Transcript conservation of momentum powerpoint

Conservation of Momentum
It’s the Law!
Momentum is neither created nor
destroyed, only transferred from one
object to another OR
The total momentum of a closed system
is a constant OR
pi = p f
Start with pi = pf
For each side, include a term for each
separate object
Each term is mv – use subscripts to tell
them apart
Sample Problem 1
A bullet of mass 0.050 kg
leaves the muzzle of a
gun of mass 4.0 kg with
a velocity of 400 m/s.
What is the recoil
velocity of the gun?
Sample Problem 1 Solution
mvbgi=mvbf+mvgf
0=(.050kg)(400m/s)+(4.0kg)vgf
vgf= -5.0 m/s
Sample Problem 2
A model railroad engine of mass 1.0 kg
and a speed of 2.0 m/s collides with an
identical engine which is at rest. On
colliding, the two engines lock together
and move away. What is the velocity of
the two after the collision?
Sample Problem 2 Solution
mvAi+mvBi=mvABf
(1.0kg)(2.0m/s)+0=(1.0 kg+1.0 kg)vABf
vABf=1.0m/s
Sample Problem 3
A skater with a mass of 60.0 kg is
moving at 3.0 m/s to the right. Another
skater of mass 40.0 kg is moving at 4.0
m/s to the left (negative 4.0 m/s).
They collide and grab onto each other
for support. What is their velocity and
direction after the collision?
Sample Problem 3 Solution
mvAi+mvBi=mvABf
(60.0kg)(3.0m/s)+(40.0kg)(-4.0m/s)
=(60.0 kg + 40.0 kg) vABf
vABf=.20m/s (right)
Sample Problem 4
A 25.0 kg cart moves to the right at
5.00 m/s. It overtakes and collides with
a 35.0 kg cart moving to the right at
2.00 m/s. After the elastic collision, the
25.0 kg cart slows to 1.50 m/s. What is
the final velocity of the 35.0 kg cart?
Sample Problem 4 solution
mvai + mvbi = mvaf + mvbf
(25.0kg)(5.00m/s) + (35.0kg)(2.00m/s)
= (25.0kg)(1.50m/s) + (35.0kg)vbf
195 kgm/s = 37.5 kgm/s + (35.0kg)vbf
vbf = 4.5 m/s
Sample Problem 5
A bomb with a mass of 8.0 kg explodes
and breaks into two large fragments.
The first piece has a mass of 3.0 kg and
moves to the left at 10.0 m/s. How fast
must the other piece be moving?
Sample Problem 5 Solution
mvABi=mvAf+mvBf
0=(3.0kg)(-10.0m/s)+(5.0kg)(vB)
vBf=6.0m/s
Sample Problem 6
A lumberjack standing on a log floating
in calm water, begins to walk along the
log toward the east at 1.5 m/s. The
lumberjack has a mass of 85 kg (and
he’s okay, for all you Monty Python
fans), and the log has a mass of
145 kg. How fast will the
log move?
Sample Problem 6 Solution
mvABi=mvAf+mvBf
0=(85kg)(1.5m/s)+(145kg)vBf
vBf= -.879m/s (west)
Sample Problem 7
During a snowball fight, a little girl of
mass 14.6 kg is moving across nearly
frictionless ice at 3.0 m/s when a 0.40
kg snowball moving at 15 m/s hits her
in the back and sticks. How fast is she
now moving along the ice?
Sample Problem 7 Solution
mvAi+mvBi=mvABf
(14.6kg)(3.0m/s)+(.40kg)(15m/s)=
(14.6 + 0.40 kg)vABf
vABf=3.3m/s
Sample Problem 8
At the ice show, a 75 kg clown is
skating along at 3.0 m/s to the right
while holding a 25 kg clown in his arms.
He suddenly throws the little clown
ahead of him. After the toss, the big
clown is now moving at -1.0 m/s (that
is, to the left). How fast is the little
clown skating along the ice?
Sample Problem 8 Solution
mvABi=mvAf+mvBf
(75 kg + 25 kg)(3.0m/s)=
(75kg)(-1.0m/s)+(25kg) vBf
375kgm/s=25kg vBf
vBf=15m/s
Back to Unit 5 again
Newton’s 3rd Law – every action force has an equal
and opposite reaction force
That’s another way of saying conservation of
momentum:
mvai + mvbi = mvaf + mvbf
mvai - mvaf = mvbf – mvbi
-Δpa = Δpb
-Fta = Ftb
-Fa = Fb
the forces are equal and opposite!
Elastic and Inelastic Collisions
Perfectly elastic collisions – collisions in
which the objects rebound perfectly –
with no loss of EK

Ideal situation!
Perfectly inelastic collisions – the
objects stick together – always some EK
loss because of shape change
Most Collisions…
Are somewhere in between – the
objects don’t stick, but they don’t
bounce off perfectly either
“Coefficient of restitution” helps
quantify how elastic the collision is