Conservation of Momentum in Physics
undefined
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
p
i
= p
f
Start with p
i
= p
f
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 40
0
m/s.
What is the recoil
velocity of the gun?
Sample Problem 1 Solution
mv
bgi
=mv
bf
+mv
gf
0=(.050kg)(400m/s)+(4.0kg)v
gf
v
gf
= -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
mv
Ai
+mv
Bi
=mv
ABf
(1.0kg)(2.0m/s)+0=(1.0 kg+1.0 kg)v
ABf
v
ABf
=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
mv
Ai
+mv
Bi
=mv
ABf
(60.0kg)(3.0m/s)+(40.0kg)(-4.0m/s)
=(60.0 kg + 40.0 kg) v
ABf
v
ABf
=.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
mv
ai
+ mv
bi
= mv
af
+ mv
bf
(25.0kg)(5.00m/s) + (35.0kg)(2.00m/s)
= (25.0kg)(1.50m/s) + (35.0kg)v
bf
195 kgm/s = 37.5 kgm/s + (35.0kg)v
bf
v
bf
= 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
mv
ABi
=mv
Af
+mv
Bf
0=(3.0kg)(-10.0m/s)+(5.0kg)(v
B
)
v
Bf
=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
mv
ABi
=mv
Af
+mv
Bf
0=(85kg)(1.5m/s)+(145kg)v
Bf
v
Bf
= -.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
mv
Ai
+mv
Bi
=mv
ABf
(14.6kg)(3.0m/s)+(.40kg)(15m/s)=
(14.6 + 0.40 kg)v
ABf
v
ABf
=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
mv
ABi
=mv
Af
+mv
Bf
(75 kg + 25 kg)(3.0m/s)=
(75kg)(-1.0m/s)+(25kg) v
Bf
375kgm/s=25kg v
Bf
v
Bf
=15m/s
Back to Unit 5 again
Newton’s 3
rd
Law – every action force has an equal
and opposite reaction force
That’s another way of saying conservation of
momentum:
mv
ai
+ mv
bi
= mv
af
+ mv
bf
mv
ai
- mv
af
= mv
bf
– mv
bi
-
Δ
p
a
=
Δ
p
b
-Ft
a
= Ft
b
-F
a
= F
b
the forces are equal and opposite!
Elastic and Inelastic Collisions
Perfectly elastic collisions – collisions in
which the objects rebound
perfectly
–
with no loss of E
K
Ideal situation!
Perfectly inelastic collisions – the
objects stick together – always some E
K
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
Conservation of momentum is a fundamental law stating that momentum is conserved in a closed system where it is neither created nor destroyed, but rather transferred between objects. This principle is applied in various scenarios, such as bullet-gun recoil, collision of objects, and more, to determine velocities and directions post-interaction.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
E N D
Presentation Transcript
Its 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= pf
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 3rdLaw 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 - pa= pb -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
