Download Conservation of Momentum

Conservation of
Momentum
It’s the Law!!!!!!!!!
We have already learned that the momentum of an object is equal to the mass of the
object multiplied by its velocity:
𝐩 = 𝐦𝐯
We also know that if we want the momentum of a system we must combine the
momenta of each object in the system, remembering that momentum is a vector.
We also learned that the impulse on an object was equal to the net external
force on the object multiplied by the time the net external force is applied to
the object:
𝐉 = 𝐅∆𝐭
Finally we also learned that impulse on a system is equal to the change in momentum
of the system. As an equation this looks like:
𝐅∆𝐭 = ∆𝐩
What happens if there is no net external force acting on the object?
𝐅∆𝐭 = ∆𝐩
(𝟎)∆𝐭 = ∆𝐩
𝟎 = ∆𝐩
∆𝐩 = 𝟎
This means that if there is no net external force on a system, then:
1)
The momentum of the system doesn’t change (It is conserved)
2)
The initial momentum of the system equals the final momentum
of the system. (𝐩𝐨 = 𝐩)
THIS IS THE LAW OF CONSERVATION
OF MOMENTUM
Momentum Example 7: Rifle Recoil
Calculate the recoil velocity of a 6.5 kg rifle that fires a 0.025 kg bullet at 525 m/s.
Before the rifle is fired the system is at rest so:
𝐩𝐨 = 𝟎
After the rifle is fired the momentum of the system is still zero since the forces involved
are only internal. So:
𝐩=𝟎
𝐩𝐫𝐢𝐟𝐥𝐞 + 𝐩𝐛𝐮𝐥𝐥𝐞𝐭 = 𝟎
𝟔. 𝟓𝐯 + 𝟎. 𝟎𝟐𝟓 𝟓𝟐𝟓 = 𝟎
𝟔. 𝟓𝐯 = −𝟏𝟑. 𝟏𝟐𝟓
𝐯 = −𝟐. 𝟎 𝐦/𝐬
Momentum Example 8: Blobs of Putty
Two blobs of putty are traveling towards each other at the 10-m/s. If one blob of putty has
twice the mass of the other blob of putty, how fast would the blobs be traveling after the
collision if they stick together? {This is called an inelastic collision}
2m
Before Collision
m
10 m/s
10 m/s
𝐩𝐨 = 𝟐𝐦 𝟏𝟎 − 𝐦 𝟏𝟎 = 𝟏𝟎𝐦
After Collision
3m
v
𝐩 = (𝟑𝐦)(𝐯)
Since momentum is conserved (all forces are internal):
𝐩 = 𝐩𝐨
𝟑𝐦𝐯 = 𝟏𝟎𝐦
𝟏𝟎
𝐯=
𝐦/𝐬 = 𝟑. 𝟑 𝐦/𝐬
𝟑
Now, print off your online
assignment and start it. You will not
be able to get to it this weekend due
to maintenance by UT!!!!!!!!
Momentum Example 9: Elastic Collision
A 2.0-kg ball is moving towards a 3.0-kg ball at 3.0 m/s. The 3-kg ball is moving towards the
2.0 kg ball at 4.0 m/s. If the two balls hit each other, what is the speed of the 3.0-kg ball after
the collision if the 2.0-kg ball bounces back at 2.0-m/s?
3.0 kg
Before Collision
2.0-kg
4.0 m/s
3.0 m/s
𝐩𝐨 = 𝟑 𝟒 − 𝟐 𝟑 = 𝟔𝐤𝐠𝐦/𝐬
3.0 kg
After Collision
2.0-kg
v
2.0 m/s
𝐩 = 𝟐 𝟐 − 𝟑 𝐯 = 𝟒 − 𝟑𝐯
Since momentum is conserved (all forces are internal):
𝐩 = 𝐩𝐨
𝟒 − 𝟑𝐯 = 𝟔
−𝟑𝐯 = 𝟐
𝐯 = −𝟎. 𝟔𝟕𝐦/𝐬