1. Physics 1D03 - Lecture 25 Momentum and Impulse Newton’s original “quantity of motion” a conserved quantity a vector Serway & Jewett 9.1 – 9.3 -Newton’s Second Law in another form -momentum and impulse Today:

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

3. Physics 1D03 - Lecture 25 p total = p 1 + p 2 +... = m 1 v 1 + m 2 v 2 +... The total momentum of a system of particles is the vector sum of the momenta of the individual particles: Since we are adding vectors, we can break this up into components so that: p x,Tot = p 1x + p 2x + …. Etc.

4. Physics 1D03 - Lecture 25 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 This is how Newton wrote the Second Law. It remains true in cases where the mass is not constant. net external force = rate of change of momentum

5. Physics 1D03 - Lecture 25 Relation between kinetic energy K and momentum p: (practice problem)

6. Physics 1D03 - Lecture 25 Example 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). v F Calculate this extra force if: v = 20 m/s The water collects in the car at the rate of 6 kg per minute

7. Physics 1D03 - Lecture 25 Solution Plan: The momentum of the car increases as it gains mass (water). Use Newton’s second law to find F. v 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.

8. Physics 1D03 - Lecture 25 Impulse Newton #2:, or dp = F dt (Extra) In general (force not constant), we integrate: For a constant force,  p = F  t. The vector quantity F  t is called the Impulse: I = FΔt = Δp (change in p) = (total impulse from external forces) (Newton’s Second Law again) Recall, the integral gives the area under a curve…

9. Physics 1D03 - Lecture 25 t Area = t Impulse is the area under the curve. The average force is the constant force which would give the same impulse. Compare with work: W = F  x ; so the work-energy theorem (derived from Newton #2) is  K = F  x.

10. Physics 1D03 - Lecture 25 Example A golf ball is launched with a velocity of 44 m/s. The ball has a mass of 50g. Determine the average force on the ball during the collision with the club, if the collision lasted 0.01 s.

11. Physics 1D03 - Lecture 25 Quiz A rubber ball and a snowball, each 100 grams, are thrown at a school bus window at identical speeds. The snowball sticks, the rubber ball bounces off. Which one transfers the larger impulse to the window? A)The rubber ball B)The snowball C)They both transfer the same amount

12. Physics 1D03 - Lecture 25 Example Figure below shows an approximate plot of force magnitude F versus time t during the collision of a 58 g ball with a wall. The initial velocity of the ball is 34 m/s perpendicular to the wall. The ball rebounds directly back with approximately the same speed, also perpendicular to the wall. What is F max, the maximum magnitude of the force on the ball from the wall during the collision?

13. Physics 1D03 - Lecture 25 Newton #3 and Momentum Conservation Two particles interact: Newton’s 3 rd Law: F 21 = -F 12 The momentum changes are equal and opposite; the total momentum: p = p 1 + p 2 =0 doesn’t change. The fine print: Only internal forces act. External forces would transfer momentum into or out of the system.  p 1 = F 12  t  p 2 = F 21  t F 12 F 21 = -F 12 m1m1 m2m2

14. Physics 1D03 - Lecture 25 Since momentum is a vector, we can also express it in terms of the components. These are independently conserved: p ix =p fx p iy =p fy p iz =p fz

15. Physics 1D03 - Lecture 25 Application A subatomic particle may decay into two different particles. If the momentum before is zero, it must also be zero after, so: p i =0 p f =0 Given the mass and velocity of one of the particles, if we measure the velocity of the other one we can figure out it’s momentum and hence the mass.

16. Physics 1D03 - Lecture 25 Example: the fire extinguisher How can we calculate the thrust on a fire extinguisher when gas of mass  m is ejected out of the nozzle in time  t ? gas Plan: Calculate the change in momentum of the gas; this is equal to the impulse the gas gets from the extinguisher nozzle. From Newton’s Third Law, the nozzle gets an equal but opposite impulse from the gas. Since impulse is thrust times  t, divide by  t to get the average thrust (force). Result: thrust = (speed of gas relative to nozzle) times (mass of gas ejected per unit time) thrust

17. Physics 1D03 - Lecture 25 Summary Definitions: momentum, impulse Newton’s Second Law, in terms of momentum and impulse conservation of momentum