Momentum and Impulse

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

2. Definition: The linear momentump 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

3. The total momentum of a system of particles is the vector sum of the momenta of the individual particles: ptotal = p1 + p2 + ... = m1v1 + m2v2 + ... Since we are adding vectors, we can break this up into components so that: px,Tot = p1x + p2x + ….Etc. Physics 1D03 - Lecture 25

4. 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

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

6. v F 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). 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

7. v F Solution 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

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

9. Area = t 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 = FDx ; so the work-energy theorem (derived from Newton #2) is DK = FDx. Physics 1D03 - Lecture 25

10. 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. Physics 1D03 - Lecture 25

11. 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? • The rubber ball • The snowball • They both transfer the same amount Physics 1D03 - Lecture 25

12. 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 Fmax, the maximum magnitude of the force on the ball from the wall during the collision? Physics 1D03 - Lecture 25

13. F12 m1 F21 = -F12 m2 Newton #3 and Momentum Conservation Two particles interact: Dp1= F12Dt Dp2= F21Dt Newton’s 3rd Law: F21= -F12 The momentum changes are equal and opposite; the total momentum: p = p1 + p2=0doesn’t change. The fine print:Only internal forces act. External forces would transfer momentum into or out of the system. Physics 1D03 - Lecture 25

14. 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

15. Application • A subatomic particle may decay into two different particles. If the momentum before is zero, it must also be zero after, so: pi=0 pf=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. Physics 1D03 - Lecture 25

16. Example: the fire extinguisher How can we calculate the thrust on a fire extinguisher when gas of mass Dm is ejected out of the nozzle in time Dt ? Plan: Calculate the change in momentum of the gas; this is equal to the impulse the gas gets from the extinguisher nozzle. gas thrust From Newton’s Third Law, the nozzle gets an equal but opposite impulse from the gas. Since impulse is thrust times Dt, divide by Dt to get the average thrust (force). Result: thrust = (speed of gas relative to nozzle) times (mass of gas ejected per unit time) Physics 1D03 - Lecture 25

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