1. Test – What to study
The test will cover things we have done in class up to Lecture 9
Momentum will NOT be on the test
Please look over old test (available online)
Make sure you’re up on your reading !
Review error calculations (done in lab)
Physics 1D03 - Lecture 25

2. Momentum Newton’s original “quantity of motion” a conserved quantity
a vector
Today:
Newton’s Second Law in another form
Momentum and Momentum Conservation
Physics 1D03 - Lecture 25

3. Definition: The linear momentum p of a particle is its mass times its velocity:
p  mv
Momentum is a vector, since velocity is a vector.
Units: kg m/s (no special name).
(We say “linear” momentum to distinguish it from angular momentum, a different physical quantity.)
Physics 1D03 - Lecture 25

4. Example 1 & 2
A car of mass 1500kg is moving with a velocity of 72km/h. What is its momentum ?
Two cars, one of mass 1000kg is moving at 36km/h and the other of mass 1200kg is moving at 72km/h in the opposite direction. What is the momentum of the system ?
Physics 1D03 - Lecture 25

5. Concept Quiz Which object will have the smallest momentum ?
A 1.6x10-19kg particle moving at 1x103m/s
A 1000kg car moving at 2m/s
A 100kg person moving at 10m/s
A 5000kg truck at rest
Physics 1D03 - Lecture 25

6. Net external force = rate of change of momentum
Newton’s Second Law
If mass is constant, then the rate of change of (mv) is equal to m times the rate of change of v. We can rewrite Newton’s Second Law: or
Net external force = rate of change of momentum
This is how Newton wrote the Second Law. It remains true in cases where the mass is not constant.
Physics 1D03 - Lecture 25

7. Example 3 v F
Rain is falling vertically into an open railroad car which moves along a horizontal track at a constant speed. The engine must exert an extra force on the car as the water collects in it (the water is initially stationary, and must be brought up to the speed of the train).
Calculate this extra force if:
v = 20 m/s
The water collects in the car at the rate of 6 kg per minute
Physics 1D03 - Lecture 25

8. Solution v F
Plan: The momentum of the car increases as it gains mass (water). Use Newton’s second law to find F.
v is constant, and dm/dt is 6 kg/min or 0.1 kg/s (change to SI units!),
so F = (0.1 kg/s) (20 m/s) = 2.0 N
F and p are vectors; we get the horizontal force from the rate of increase of the horizontal component of momentum.
Physics 1D03 - Lecture 25

9. The total momentum of a system of particles is the vector sum of the momenta of the individual particles:
ptotal = p1 + p = m1v1 + m2v
Since we are adding vectors, we can break this up into components so that:
px,Tot = p1x + p2x + …. Etc.
Physics 1D03 - Lecture 25

10. Example 4
A particle of mass 2kg is moving with a velocity of 5m/s and an angle of 45o to the horizontal. Determine the components of its momentum.
Physics 1D03 - Lecture 25

11. Newton’s 3rd Law and Momentum Conservation
Two particles interact:
F12
F21 = -F12 m1 m2
Dp1= F12 Dt Dp2= F21 Dt
Newton’s 3rd Law: F21 = -F12
The momentum changes are equal and opposite; the total momentum:
p = p1 + p2=0
doesn’t change.
The fine print: Only internal forces act. External forces would transfer momentum into or out of the system.
eg: particles moving through a cloud of gas
Physics 1D03 - Lecture 25

12. Conservation of momentum simply says that the initial and final momenta are equal: pi=pf
Since momentum is a vector, we can also express it in terms of the components. These are independently conserved: pix=pfx piy=pfy piz=pfz
Physics 1D03 - Lecture 25

13. Concept Quiz
A subatomic particle may decay into two different particles. If the total momentum before is zero before the decay before, what is the total after ?
0 kg m/s
depends on the masses
depends on the final velocities
depends on b) and c) together
Before
After
Physics 1D03 - Lecture 25