 = 5 m/s
You do this one
Elastic or inelastic?
(Work on board – calculate the total momentum before 1st.)
You do this one
MVb + mvb = MVa + mVa  (3)(2) + (1)(0) = 3V + 1V
6 + 0 = 6 = 4V  V = 6/4 = 1.5 m/s
You do this one
A 5-kg ball traveling at 6.0 m/s collides head-on with a
10-kg ball moving in the opposite direction at a velocity
of -12.0 m/s. The 5-kg ball moves away at
-14 m/s after the collision. Find the velocity of the
second ball (orange) after the collision.
Elastic or inelastic?
(Work on board – first calculate the total momentum before.)
mrvr + movo = mrv’r + mov’o
First calculate the total momentum before. (30 – 120 = -90 kg m/s)
(5)(6) + (10)(-12) = (5)(-14) + (10)V
30 – 120 = -90 = -70 + 10V 
-90 = -70 + 10 V  add 70 to both sides…
-20 = 10V  V = -2 m/s to the left.
Impulse causes a change in momentum
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Combinations of Force and Time Required to
Produce 100 units of Impulse:
Force
Time
Impulse
100 N
1 sec
100 Ns
50 N
2 sec
100 Ns
25 N
4 sec
100 Ns
10 N
10 sec
100 Ns
4N
25 sec
100 Ns
2N
50 sec
100 Ns
1N
100 sec
100 Ns
0.1 N
1000 sec
100 Ns
In each instance, the product of the force applied and the time period it
was applied is the same… giving us 100Ns of impulse in each case.
Calculating impulse
Impulse results in a change in momentum.
Impulse is a force applied for a period of time…
Impulse  F∆t
If this guy applies a force of 500N for 0.5 seconds, what is the
impulse?
F∆t = (500N)(0.5s) = 250 Ns
Determining force for an impulse
F∆t = m∆v
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A player swings his bat and hits a baseball with a
weight of 1 kg and thrown at 40 m/s such that it
leaves the bat moving in the opposite direction at
-35 m/s.
If the ball stayed in contact with the bat for 0.5
seconds, what force did the bat apply?
Impulse 
F∆t = m∆v
What is the change in velocity (∆v)?
(40 m/s - -35 m/s = 75 m/s) 
The change in momentum = m∆v = (1)(75) = 75 kg m/s
Since the impulse caused the change in momentum…
F(0.5s) = 75 kg m/s  F = 75 kg m/(s2 (0.5)) = 150 N